Vector Spaces

The first chapter finished with a fair understanding of how Gauss’s Method solves a linear system. It systematically takes linear combinations of the rows. Here we move to a general study of linear combinations.

We need a setting. At times in the first chapter we’ve combined vectors from $ℝ^{2}$ , at other times vectors from $ℝ^{3}$ , and at other times vectors from higher-dimensional spaces. So our first impulse might be to work in $ℝ^{n}$ , leaving $n$ unspecified. This would have the advantage that any of the results would hold for $ℝ^{2}$ and for $ℝ^{3}$ and for many other spaces, simultaneously.

But if having the results apply to many spaces at once is advantageous then sticking only to $ℝ^{n}$ ’s is restrictive. We’d like our results to apply to combinations of row vectors, as in the final section of the first chapter. We’ve even seen some spaces that are not simply a collection of all of the same-sized column vectors or row vectors. For instance, we’ve seen a homogeneous system’s solution set that is a plane inside of $ℝ^{3}$ . This set is a closed system in that a linear combination of these solutions is also a solution. But it does not contain all of the three-tall column vectors, only some of them.

We want the results about linear combinations to apply anywhere that linear combinations make sense. We shall call any such set a vector space. Our results, instead of being phrased as “Whenever we have a collection in which we can sensibly take linear combinations …”, will be stated “In any vector space …”

Such a statement describes at once what happens in many spaces. To understand the advantages of moving from studying a single space to studying a class of spaces, consider this analogy. Imagine that the government made laws one person at a time: “Leslie Jones can’t jay walk.” That would be bad; statements have the virtue of economy when they apply to many cases at once. Or suppose that they said, “Kim Ke must stop when passing an accident.” Contrast that with, “Any doctor must stop when passing an accident.” More general statements, in some ways, are clearer.

Definition of Vector Space

We shall study structures with two operations, an addition and a scalar multiplication, that are subject to some simple conditions. We will reflect more on the conditions later but on first reading notice how reasonable they are. For instance, surely any operation that can be called an addition (e.g., column vector addition, row vector addition, or real number addition) will satisfy conditions (1) through (5) below.

Definition and Examples

Definition 1.1 A vector space (over $ℝ$ ) consists of a set $V$ along with two operations ‘+’ and ‘ $\cdot$ ’ subject to the conditions that for all vectors $\vec{v}, \vec{w}, \vec{u} \in V$ and all scalars $r, s \in ℝ$ :

the set $V$ is closed under vector addition, that is, $\vec{v} + \vec{w} \in V$
vector addition is commutative, $\vec{v} + \vec{w} = \vec{w} + \vec{v}$
vector addition is associative, $(\vec{v} + \vec{w}) + \vec{u} = \vec{v} + (\vec{w} + \vec{u})$
there is a zero vector $\vec{0} \in V$ such that $\vec{v} + \vec{0} = \vec{v}$ for all $\vec{v} \in V$
each $\vec{v} \in V$ has an additive inverse $\vec{w} \in V$ such that $\vec{w} + \vec{v} = \vec{0}$
the set $V$ is closed under scalar multiplication, that is, $r \cdot \vec{v} \in V$
scalar multiplication distributes over scalar addition, $(r + s) \cdot \vec{v} = r \cdot \vec{v} + s \cdot \vec{v}$
scalar multiplication distributes over vector addition, $r \cdot (\vec{v} + \vec{w}) = r \cdot \vec{v} + r \cdot \vec{w}$
ordinary multiplication of scalars associates with scalar multiplication, $(r s) \cdot \vec{v} = r \cdot (s \cdot \vec{v})$
multiplication by the scalar $1$ is the identity operation, $1 \cdot \vec{v} = \vec{v}$ .

Remark 1.2 The definition involves two kinds of addition and two kinds of multiplication, and so may at first seem confused. For instance, in condition (7) the ‘ $+$ ’ on the left is addition of two real numbers while the ‘ $+$ ’ on the right is addition of two vectors in $V$ . These expressions aren’t ambiguous because of context; for example, $r$ and $s$ are real numbers so ‘ $r + s$ ’ can only mean real number addition. In the same way, item (9)’s left side ‘ $r s$ ’ is ordinary real number multiplication, while its right side ‘ $s \cdot \vec{v}$ ’ is the scalar multiplication defined for this vector space.

The best way to understand the definition is to go through the examples below and for each, check all ten conditions. The first example includes that check, written out at length. Use it as a model for the others. Especially important are the closure conditions, (1) and (6). They specify that the addition and scalar multiplication operations are always sensible—they are defined for every pair of vectors and every scalar and vector, and the result of the operation is a member of the set.

Example 1.3 This subset of $ℝ^{2}$ is a line through the origin.

$L = {(\begin{matrix} x \\ y \end{matrix}) ∣ y = 3 x}$

We shall verify that it is a vector space under the usual meaning of ‘+’ and ‘ $\cdot$ ’.

$(\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix}) r \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} r x \\ r y \end{matrix})$

These operations are just the ordinary ones, reused on its subset $L$ . We say that $L$ inherits these operations from $ℝ^{2}$ .

We shall check all ten conditions. The paragraph having to do with addition has five conditions. For condition (1), closure under addition, suppose that we start with two vectors from the line $L$ ,

${\vec{v}}_{1} = (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) {\vec{v}}_{2} = (\begin{matrix} x_{2} \\ y_{2} \end{matrix})$

so that they satisfy the restrictions that $y_{1} = 3 x_{1}$ and $y_{2} = 3 x_{2}$ . Their sum

${\vec{v}}_{1} + {\vec{v}}_{2} = (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix})$

is also a member of the line $L$ because the fact that its second component is three times its first $y_{1} + y_{2} = 3 (x_{1} + x_{2})$ follows from the restrictions on ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ . For (2), that addition of vectors commutes, just compare

${\vec{v}}_{1} + {\vec{v}}_{2} = (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix}) {\vec{v}}_{2} + {\vec{v}}_{1} = (\begin{matrix} x_{2} + x_{1} \\ y_{2} + y_{1} \end{matrix})$

and note that they are equal since their entries are real numbers and real numbers commute. (That the vectors satisfy the restriction of lying in the line is not relevant for this condition; they commute just because all vectors in the plane commute.) Condition (3), associativity of vector addition, is similar.

$\begin{aligned} ((\begin{array}{c} x_{1} \\ y_{1} \end{array}) + (\begin{array}{c} x_{2} \\ y_{2} \end{array})) + (\begin{array}{c} x_{3} \\ y_{3} \end{array}) & = (\begin{array}{c} (x_{1} + x_{2}) + x_{3} \\ (y_{1} + y_{2}) + y_{3} \end{array}) \\ = (\begin{array}{c} x_{1} + (x_{2} + x_{3}) \\ y_{1} + (y_{2} + y_{3}) \end{array}) \\ = (\begin{array}{c} x_{1} \\ y_{1} \end{array}) + ((\begin{array}{c} x_{2} \\ y_{2} \end{array}) + (\begin{array}{c} x_{3} \\ y_{3} \end{array})) \end{aligned}$

For the fourth condition we must produce a vector that acts as the zero element. The vector of zero entries will do.

$(\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 0 \\ 0 \end{matrix}) = (\begin{matrix} x \\ y \end{matrix})$

Note that $\vec{0} \in L$ as its second component is triple its first. For (5), that given any $\vec{v} \in L$ we can produce an additive inverse, we have

$(\begin{matrix} - x \\ - y \end{matrix}) + (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$

and so the vector $- \vec{v}$ is the desired inverse. As with the prior condition, observe here that if $\vec{v} \in L$ , so that $y = 3 x$ , then $- \vec{v} \in L$ also, since $- y = 3 (- x)$ .

The checks for the five conditions having to do with scalar multiplication are similar. For (6), closure under scalar multiplication, suppose that $r \in ℝ$ and $\vec{v} \in L$ , that is,

$\vec{v} = (\begin{matrix} x \\ y \end{matrix})$

satisfies that $y = 3 x$ . Then

$r \cdot \vec{v} = r \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} r x \\ r y \end{matrix})$

is also a member of $L$ : the relation $r y = 3 \cdot r x$ holds because $y = 3 x$ . Next, this checks (7).

$(r + s) \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} (r + s) x \\ (r + s) y \end{matrix}) = (\begin{matrix} r x + s x \\ r y + s y \end{matrix}) = r \cdot (\begin{matrix} x \\ y \end{matrix}) + s \cdot (\begin{matrix} x \\ y \end{matrix})$

For (8) we have this.

$r \cdot ((\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \end{matrix})) = (\begin{matrix} r (x_{1} + x_{2}) \\ r (y_{1} + y_{2}) \end{matrix}) = (\begin{matrix} r x_{1} + r x_{2} \\ r y_{1} + r y_{2} \end{matrix}) = r \cdot (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + r \cdot (\begin{matrix} x_{2} \\ y_{2} \end{matrix})$

The ninth

$(r s) \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} (r s) x \\ (r s) y \end{matrix}) = (\begin{matrix} r (s x) \\ r (s y) \end{matrix}) = r \cdot (s \cdot (\begin{matrix} x \\ y \end{matrix}))$

and tenth conditions are also straightforward.

$1 \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 1 x \\ 1 y \end{matrix}) = (\begin{matrix} x \\ y \end{matrix})$

Example 1.4 The whole plane, the set $ℝ^{2}$ , is a vector space where the operations ‘ $+$ ’ and ‘ $\cdot$ ’ have their usual meaning.

We shall check just two of the conditions, the closure conditions.

For (1) observe that the result of the vector sum

$(\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix})$

is a column array with two real entries, and so is a member of the plane $ℝ^{2}$ . In contrast with the prior example, here there is no restriction on the first and second components of the vectors.

Condition (6) is similar. The vector

$r \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} r x \\ r y \end{matrix})$

has two real entries, and so is a member of $ℝ^{2}$ .

In a similar way, each $ℝ^{n}$ is a vector space with the usual operations of vector addition and scalar multiplication. (In $ℝ^{1}$ , we usually do not write the members as column vectors, i.e., we usually do not write ‘ $(π)$ ’. Instead we just write ‘ $π$ ’.)

Example 1.5 Example 1.3 gives a subset of $ℝ^{2}$ that is a vector space. For contrast, consider the set of two-tall columns with entries that are integers, under the same operations of component-wise addition and scalar multiplication. This is a subset of $ℝ^{2}$ but it is not a vector space: it is not closed under scalar multiplication, that is, it does not satisfy condition (6). For instance, on the left below is a vector with integer entries, and a scalar.

$0.5 \cdot (\begin{matrix} 4 \\ 3 \end{matrix}) = (\begin{matrix} 2 \\ 1.5 \end{matrix})$

On the right is a column vector that is not a member of the set, since its entries are not all integers.

Example 1.6 The one-element set

${(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix})}$

is a vector space under the operations

$(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) r \cdot (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix})$

that it inherits from $ℝ^{4}$ .

A vector space must have at least one element, its zero vector. Thus a one-element vector space is the smallest possible.

Definition 1.7 A one-element vector space is a trivial space.

The examples so far involve sets of column vectors with the usual operations. But vector spaces need not be collections of column vectors, or even of row vectors. Below are some other types of vector spaces. The term ‘vector space’ does not mean ‘collection of columns of reals’. It means something more like ‘collection in which any linear combination is sensible’.

Example 1.8 Consider $𝒫_{3} = {a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} ∣ a_{0}, \dots, a_{3} \in ℝ}$ , the set of polynomials of degree three or less (in this book, we’ll take constant polynomials, including the zero polynomial, to be of degree zero). It is a vector space under the operations

$\begin{matrix} (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}) + (b_{0} + b_{1} x + b_{2} x^{2} + b_{3} x^{3}) \\ = (a_{0} + b_{0}) + (a_{1} + b_{1}) x + (a_{2} + b_{2}) x^{2} + (a_{3} + b_{3}) x^{3} \end{matrix}$

and

$r \cdot (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}) = (r a_{0}) + (r a_{1}) x + (r a_{2}) x^{2} + (r a_{3}) x^{3}$

(the verification is easy). This vector space is worthy of attention because these are the polynomial operations familiar from high school algebra. For instance, $3 \cdot (1 - 2 x + 3 x^{2} - 4 x^{3}) - 2 \cdot (2 - 3 x + x^{2} - (1 / 2) x^{3}) = - 1 + 7 x^{2} - 11 x^{3}$ .

Although this space is not a subset of any $ℝ^{n}$ , there is a sense in which we can think of $𝒫_{3}$ as “the same” as $ℝ^{4}$ . If we identify these two space’s elements in this way

$a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} corresponds to (\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix})$

then the operations also correspond. Here is an example of corresponding additions.

$\begin{array}{lr} 1 - 2 x + 0 x^{2} + 1 x^{3} \\ + & 2 + 3 x + 7 x^{2} - 4 x^{3} \\ 3 + 1 x + 7 x^{2} - 3 x^{3} \end{array} corresponds to (\begin{matrix} 1 \\ - 2 \\ 0 \\ 1 \end{matrix}) + (\begin{matrix} 2 \\ 3 \\ 7 \\ - 4 \end{matrix}) = (\begin{matrix} 3 \\ 1 \\ 7 \\ - 3 \end{matrix})$

Things we are thinking of as “the same” add to “the same” sum. Chapter Three makes precise this idea of vector space correspondence. For now we shall just leave it as an intuition.

In general we write $𝒫_{n}$ for the vector space of polynomials of degree $n$ or less ${a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} ∣ a_{0}, \dots, a_{n} \in ℝ}$ , under the operations of the usual polynomial addition and scalar multiplication. We will often use these spaces as examples.

Example 1.9 The set $ℳ_{2 \times 2}$ of $2 \times 2$ matrices with real number entries is a vector space under the natural entry-by-entry operations.

$(\begin{matrix} a & b \\ c & d \end{matrix}) + (\begin{matrix} w & x \\ y & z \end{matrix}) = (\begin{matrix} a + w & b + x \\ c + y & d + z \end{matrix}) r \cdot (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} r a & r b \\ r c & r d \end{matrix})$

As in the prior example, we can think of this space as “the same” as $ℝ^{4}$ .

We write $ℳ_{n \times m}$ for the vector space of $n \times m$ matrices under the natural operations of matrix addition and scalar multiplication. As with the polynomial spaces, we will often use these as examples.

Example 1.10 The set ${f ∣ f : ℕ \to ℝ}$ of all real-valued functions of one natural number variable is a vector space under the operations

$(f_{1} + f_{2}) (n) = f_{1} (n) + f_{2} (n) (r \cdot f) (n) = r f (n)$

so that if, for example, $f_{1} (n) = n^{2} + 2 \sin (n)$ and $f_{2} (n) = - \sin (n) + 0.5$ then $(f_{1} + 2 f_{2}) (n) = n^{2} + 1$ .

We can view this space as a generalization of Example 1.4—instead of $2$ -tall vectors, these functions are like infinitely-tall vectors.

$\begin{array}{cc} n & f (n) = n^{2} + 1 \\ 0 & 1 \\ 1 & 2 \\ 2 & 5 \\ 3 & 10 \\ ⋮ & ⋮ \end{array} corresponds to (\begin{matrix} 1 \\ 2 \\ 5 \\ 10 \\ ⋮ \end{matrix})$

Addition and scalar multiplication are component-wise, as in Example 1.4. (We can formalize “infinitely-tall” by saying that it means an infinite sequence, or that it means a function from $ℕ$ to $ℝ$ .)

Example 1.11 The set of polynomials with real coefficients

${a_{0} + a_{1} x + \dots + a_{n} x^{n} ∣ n \in ℕ and a_{0}, \dots, a_{n} \in ℝ}$

makes a vector space when given the natural ‘ $+$ ’

$\begin{matrix} (a_{0} + a_{1} x + \dots + a_{n} x^{n}) + (b_{0} + b_{1} x + \dots + b_{n} x^{n}) \\ = (a_{0} + b_{0}) + (a_{1} + b_{1}) x + \dots + (a_{n} + b_{n}) x^{n} \end{matrix}$

and ‘ $\cdot$ ’.

$r \cdot (a_{0} + a_{1} x + \dots a_{n} x^{n}) = (r a_{0}) + (r a_{1}) x + \dots (r a_{n}) x^{n}$

This space differs from the space $𝒫_{3}$ of Example 1.8. This space contains not just degree three polynomials, but degree thirty polynomials and degree three hundred polynomials, too. Each individual polynomial of course is of a finite degree, but the set has no single bound on the degree of all of its members.

We can think of this example, like the prior one, in terms of infinite-tuples. For instance, we can think of $1 + 3 x + 5 x^{2}$ as corresponding to $(1, 3, 5, 0, 0, \dots)$ . However, this space differs from the one in Example 1.10. Here, each member of the set has a finite degree, that is, under the correspondence there is no element from this space matching $(1, 2, 5, 10, \dots)$ . Vectors in this space correspond to infinite-tuples that end in zeroes.

Example 1.12 The set ${f ∣ f : ℝ \to ℝ}$ of all real-valued functions of one real variable is a vector space under these.

$(f_{1} + f_{2}) (x) = f_{1} (x) + f_{2} (x) (r \cdot f) (x) = r f (x)$

The difference between this and Example 1.10 is the domain of the functions.

Example 1.13 The set $F = {a \cos θ + b \sin θ ∣ a, b \in ℝ}$ of real-valued functions of the real variable $θ$ is a vector space under the operations

$(a_{1} \cos θ + b_{1} \sin θ) + (a_{2} \cos θ + b_{2} \sin θ) = (a_{1} + a_{2}) \cos θ + (b_{1} + b_{2}) \sin θ$

and

$r \cdot (a \cos θ + b \sin θ) = (r a) \cos θ + (r b) \sin θ$

inherited from the space in the prior example. (We can think of $F$ as “the same” as $ℝ^{2}$ in that $a \cos θ + b \sin θ$ corresponds to the vector with components $a$ and $b$ .)

Example 1.14 The set

${f : ℝ \to ℝ ∣ \frac{d^{2} f}{d x^{2}} + f = 0}$

is a vector space under the, by now natural, interpretation.

$(f + g) (x) = f (x) + g (x) (r \cdot f) (x) = r f (x)$

In particular, notice that basic Calculus gives

$\frac{d^{2} (f + g)}{d x^{2}} + (f + g) = (\frac{d^{2} f}{d x^{2}} + f) + (\frac{d^{2} g}{d x^{2}} + g)$

and

$\frac{d^{2} (r f)}{d x^{2}} + (r f) = r (\frac{d^{2} f}{d x^{2}} + f)$

and so the space is closed under addition and scalar multiplication. This turns out to equal the space from the prior example—functions satisfying this differential equation have the form $a \cos θ + b \sin θ$ —but this description suggests an extension to solutions sets of other differential equations.

Example 1.15 The set of solutions of a homogeneous linear system in $n$ variables is a vector space under the operations inherited from $ℝ^{n}$ . For example, for closure under addition consider a typical equation in that system $c_{1} x_{1} + \dots + c_{n} x_{n} = 0$ and suppose that both these vectors

$\vec{v} = (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) \vec{w} = (\begin{matrix} w_{1} \\ ⋮ \\ w_{n} \end{matrix})$

satisfy the equation. Then their sum $\vec{v} + \vec{w}$ also satisfies that equation: $c_{1} (v_{1} + w_{1}) + \dots + c_{n} (v_{n} + w_{n}) = (c_{1} v_{1} + \dots + c_{n} v_{n}) + (c_{1} w_{1} + \dots + c_{n} w_{n}) = 0$ . The checks of the other vector space conditions are just as routine.

We often omit the multiplication symbol ‘ $\cdot$ ’ between the scalar and the vector. We distinguish the multiplication in $c_{1} v_{1}$ from that in $r \vec{v}$ by context, since if both multiplicands are real numbers then it must be real-real multiplication while if one is a vector then it must be scalar-vector multiplication.

Example 1.15 has brought us full circle since it is one of our motivating examples. Now, with some feel for the kinds of structures that satisfy the definition of a vector space, we can reflect on that definition. For example, why specify in the definition the condition that $1 \cdot \vec{v} = \vec{v}$ but not a condition that $0 \cdot \vec{v} = \vec{0}$ ?

One answer is that this is just a definition—it gives the rules and you need to follow those rules to continue.

Another answer is perhaps more satisfying. People in this area have worked to develop the right balance of power and generality. This definition is shaped so that it contains the conditions needed to prove all of the interesting and important properties of spaces of linear combinations. As we proceed, we shall derive all of the properties natural to collections of linear combinations from the conditions given in the definition.

The next result is an example. We do not need to include these properties in the definition of vector space because they follow from the properties already listed there.

Lemma 1.16 In any vector space $V$ , for any $\vec{v} \in V$ and $r \in ℝ$ , we have (1) $0 \cdot \vec{v} = \vec{0}$ , (2) $(- 1 \cdot \vec{v}) + \vec{v} = \vec{0}$ , and (3) $r \cdot \vec{0} = \vec{0}$ .

Proof For (1) note that $\vec{v} = (1 + 0) \cdot \vec{v} = \vec{v} + (0 \cdot \vec{v})$ . Add to both sides the additive inverse of $\vec{v}$ , the vector $\vec{w}$ such that $\vec{w} + \vec{v} = \vec{0}$ .

$\begin{aligned} \vec{w} + \vec{v} & = \vec{w} + \vec{v} + 0 \cdot \vec{v} \\ \vec{0} & = \vec{0} + 0 \cdot \vec{v} \\ \vec{0} & = 0 \cdot \vec{v} \end{aligned}$

Item (2) is easy: $(- 1 \cdot \vec{v}) + \vec{v} = (- 1 + 1) \cdot \vec{v} = 0 \cdot \vec{v} = \vec{0}$ . For (3), $r \cdot \vec{0} = r \cdot (0 \cdot \vec{0}) = (r \cdot 0) \cdot \vec{0} = \vec{0}$ will do.

QED

The second item shows that we can write the additive inverse of $\vec{v}$ as ‘ $- \vec{v}$ ’ without worrying about any confusion with $(- 1) \cdot \vec{v}$ .

A recap: our study in Chapter One of Gaussian reduction led us to consider collections of linear combinations. So in this chapter we have defined a vector space to be a structure in which we can form such combinations, subject to simple conditions on the addition and scalar multiplication operations. In a phrase: vector spaces are the right context in which to study linearity.

From the fact that it forms a whole chapter, and especially because that chapter is the first one, a reader could suppose that our purpose in this book is the study of linear systems. The truth is that we will not so much use vector spaces in the study of linear systems as we instead have linear systems start us on the study of vector spaces. The wide variety of examples from this subsection shows that the study of vector spaces is interesting and important in its own right. Linear systems won’t go away. But from now on our primary objects of study will be vector spaces.

Exercises

Exercise 1.17 Worked answer

Name the zero vector for each of these vector spaces.
1. The space of degree three polynomials under the natural operations.
2. The space of $2 \times 4$ matrices.
3. The space ${f : [0. .1] \to ℝ ∣ f is continuous}$ .
4. The space of real-valued functions of one natural number variable.
Back to Exercise 1.17

Answer.
1. $0 + 0 x + 0 x^{2} + 0 x^{3}$
2. $(\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$
3. The constant function $f (x) = 0$
4. The constant function $f (n) = 0$
Exercise 1.18 Worked answer

Recommended. Find the additive inverse, in the vector space, of the vector.
1. In $𝒫_{3}$ , the vector $- 3 - 2 x + x^{2}$ .
2. In the space $2 \times 2$ ,
  
  $(\begin{matrix} 1 & - 1 \\ 0 & 3 \end{matrix}) .$
3. In ${a e^{x} + b e^{- x} ∣ a, b \in ℝ}$ , the space of functions of the real variable $x$ under the natural operations, the vector $3 e^{x} - 2 e^{- x}$ .
Back to Exercise 1.18

Answer.
1. $3 + 2 x - x^{2}$
2. $(\begin{matrix} - 1 & + 1 \\ 0 & - 3 \end{matrix})$
3. $- 3 e^{x} + 2 e^{- x}$
Exercise 1.19 Worked answer

Recommended. For each, list three elements and then show it is a vector space.
1. The set of linear polynomials $𝒫_{1} = {a_{0} + a_{1} x ∣ a_{0}, a_{1} \in ℝ}$ under the usual polynomial addition and scalar multiplication operations.
2. The set of linear polynomials ${a_{0} + a_{1} x ∣ a_{0} - 2 a_{1} = 0}$ , under the usual polynomial addition and scalar multiplication operations.
Hint. Use Example 1.3 as a guide. Most of the ten conditions are just verifications.
Back to Exercise 1.19

Answer.
1. Three elements are: $1 + 2 x$ , $2 - 1 x$ , and $x$ . (Of course, many answers are possible.)
  
  The verification is just like Example 1.4. We first do conditions $1$ - $5$ from Definition 1.1, having to do with addition. For closure under addition, condition (1), note that where $a + b x, c + d x \in 𝒫_{1}$ we have that $(a + b x) + (c + d x) = (a + c) + (b + d) x$ is a linear polynomial with real coefficients and so is an element of $𝒫_{1}$ . Condition (2) is verified with: where $a + b x, c + d x \in 𝒫_{1}$ then $(a + b x) + (c + d x) = (a + c) + (b + d) x$ , while in the other order they are $(c + d x) + (a + b x) = (c + a) + (d + b) x$ , and both $a + c = c + a$ and $b + d = d + b$ as these are real numbers. Condition (3) is similar: suppose $a + b x, c + d x, e + f x \in 𝒫$ then $((a + b x) + (c + d x)) + (e + f x) = (a + c + e) + (b + d + f) x$ while $(a + b x) + ((c + d x) + (e + f x)) = (a + c + e) + (b + d + f) x$ , and the two are equal (that is, real number addition is associative so $(a + c) + e = a + (c + e)$ and $(b + d) + f = b + (d + f)$ ). For condition (4) observe that the linear polynomial $0 + 0 x \in 𝒫_{1}$ has the property that $(a + b x) + (0 + 0 x) = a + b x$ and $(0 + 0 x) + (a + b x) = a + b x$ . For the last condition in this paragraph, condition (5), note that for any $a + b x \in 𝒫_{1}$ the additive inverse is $- a - b x \in 𝒫_{1}$ since $(a + b x) + (- a - b x) = (- a - b x) + (a + b x) = 0 + 0 x$ .
  
  We next also check conditions (6)-(10), involving scalar multiplication. For (6), the condition that the space be closed under scalar multiplication, suppose that $r$ is a real number and $a + b x$ is an element of $𝒫_{1}$ , and then $r (a + b x) = (r a) + (r b) x$ is an element of $𝒫_{1}$ because it is a linear polynomial with real number coefficients. Condition (7) holds because $(r + s) (a + b x) = r (a + b x) + s (a + b x)$ is true from the distributive property for real number multiplication. Condition (8) is similar: $r ((a + b x) + (c + d x)) = r ((a + c) + (b + d) x) = r (a + c) + r (b + d) x = (r a + r c) + (r b + r d) x = r (a + b x) + r (c + d x)$ . For (9) we have $(r s) (a + b x) = (r s a) + (r s b) x = r (s a + s b x) = r (s (a + b x))$ . Finally, condition (10) is $1 (a + b x) = (1 a) + (1 b) x = a + b x$ .
2. Call the set $P$ . In the prior item in this exercise there was no restriction on the coefficients but here we are restricting attention to those linear polynomials where $a_{0} - 2 a_{1} = 0$ , that is, where the constant term minus twice the coefficient of the linear term is zero. Thus, three typical elements of $P$ are $2 + 1 x$ , $6 + 3 x$ , and $- 4 - 2 x$ .
  
  For condition (1) we must show that if we add two linear polynomials that satisfy the restriction then we get a linear polynomial also satisfying the restriction: here that argument is that if $a + b x, c + d x \in P$ then $(a + b x) + (c + d x) = (a + c) + (b + d) x$ is an element of $P$ because $(a + c) - 2 (b + d) = (a - 2 b) + (c - 2 d) = 0 + 0 = 0$ . We can verify condition (2) with: where $a + b x, c + d x \in 𝒫_{1}$ then $(a + b x) + (c + d x) = (a + c) + (b + d) x$ , while in the other order they are $(c + d x) + (a + b x) = (c + a) + (d + b) x$ , and both $a + c = c + a$ and $b + d = d + b$ as these are real numbers. (That is, this condition is not affected by the restriction and the verification is the same as the verification in the first item of this exercise). Condition (3) is also not affected by the extra restriction: suppose that $a + b x, c + d x, e + f x \in 𝒫$ then $((a + b x) + (c + d x)) + (e + f x) = (a + c + e) + (b + d + f) x$ while $(a + b x) + ((c + d x) + (e + f x)) = (a + c + e) + (b + d + f) x$ , and the two are equal. For condition (4) observe that the linear polynomial satisfies the restriction $0 + 0 x \in P$ because its constant term minus twice the coefficient of its linear term is zero, and then the verification from the first item of this question applies: $0 + 0 x \in 𝒫_{1}$ has the property that $(a + b x) + (0 + 0 x) = a + b x$ and $(0 + 0 x) + (a + b x) = a + b x$ . To check condition (5), note that for any $a + b x \in P$ the additive inverse is $- a - b x$ since it is an element of $P$ (because $a + b x \in P$ we know that $a - 2 b = 0$ and multiplying both sides by $- 1$ gives that $- a + 2 b = 0$ ), and as in the first item it acts as the additive inverse $(a + b x) + (- a - b x) = (- a - b x) + (a + b x) = 0 + 0 x$ .
  
  We must also check conditions (6)-(10), those for scalar multiplication. For (6), the condition that the space be closed under scalar multiplication, suppose that $r$ is a real number and $a + b x \in P$ (so that $a - 2 b = 0$ ), then $r (a + b x) = (r a) + (r b) x$ is an element of $P$ because it is a linear polynomial with real number coefficients satisfying that $(r a) - 2 (r b) = r (a - 2 b) = 0$ . Condition (7) holds for the same reason that it holds in the first item of this exercise, because $(r + s) (a + b x) = r (a + b x) + s (a + b x)$ is true from the distributive property for real number multiplication. Condition (8) is also unchanged from the first item: $r ((a + b x) + (c + d x)) = r ((a + c) + (b + d) x) = r (a + c) + r (b + d) x = (r a + r c) + (r b + r d) x = r (a + b x) + r (c + d x)$ . So is (9): $(r s) (a + b x) = (r s a) + (r s b) x = r (s a + s b x) = r (s (a + b x))$ . Finally, so is condition (10): $1 (a + b x) = (1 a) + (1 b) x = a + b x$ .
Exercise 1.20 Worked answer

For each, list three elements and then show it is a vector space.
1. The set of $2 \times 2$ matrices with real entries under the usual matrix operations.
2. The set of $2 \times 2$ matrices with real entries where the $2, 1$ entry is zero, under the usual matrix operations.
Back to Exercise 1.20

Answer. Use Example 1.4 as a guide. (Comment. Because many of the conditions are quite easy to check, sometimes a person can feel that they must have missed something. Keep in mind that easy to do, or routine, is different from not necessary to do.)
1. Here are three elements.
  
  $(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}), (\begin{matrix} - 1 & - 2 \\ - 3 & - 4 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$
  
  For (1), the sum of $2 \times 2$ real matrices is a $2 \times 2$ real matrix. For (2) we consider the sum of two matrices
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) + (\begin{matrix} e & f \\ g & h \end{matrix}) = (\begin{matrix} a + e & b + f \\ c + g & d + h \end{matrix})$
  
  and apply commutativity of real number addition
  
  $= (\begin{matrix} e + a & f + b \\ g + c & h + d \end{matrix}) = (\begin{matrix} e & f \\ g & h \end{matrix}) + (\begin{matrix} a & b \\ c & d \end{matrix})$
  
  to verify that the addition of the matrices is commutative. The verification for condition (3), associativity of matrix addition, is similar to the prior verification:
  
  $((\begin{matrix} a & b \\ c & d \end{matrix}) + (\begin{matrix} e & f \\ g & h \end{matrix})) + (\begin{matrix} i & j \\ k & l \end{matrix}) = (\begin{matrix} (a + e) + i & (b + f) + j \\ (c + g) + k & (d + h) + l \end{matrix})$
  
  while
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) + ((\begin{matrix} e & f \\ g & h \end{matrix}) + (\begin{matrix} i & j \\ k & l \end{matrix})) = (\begin{matrix} a + (e + i) & b + (f + j) \\ c + (g + k) & d + (h + l) \end{matrix})$
  
  and the two are the same entry-by-entry because real number addition is associative. For (4), the zero element of this space is the $2 \times 2$ matrix of zeroes. Condition (5) holds because for any $2 \times 2$ matrix $A$ the additive inverse is the matrix whose entries are the negative of $A$ ’s, the matrix $- 1 \cdot A$ .
  
  Condition ( $6$ ) holds because a scalar multiple of a $2 \times 2$ matrix is a $2 \times 2$ matrix. For condition (7) we have this.
  
  $\begin{matrix} (r + s) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} (r + s) a & (r + s) b \\ (r + s) c & (r + s) d \end{matrix}) \\ = (\begin{matrix} r a + s a & r b + s b \\ r c + s c & r d + s d \end{matrix}) = r (\begin{matrix} a & b \\ c & d \end{matrix}) + s (\begin{matrix} a & b \\ c & d \end{matrix}) \end{matrix}$
  
  Condition (8) goes the same way.
  
  $\begin{matrix} r ((\begin{matrix} a & b \\ c & d \end{matrix}) + (\begin{matrix} e & f \\ g & h \end{matrix})) = r (\begin{matrix} a + e & b + f \\ c + g & d + h \end{matrix}) = (\begin{matrix} r a + r e & r b + r f \\ r c + r g & r d + r h \end{matrix}) \\ = r (\begin{matrix} a & b \\ c & d \end{matrix}) + r (\begin{matrix} e & f \\ g & h \end{matrix}) = r ((\begin{matrix} a & b \\ c & d \end{matrix}) + (\begin{matrix} e & f \\ g & h \end{matrix})) \end{matrix}$
  
  For (9) we have this.
  
  $(r s) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} r s a & r s b \\ r s c & r s d \end{matrix}) = r (\begin{matrix} s a & s b \\ s c & s d \end{matrix}) = r (s (\begin{matrix} a & b \\ c & d \end{matrix}))$
  
  Condition (10) is just as easy.
  
  $1 (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} 1 \cdot a & 1 \cdot b \\ 1 \cdot c & 1 \cdot d \end{matrix}) = (\begin{matrix} s a & s b \\ s c & s d \end{matrix})$
2. This differs from the prior item in this exercise only in that we are restricting to the set $T$ of matrices with a zero in the second row and first column. Here are three elements of $T$ .
  
  $(\begin{matrix} 1 & 2 \\ 0 & 4 \end{matrix}), (\begin{matrix} - 1 & - 2 \\ 0 & - 4 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$
  
  Some of the verifications for this item are the same as for the first item in this exercise, and below we’ll just do the ones that are different.
  
  For (1), the sum of $2 \times 2$ real matrices with a zero in the $2, 1$ entry is also a $2 \times 2$ real matrix with a zero in the $2, 1$ entry.
  
  $(\begin{matrix} a & b \\ 0 & d \end{matrix}) + (\begin{matrix} e & f \\ 0 & h \end{matrix}) (\begin{matrix} a + e & b + f \\ 0 & d + h \end{matrix})$
  
  The verification for condition (2) given in the prior item works in this item also. The same holds for condition (3). For (4), note that the $2 \times 2$ matrix of zeroes is an element of $T$ . Condition (5) holds because for any $2 \times 2$ matrix $A$ the additive inverse is the matrix $- 1 \cdot A$ and so the additive inverse of a matrix with a zero in the $2, 1$ entry is also a matrix with a zero in the $2, 1$ entry.
  
  Condition $6$ holds because a scalar multiple of a $2 \times 2$ matrix with a zero in the $2, 1$ entry is a $2 \times 2$ matrix with a zero in the $2, 1$ entry. Condition (7)’s verification is the same as in the prior item. So are condition (8)’s, (9)’s, and (10)’s.
Exercise 1.21 Worked answer

Recommended. For each, list three elements and then show it is a vector space.
1. The set of three-component row vectors with their usual operations.
2. The set
  
  ${(\begin{matrix} x \\ y \\ z \\ w \end{matrix}) \in ℝ^{4} ∣ x + y - z + w = 0}$
  
  under the operations inherited from $ℝ^{4}$ .
Back to Exercise 1.21

Answer.
1. Three elements are $(\begin{matrix} 1 & 2 & 3 \end{matrix})$ , $(\begin{matrix} 2 & 1 & 3 \end{matrix})$ , and $(\begin{matrix} 0 & 0 & 0 \end{matrix})$ .
  
  We must check conditions (1)-(10) in Definition 1.1. Conditions (1)-(5) concern addition. For condition (1) recall that the sum of two three-component row vectors
  
  $(\begin{matrix} a & b & c \end{matrix}) + (\begin{matrix} d & e & f \end{matrix}) = (\begin{matrix} a + d & b + e & c + f \end{matrix})$
  
  is also a three-component row vector (all of the letters $a, \dots, f$ represent real numbers). Verification of (2) is routine
  
  $\begin{matrix} (\begin{matrix} a & b & c \end{matrix}) + (\begin{matrix} d & e & f \end{matrix}) = (\begin{matrix} a + d & b + e & c + f \end{matrix}) \\ = (\begin{matrix} d + a & e + b & f + c \end{matrix}) = (\begin{matrix} d & e & f \end{matrix}) + (\begin{matrix} a & b & c \end{matrix}) \end{matrix}$
  
  (the second equality holds because the three entries are real numbers and real number addition commutes). Condition (3)’s verification is similar.
  
  $\begin{matrix} ((\begin{matrix} a & b & c \end{matrix}) + (\begin{matrix} d & e & f \end{matrix})) + (\begin{matrix} g & h & i \end{matrix}) = (\begin{matrix} (a + d) + g & (b + e) + h & (c + f) + i \end{matrix}) \\ = (\begin{matrix} a + (d + g) & b + (e + h) & c + (f + i) \end{matrix}) = (\begin{matrix} a & b & c \end{matrix}) + ((\begin{matrix} d & e & f \end{matrix}) + (\begin{matrix} g & h & i \end{matrix})) \end{matrix}$
  
  For (4), observe that the three-component row vector $(\begin{matrix} 0 & 0 & 0 \end{matrix})$ is the additive identity: $(\begin{matrix} a & b & c \end{matrix}) + (\begin{matrix} 0 & 0 & 0 \end{matrix}) = (\begin{matrix} a & b & c \end{matrix})$ . To verify condition (5), assume we are given the element $(\begin{matrix} a & b & c \end{matrix})$ of the set and note that $(\begin{matrix} - a & - b & - c \end{matrix})$ is also in the set and has the desired property: $(\begin{matrix} a & b & c \end{matrix}) + (\begin{matrix} - a & - b & - c \end{matrix}) = (\begin{matrix} 0 & 0 & 0 \end{matrix})$ .
  
  Conditions (6)-(10) involve scalar multiplication. To verify (6), that the space is closed under the scalar multiplication operation that was given, note that $r (\begin{matrix} a & b & c \end{matrix}) = (\begin{matrix} r a & r b & r c \end{matrix})$ is a three-component row vector with real entries. For (7) we compute.
  
  $\begin{matrix} (r + s) (\begin{matrix} a & b & c \end{matrix}) = (\begin{matrix} (r + s) a & (r + s) b & (r + s) c \end{matrix}) = (\begin{matrix} r a + s a & r b + s b & r c + s c \end{matrix}) \\ = (\begin{matrix} r a & r b & r c \end{matrix}) + (\begin{matrix} s a & s b & s c \end{matrix}) = r (\begin{matrix} a & b & c \end{matrix}) + s (\begin{matrix} a & b & c \end{matrix}) \end{matrix}$
  
  Condition (8) is very similar.
  
  $\begin{matrix} r ((\begin{matrix} a & b & c \end{matrix}) + (\begin{matrix} d & e & f \end{matrix})) \\ \begin{aligned} = r (\begin{array}{c} a + d & b + e & c + f \end{array}) = (\begin{array}{c} r (a + d) & r (b + e) & r (c + f) \end{array}) \\ = (\begin{array}{c} r a + r d & r b + r e & r c + r f \end{array}) = (\begin{array}{c} r a & r b & r c \end{array}) + (\begin{array}{c} r d & r e & r f \end{array}) \\ = r (\begin{array}{c} a & b & c \end{array}) + r (\begin{array}{c} d & e & f \end{array}) \end{aligned} \end{matrix}$
  
  So is the computation for condition (9).
  
  $(r s) (\begin{matrix} a & b & c \end{matrix}) = (\begin{matrix} r s a & r s b & r s c \end{matrix}) = r (\begin{matrix} s a & s b & s c \end{matrix}) = r (s (\begin{matrix} a & b & c \end{matrix}))$
  
  Condition (10) is just as routine $1 (\begin{matrix} a & b & c \end{matrix}) = (\begin{matrix} 1 \cdot a & 1 \cdot b & 1 \cdot c \end{matrix}) = (\begin{matrix} a & b & c \end{matrix})$ .
2. Call the set $L$ . Closure of addition, condition (1), involves checking that if the summands are members of $L$ then the sum
  
  $(\begin{matrix} a \\ b \\ c \\ d \end{matrix}) + (\begin{matrix} e \\ f \\ g \\ h \end{matrix}) = (\begin{matrix} a + e \\ b + f \\ c + g \\ d + h \end{matrix})$
  
  is also a member of $L$ , which is true because it satisfies the criteria for membership in $L$ : $(a + e) + (b + f) - (c + g) + (d + h) = (a + b - c + d) + (e + f - g + h) = 0 + 0$ . The verifications for conditions (2), (3), and (5) are similar to the ones in the first part of this exercise. For condition (4) note that the vector of zeroes is a member of $L$ because its first component plus its second, minus its third, and plus its fourth, totals to zero.
  
  Condition (6), closure of scalar multiplication, is similar: where the vector is an element of $L$ ,
  
  $r (\begin{matrix} a \\ b \\ c \\ d \end{matrix}) = (\begin{matrix} r a \\ r b \\ r c \\ r d \end{matrix})$
  
  is also an element of $L$ because $r a + r b - r c + r d = r (a + b - c + d) = r \cdot 0 = 0$ . The verification for conditions (7), (8), (9), and (10) are as in the prior item of this exercise.
Exercise 1.22 Worked answer

Recommended. Show that each of these is not a vector space. (Hint. Check closure by listing two members of each set and trying some operations on them.)
1. Under the operations inherited from $ℝ^{3}$ , this set
  
  ${(\begin{matrix} x \\ y \\ z \end{matrix}) \in ℝ^{3} ∣ x + y + z = 1}$
2. Under the operations inherited from $ℝ^{3}$ , this set
  
  ${(\begin{matrix} x \\ y \\ z \end{matrix}) \in ℝ^{3} ∣ x^{2} + y^{2} + z^{2} = 1}$
3. Under the usual matrix operations,
  
  ${(\begin{matrix} a & 1 \\ b & c \end{matrix}) ∣ a, b, c \in ℝ}$
4. Under the usual polynomial operations,
  
  ${a_{0} + a_{1} x + a_{2} x^{2} ∣ a_{0}, a_{1}, a_{2} \in ℝ^{+}}$
  
  where $ℝ^{+}$ is the set of reals greater than zero
5. Under the inherited operations,
  
  ${(\begin{matrix} x \\ y \end{matrix}) \in ℝ^{2} ∣ x + 3 y = 4 and 2 x - y = 3 and 6 x + 4 y = 10}$
Back to Exercise 1.22

Answer. In each item the set is called $Q$ . For some items, there are other correct ways to show that $Q$ is not a vector space.
1. It is not closed under addition; it fails to meet condition (1).
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \in Q (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) \notin Q$
2. It is not closed under addition.
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \in Q (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) \notin Q$
3. It is not closed under addition.
  
  $(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}) \in Q (\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}) \notin Q$
4. It is not closed under scalar multiplication.
  
  $1 + 1 x + 1 x^{2} \in Q - 1 \cdot (1 + 1 x + 1 x^{2}) \notin Q$
5. It is empty, violating condition (4).
Exercise 1.23 Worked answer

Define addition and scalar multiplication operations to make the complex numbers a vector space over $ℝ$ .

Back to Exercise 1.23

Answer. The usual operations $(v_{0} + v_{1} i) + (w_{0} + w_{1} i) = (v_{0} + w_{0}) + (v_{1} + w_{1}) i$ and $r (v_{0} + v_{1} i) = (r v_{0}) + (r v_{1}) i$ suffice. The check is easy.
Exercise 1.24 Worked answer

Is the set of rational numbers a vector space over $ℝ$ under the usual addition and scalar multiplication operations?

Back to Exercise 1.24

Answer. No, it is not closed under scalar multiplication since, e.g., $π \cdot (1)$ is not a rational number.
Exercise 1.25 Worked answer

Show that the set of linear combinations of the variables $x, y, z$ is a vector space under the natural addition and scalar multiplication operations.

Back to Exercise 1.25

Answer. The natural operations are $(v_{1} x + v_{2} y + v_{3} z) + (w_{1} x + w_{2} y + w_{3} z) = (v_{1} + w_{1}) x + (v_{2} + w_{2}) y + (v_{3} + w_{3}) z$ and $r \cdot (v_{1} x + v_{2} y + v_{3} z) = (r v_{1}) x + (r v_{2}) y + (r v_{3}) z$ . The check that this is a vector space is easy; use Example 1.4 as a guide.
Exercise 1.26 Worked answer

Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations.

$(\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \end{matrix}) = (\begin{matrix} x_{1} - x_{2} \\ y_{1} - y_{2} \end{matrix}) r \cdot (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} r x \\ r y \end{matrix})$

Back to Exercise 1.26

Answer. The ‘ $+$ ’ operation is not commutative (that is, condition (2) is not met); producing two members of the set witnessing this assertion is easy.
Exercise 1.27 Worked answer

Prove or disprove that $ℝ^{3}$ is a vector space under these operations.
1. $(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) and r (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r x \\ r y \\ r z \end{matrix})$
2. $(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) and r (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
Back to Exercise 1.27

Answer.
1. It is not a vector space.
  
  $(1 + 1) \cdot (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \neq (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})$
2. It is not a vector space.
  
  $1 \cdot (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \neq (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})$
Exercise 1.28 Worked answer

Recommended. For each, decide if it is a vector space; the intended operations are the natural ones.
1. The diagonal $2 \times 2$ matrices
  
  ${(\begin{matrix} a & 0 \\ 0 & b \end{matrix}) ∣ a, b \in ℝ}$
2. This set of $2 \times 2$ matrices
  
  ${(\begin{matrix} x & x + y \\ x + y & y \end{matrix}) ∣ x, y \in ℝ}$
3. This set
  
  ${(\begin{matrix} x \\ y \\ z \\ w \end{matrix}) \in ℝ^{4} ∣ x + y + z + w = 1}$
4. The set of functions ${f : ℝ \to ℝ ∣ d f / d x + 2 f = 0}$
5. The set of functions ${f : ℝ \to ℝ ∣ d f / d x + 2 f = 1}$
Back to Exercise 1.28

Answer. For each “yes” answer, you can just check a couple of linear combinations in the space. For each “no” answer, give a specific example of the failure of one of the conditions.
1. Yes. One linear combination picked out of the air is this.
  
  $2 \cdot (\begin{matrix} 3 & 0 \\ 0 & 4 \end{matrix}) + 5 \cdot (\begin{matrix} 6 & 0 \\ 0 & 7 \end{matrix})$
  
  The key point about the collection being a vector space is that the result is a diagonal matrix.
  
  $= (\begin{matrix} 36 & 0 \\ 0 & 43 \end{matrix})$
2. Yes. Here is a linear combination.
  
  $10 \cdot (\begin{matrix} 1 & 3 \\ 3 & 2 \end{matrix}) + 11 \cdot (\begin{matrix} 3 & 7 \\ 7 & 4 \end{matrix})$
  
  We look for the result to follow the form that the upper left and lower right entries add to make the upper right entry, and also the lower left.
  
  $= (\begin{matrix} 43 & 107 \\ 107 & 64 \end{matrix})$
3. No, this set is not closed under the natural addition operation. The vector of all $1 / 4$ ’s is a member of this set but when added to itself the result, the vector of all $1 / 2$ ’s, is a nonmember.
4. Yes.
5. No, $f (x) = e^{- 2 x} + (1 / 2)$ is in the set but $2 \cdot f$ is not (that is, condition (6) fails).
Exercise 1.29 Worked answer

Recommended. Prove or disprove that this is a vector space: the real-valued functions $f$ of one real variable such that $f (7) = 0$ .

Back to Exercise 1.29

Answer. It is a vector space. Most conditions of the definition of vector space are routine; we here check only closure. For addition, $(f_{1} + f_{2}) (7) = f_{1} (7) + f_{2} (7) = 0 + 0 = 0$ . For scalar multiplication, $(r \cdot f) (7) = r f (7) = r 0 = 0$ .
Exercise 1.30 Worked answer

Recommended. Show that the set $ℝ^{+}$ of positive reals is a vector space when we interpret ‘ $x + y$ ’ to mean the product of $x$ and $y$ (so that $2 + 3$ is $6$ ), and we interpret ‘ $r \cdot x$ ’ as the $r$ -th power of $x$ .

Back to Exercise 1.30

Answer. We check Definition 1.1.

First, closure under ‘ $+$ ’ holds because the product of two positive reals is a positive real. The second condition is satisfied because real multiplication commutes. Similarly, as real multiplication associates, the third checks. For the fourth condition, observe that multiplying a number by $1 \in ℝ^{+}$ won’t change the number. Fifth, any positive real has a reciprocal that is a positive real.

The sixth, closure under ‘ $\cdot$ ’, holds because any power of a positive real is a positive real. The seventh condition is just the rule that $v^{r + s}$ equals the product of $v^{r}$ and $v^{s}$ . The eight condition says that $(v w)^{r} = v^{r} w^{r}$ . The ninth condition asserts that $(v^{r})^{s} = v^{r s}$ . The final condition says that $v^{1} = v$ .
Exercise 1.31 Worked answer

Is ${(x, y) ∣ x, y \in ℝ}$ a vector space under these operations?
1. $(x_{1}, y_{1}) + (x_{2}, y_{2}) = (x_{1} + x_{2}, y_{1} + y_{2})$ and $r \cdot (x, y) = (r x, y)$
2. $(x_{1}, y_{1}) + (x_{2}, y_{2}) = (x_{1} + x_{2}, y_{1} + y_{2})$ and $r \cdot (x, y) = (r x, 0)$
Back to Exercise 1.31

Answer.
1. No: $1 \cdot (0, 1) + 1 \cdot (0, 1) \neq (1 + 1) \cdot (0, 1)$ .
2. No; the same calculation as the prior answer shows a condition in the definition of a vector space that is violated. Another example of a violation of the conditions for a vector space is that $1 \cdot (0, 1) \neq (0, 1)$ .
Exercise 1.32 Worked answer

Prove or disprove that this is a vector space: the set of polynomials of degree greater than or equal to two, along with the zero polynomial.

Back to Exercise 1.32

Answer. It is not a vector space since it is not closed under addition, as $(x^{2}) + (1 + x - x^{2})$ is not in the set.
Exercise 1.33 Worked answer

At this point “the same” is only an intuition, but nonetheless for each vector space identify the $k$ for which the space is “the same” as $ℝ^{k}$ .
1. The $2 \times 3$ matrices under the usual operations
2. The $n \times m$ matrices (under their usual operations)
3. This set of $2 \times 2$ matrices
  
  ${(\begin{matrix} a & 0 \\ b & c \end{matrix}) ∣ a, b, c \in ℝ}$
4. This set of $2 \times 2$ matrices
  
  ${(\begin{matrix} a & 0 \\ b & c \end{matrix}) ∣ a + b + c = 0}$
Back to Exercise 1.33

Answer.
1. $6$
2. $n m$
3. $3$
4. To see that the answer is $2$ , rewrite it as
  
  ${(\begin{matrix} a & 0 \\ b & - a - b \end{matrix}) ∣ a, b \in ℝ}$
  
  so that there are two parameters.
Exercise 1.34 Worked answer

Using $\vec{+}$ to represent vector addition and $\vec{\cdot}$ for scalar multiplication, restate the definition of vector space.

Back to Exercise 1.34

Answer. A vector space (over $ℝ$ ) consists of a set $V$ along with two operations ‘ $\vec{+}$ ’ and ‘ $\vec{\cdot}$ ’ subject to these conditions. Where $\vec{v}, \vec{w} \in V$ , (1) their vector sum $\vec{v} \vec{+} \vec{w}$ is an element of $V$ . If $\vec{u}, \vec{v}, \vec{w} \in V$ then (2) $\vec{v} \vec{+} \vec{w} = \vec{w} \vec{+} \vec{v}$ and (3) $(\vec{v} \vec{+} \vec{w}) \vec{+} \vec{u} = \vec{v} \vec{+} (\vec{w} \vec{+} \vec{u})$ . (4) There is a zero vector $\vec{0} \in V$ such that $\vec{v} \vec{+} \vec{0} = \vec{v}$ for all $\vec{v} \in V$ . (5) Each $\vec{v} \in V$ has an additive inverse $\vec{w} \in V$ such that $\vec{w} \vec{+} \vec{v} = \vec{0}$ . If $r, s$ are scalars, that is, members of $ℝ$ ), and $\vec{v}, \vec{w} \in V$ then (6) each scalar multiple $r \cdot \vec{v}$ is in $V$ . If $r, s \in ℝ$ and $\vec{v}, \vec{w} \in V$ then (7) $(r + s) \cdot \vec{v} = r \cdot \vec{v} \vec{+} s \cdot \vec{v}$ , and (8) $r \vec{\cdot} (\vec{v} + \vec{w}) = r \vec{\cdot} \vec{v} + r \vec{\cdot} \vec{w}$ , and (9) $(r s) \vec{\cdot} \vec{v} = r \vec{\cdot} (s \vec{\cdot} \vec{v})$ , and (10) $1 \vec{\cdot} \vec{v} = \vec{v}$ .
Exercise 1.35 Worked answer

Prove these.
1. For any $\vec{v} \in V$ , if $\vec{w} \in V$ is an additive inverse of $\vec{v}$ , then $\vec{v}$ is an additive inverse of $\vec{w}$ . So a vector is an additive inverse of any additive inverse of itself.
2. Vector addition left-cancels: if $\vec{v}, \vec{s}, \vec{t} \in V$ then $\vec{v} + \vec{s} = \vec{v} + \vec{t}$ implies that $\vec{s} = \vec{t}$ .
Back to Exercise 1.35

Answer.
1. Let $V$ be a vector space, let $\vec{v} \in V$ , and assume that $\vec{w} \in V$ is an additive inverse of $\vec{v}$ so that $\vec{w} + \vec{v} = \vec{0}$ . Because addition is commutative, $\vec{0} = \vec{w} + \vec{v} = \vec{v} + \vec{w}$ , so therefore $\vec{v}$ is also the additive inverse of $\vec{w}$ .
2. Let $V$ be a vector space and suppose $\vec{v}, \vec{s}, \vec{t} \in V$ . The additive inverse of $\vec{v}$ is $- \vec{v}$ so $\vec{v} + \vec{s} = \vec{v} + \vec{t}$ gives that $- \vec{v} + \vec{v} + \vec{s} = - \vec{v} + \vec{v} + \vec{t}$ , which says that $\vec{0} + \vec{s} = \vec{0} + \vec{t}$ and so $\vec{s} = \vec{t}$ .
Exercise 1.36 Worked answer

The definition of vector spaces does not explicitly say that $\vec{0} + \vec{v} = \vec{v}$ (it instead says that $\vec{v} + \vec{0} = \vec{v}$ ). Show that it must nonetheless hold in any vector space.

Back to Exercise 1.36

Answer. Addition is commutative, so in any vector space, for any vector $\vec{v}$ we have that $\vec{v} = \vec{v} + \vec{0} = \vec{0} + \vec{v}$ .
Exercise 1.37 Worked answer

Recommended. Prove or disprove that this is a vector space: the set of all matrices, under the usual operations.

Back to Exercise 1.37

Answer. It is not a vector space since addition of two matrices of unequal sizes is not defined, and thus the set fails to satisfy the closure condition.
Exercise 1.38 Worked answer

In a vector space every element has an additive inverse. Can some elements have two or more?

Back to Exercise 1.38

Answer. Each element of a vector space has one and only one additive inverse.

For, let $V$ be a vector space and suppose that $\vec{v} \in V$ . If ${\vec{w}}_{1}, {\vec{w}}_{2} \in V$ are both additive inverses of $\vec{v}$ then consider ${\vec{w}}_{1} + \vec{v} + {\vec{w}}_{2}$ . On the one hand, we have that it equals ${\vec{w}}_{1} + (\vec{v} + {\vec{w}}_{2}) = {\vec{w}}_{1} + \vec{0} = {\vec{w}}_{1}$ . On the other hand we have that it equals $({\vec{w}}_{1} + \vec{v}) + {\vec{w}}_{2} = \vec{0} + {\vec{w}}_{2} = {\vec{w}}_{2}$ . Therefore, ${\vec{w}}_{1} = {\vec{w}}_{2}$ .
Exercise 1.39 Worked answer
1. Prove that every point, line, or plane through the origin in $ℝ^{3}$ is a vector space under the inherited operations.
2. What if it doesn’t contain the origin?
Back to Exercise 1.39

Answer.
1. Every such set has the form ${r \cdot \vec{v} + s \cdot \vec{w} ∣ r, s \in ℝ}$ where either or both of $\vec{v}, \vec{w}$ may be $\vec{0}$ . With the inherited operations, closure of addition $(r_{1} \vec{v} + s_{1} \vec{w}) + (r_{2} \vec{v} + s_{2} \vec{w}) = (r_{1} + r_{2}) \vec{v} + (s_{1} + s_{2}) \vec{w}$ and scalar multiplication $c (r \vec{v} + s \vec{w}) = (c r) \vec{v} + (c s) \vec{w}$ are easy. The other conditions are also routine.
2. No such set can be a vector space under the inherited operations because it does not have a zero element.
Exercise 1.40 Worked answer

Using the idea of a vector space we can easily reprove that the solution set of a homogeneous linear system has either one element or infinitely many elements. Assume that $\vec{v} \in V$ is not $\vec{0}$ .
1. Prove that $r \cdot \vec{v} = \vec{0}$ if and only if $r = 0$ .
2. Prove that $r_{1} \cdot \vec{v} = r_{2} \cdot \vec{v}$ if and only if $r_{1} = r_{2}$ .
3. Prove that any nontrivial vector space is infinite.
4. Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion.
Back to Exercise 1.40

Answer. Assume that $\vec{v} \in V$ is not $\vec{0}$ .
1. One direction of the if and only if is clear: if $r = 0$ then $r \cdot \vec{v} = \vec{0}$ . For the other way, let $r$ be a nonzero scalar. If $r \vec{v} = \vec{0}$ then $(1 / r) \cdot r \vec{v} = (1 / r) \cdot \vec{0}$ shows that $\vec{v} = \vec{0}$ , contrary to the assumption.
2. Where $r_{1}, r_{2}$ are scalars, $r_{1} \vec{v} = r_{2} \vec{v}$ holds if and only if $(r_{1} - r_{2}) \vec{v} = \vec{0}$ . By the prior item, then $r_{1} - r_{2} = 0$ .
3. A nontrivial space has a vector $\vec{v} \neq \vec{0}$ . Consider the set ${k \cdot \vec{v} ∣ k \in ℝ}$ . By the prior item this set is infinite.
4. The solution set is either trivial, or nontrivial. In the second case, it is infinite.
Exercise 1.41 Worked answer

Is this a vector space under the natural operations: the real-valued functions of one real variable that are differentiable?

Back to Exercise 1.41

Answer. Yes. A theorem of first semester calculus says that a sum of differentiable functions is differentiable and that $(f + g)^{'} = f^{'} + g^{'}$ , and that a multiple of a differentiable function is differentiable and that $(r \cdot f)^{'} = r f^{'}$ .
Exercise 1.42 Worked answer

A vector space over the complex numbers $ℂ$ has the same definition as a vector space over the reals except that scalars are drawn from $ℂ$ instead of from $ℝ$ . Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply: $(a_{0} + a_{1} i) + (b_{0} + b_{1} i) = (a_{0} + b_{0}) + (a_{1} + b_{1}) i$ and $(a_{0} + a_{1} i) (b_{0} + b_{1} i) = (a_{0} b_{0} - a_{1} b_{1}) + (a_{0} b_{1} + a_{1} b_{0}) i$ .)
1. The set of degree two polynomials with complex coefficients
2. This set
  
  ${(\begin{matrix} 0 & a \\ b & 0 \end{matrix}) ∣ a, b \in ℂ and a + b = 0 + 0 i}$
Back to Exercise 1.42

Answer. The check is routine. Note that ‘ $1$ ’ is $1 + 0 i$ and the zero elements are these.
1. $(0 + 0 i) + (0 + 0 i) x + (0 + 0 i) x^{2}$
2. $(\begin{matrix} 0 + 0 i & 0 + 0 i \\ 0 + 0 i & 0 + 0 i \end{matrix})$
Exercise 1.43 Worked answer

Name a property shared by all of the $ℝ^{n}$ ’s but not listed as a requirement for a vector space.

Back to Exercise 1.43

Answer. Notably absent from the definition of a vector space is a distance measure.
Exercise 1.44 Worked answer
1. Prove that for any four vectors ${\vec{v}}_{1}, \dots, {\vec{v}}_{4} \in V$ we can associate their sum in any way without changing the result.
  
  $\begin{matrix} (({\vec{v}}_{1} + {\vec{v}}_{2}) + {\vec{v}}_{3}) + {\vec{v}}_{4} = ({\vec{v}}_{1} + ({\vec{v}}_{2} + {\vec{v}}_{3})) + {\vec{v}}_{4} = ({\vec{v}}_{1} + {\vec{v}}_{2}) + ({\vec{v}}_{3} + {\vec{v}}_{4}) \\ = {\vec{v}}_{1} + (({\vec{v}}_{2} + {\vec{v}}_{3}) + {\vec{v}}_{4}) = {\vec{v}}_{1} + ({\vec{v}}_{2} + ({\vec{v}}_{3} + {\vec{v}}_{4})) \end{matrix}$
  
  This allows us to write ‘ ${\vec{v}}_{1} + {\vec{v}}_{2} + {\vec{v}}_{3} + {\vec{v}}_{4}$ ’ without ambiguity.
2. Prove that any two ways of associating a sum of any number of vectors give the same sum. (Hint. Use induction on the number of vectors.)
Back to Exercise 1.44

Answer.
1. A small rearrangement does the trick.
  
  $\begin{aligned} ({\vec{v}}_{1} + ({\vec{v}}_{2} + {\vec{v}}_{3})) + {\vec{v}}_{4} & = (({\vec{v}}_{1} + {\vec{v}}_{2}) + {\vec{v}}_{3}) + {\vec{v}}_{4} \\ = ({\vec{v}}_{1} + {\vec{v}}_{2}) + ({\vec{v}}_{3} + {\vec{v}}_{4}) \\ = {\vec{v}}_{1} + ({\vec{v}}_{2} + ({\vec{v}}_{3} + {\vec{v}}_{4})) \\ = {\vec{v}}_{1} + (({\vec{v}}_{2} + {\vec{v}}_{3}) + {\vec{v}}_{4}) \end{aligned}$
  
  Each equality above follows from the associativity of three vectors that is given as a condition in the definition of a vector space. For instance, the second ‘ $=$ ’ applies the rule $({\vec{w}}_{1} + {\vec{w}}_{2}) + {\vec{w}}_{3} = {\vec{w}}_{1} + ({\vec{w}}_{2} + {\vec{w}}_{3})$ by taking ${\vec{w}}_{1}$ to be ${\vec{v}}_{1} + {\vec{v}}_{2}$ , taking ${\vec{w}}_{2}$ to be ${\vec{v}}_{3}$ , and taking ${\vec{w}}_{3}$ to be ${\vec{v}}_{4}$ .
2. The base case for induction is the three vector case. This case ${\vec{v}}_{1} + ({\vec{v}}_{2} + {\vec{v}}_{3}) = ({\vec{v}}_{1} + {\vec{v}}_{2}) + {\vec{v}}_{3}$ is one of the conditions in the definition of a vector space.
  
  For the inductive step, assume that any two sums of three vectors, any two sums of four vectors, …, any two sums of $k$ vectors are equal no matter how we parenthesize the sums. We will show that any sum of $k + 1$ vectors equals this one $((\dots (({\vec{v}}_{1} + {\vec{v}}_{2}) + {\vec{v}}_{3}) + \dots) + {\vec{v}}_{k}) + {\vec{v}}_{k + 1}$ .
  
  Any parenthesized sum has an outermost ‘ $+$ ’. Assume that it lies between ${\vec{v}}_{m}$ and ${\vec{v}}_{m + 1}$ so the sum looks like this.
  
  $(\dots {\vec{v}}_{1} \dots {\vec{v}}_{m} \dots) + (\dots {\vec{v}}_{m + 1} \dots {\vec{v}}_{k + 1} \dots)$
  
  The second half involves fewer than $k + 1$ additions, so by the inductive hypothesis we can re-parenthesize it so that it reads left to right from the inside out, and in particular, so that its outermost ‘ $+$ ’ occurs right before ${\vec{v}}_{k + 1}$ .
  
  $= (\dots {\vec{v}}_{1} \dots {\vec{v}}_{m} \dots) + ((\dots ({\vec{v}}_{m + 1} + {\vec{v}}_{m + 2}) + \dots + {\vec{v}}_{k}) + {\vec{v}}_{k + 1})$
  
  Apply the associativity of the sum of three things
  
  $= ((\dots {\vec{v}}_{1} \dots {\vec{v}}_{m} \dots) + (\dots ({\vec{v}}_{m + 1} + {\vec{v}}_{m + 2}) + \dots {\vec{v}}_{k})) + {\vec{v}}_{k + 1}$
  
  and finish by applying the inductive hypothesis inside these outermost parenthesis.
Exercise 1.45 Worked answer

Example 1.5 gives a subset of $ℝ^{2}$ that is not a vector space, under the obvious operations, because while it is closed under addition, it is not closed under scalar multiplication. Consider the set of vectors in the plane whose components have the same sign or are $0$ . Show that this set is closed under scalar multiplication but not addition.

Back to Exercise 1.45

Answer. Let $\vec{v}$ be a member of $ℝ^{2}$ with components $v_{1}$ and $v_{2}$ . We can abbreviate the condition that both components have the same sign or are $0$ by $v_{1} v_{2} \geq 0$ .

To show the set is closed under scalar multiplication, observe that the components of $r \vec{v}$ satisfy $(r v_{1}) (r v_{2}) = r^{2} (v_{1} v_{2})$ and $r^{2} \geq 0$ so $r^{2} v_{1} v_{2} \geq 0$ .

To show the set is not closed under addition we need only produce one example. The vector with components $- 1$ and $0$ , when added to the vector with components $0$ and $1$ makes a vector with mixed-sign components of $- 1$ and $1$ .

Subspaces and Spanning Sets

In Example 1.3 we saw a vector space that is a subset of $ℝ^{2}$ , a line through the origin. There, the vector space $ℝ^{2}$ contains inside it another vector space, the line.

Definition 2.1 For any vector space, a subspace is a subset that is itself a vector space, under the inherited operations.

Example 2.2 This plane through the origin

$P = {(\begin{matrix} x \\ y \\ z \end{matrix}) ∣ x + y + z = 0}$

is a subspace of $ℝ^{3}$ . As required by the definition the plane’s operations are inherited from the larger space, that is, vectors add in $P$ as they add in $ℝ^{3}$

$(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix})$

and scalar multiplication is also the same as in $ℝ^{3}$ . To show that $P$ is a subspace we need only note that it is a subset and then verify that it is a space. We won’t check all ten conditions, just the two closure ones. For closure under addition, note that if the summands satisfy that $x_{1} + y_{1} + z_{1} = 0$ and $x_{2} + y_{2} + z_{2} = 0$ then the sum satisfies that $(x_{1} + x_{2}) + (y_{1} + y_{2}) + (z_{1} + z_{2}) = (x_{1} + y_{1} + z_{1}) + (x_{2} + y_{2} + z_{2}) = 0$ . For closure under scalar multiplication, if $x + y + z = 0$ then the scalar multiple has $r x + r y + r z = r (x + y + z) = 0$ .

Example 2.3 The $x$ -axis in $ℝ^{2}$ is a subspace, where the addition and scalar multiplication operations are the inherited ones.

$(\begin{matrix} x_{1} \\ 0 \end{matrix}) + (\begin{matrix} x_{2} \\ 0 \end{matrix}) = (\begin{matrix} x_{1} + x_{2} \\ 0 \end{matrix}) r \cdot (\begin{matrix} x \\ 0 \end{matrix}) = (\begin{matrix} r x \\ 0 \end{matrix})$

As in the prior example, to verify directly from the definition that this is a subspace we simply note that it is a subset and then check that it satisfies the conditions in definition of a vector space. For instance the two closure conditions are satisfied: adding two vectors with a second component of zero results in a vector with a second component of zero and multiplying a scalar times a vector with a second component of zero results in a vector with a second component of zero.

Example 2.4 Another subspace of $ℝ^{2}$ is its trivial subspace.

${(\begin{matrix} 0 \\ 0 \end{matrix})}$

Any vector space has a trivial subspace ${\vec{0}}$ . At the opposite extreme, any vector space has itself for a subspace. A subspace that is not the entire space is a proper subspace.

Example 2.5 Vector spaces that are not $ℝ^{n}$ ’s also have subspaces. The space of cubic polynomials ${a + b x + c x^{2} + d x^{3} ∣ a, b, c, d \in ℝ}$ has a subspace comprised of all linear polynomials ${m + n x ∣ m, n \in ℝ}$ .

Example 2.6 Another example of a subspace that is not a subset of an $ℝ^{n}$ followed the definition of a vector space. The space in Example 1.12 of all real-valued functions of one real variable ${f ∣ f : ℝ \to ℝ}$ has the subspace in Example 1.14 of functions satisfying the restriction $(d^{2} f / d x^{2}) + f = 0$ .

Example 2.7 The definition requires that the addition and scalar multiplication operations must be the ones inherited from the larger space. The set $S = {1}$ is a subset of $ℝ^{1}$ . And, under the operations $1 + 1 = 1$ and $r \cdot 1 = 1$ the set $S$ is a vector space, specifically, a trivial space. However, $S$ is not a subspace of $ℝ^{1}$ because those aren’t the inherited operations, since of course $ℝ^{1}$ has $1 + 1 = 2$ .

Example 2.8 Being vector spaces themselves, subspaces must satisfy the closure conditions. The set $ℝ^{+}$ is not a subspace of the vector space $ℝ^{1}$ because with the inherited operations it is not closed under scalar multiplication: if $\vec{v} = 1$ then $- 1 \cdot \vec{v} \notin ℝ^{+}$ .

The next result says that Example 2.8 is prototypical. The only way that a subset can fail to be a subspace, if it is nonempty and uses the inherited operations, is if it isn’t closed.

Lemma 2.9 For a nonempty subset $S$ of a vector space, under the inherited operations the following are equivalent statements.¹

$S$ is a subspace of that vector space
$S$ is closed under linear combinations of pairs of vectors: for any vectors ${\vec{s}}_{1}, {\vec{s}}_{2} \in S$ and scalars $r_{1}, r_{2}$ the vector $r_{1} {\vec{s}}_{1} + r_{2} {\vec{s}}_{2}$ is in $S$
$S$ is closed under linear combinations of any number of vectors: for any vectors ${\vec{s}}_{1}, \dots, {\vec{s}}_{n} \in S$ and scalars $r_{1}, \dots, r_{n}$ the vector $r_{1} {\vec{s}}_{1} + \dots + r_{n} {\vec{s}}_{n}$ is an element of $S$ .

Briefly, a subset is a subspace if and only if it is closed under linear combinations.

Proof ‘The following are equivalent’ means that each pair of statements are equivalent.

$(1) ⟺ (2) (2) ⟺ (3) (3) ⟺ (1)$

We will prove the equivalence by establishing that $(1) ⟹ (3) ⟹ (2) ⟹ (1)$ . This strategy is suggested by the observation that the implications $(1) ⟹ (3)$ and $(3) ⟹ (2)$ are easy and so we need only argue that $(2) ⟹ (1)$ .

Assume that $S$ is a nonempty subset of a vector space $V$ that is closed under combinations of pairs of vectors. We will show that $S$ is a vector space by checking the conditions.

The vector space definition has five conditions on addition. First, for closure under addition, if ${\vec{s}}_{1}, {\vec{s}}_{2} \in S$ then ${\vec{s}}_{1} + {\vec{s}}_{2} \in S$ , as it is a combination of a pair of vectors and we are assuming that $S$ is closed under those. Second, for any ${\vec{s}}_{1}, {\vec{s}}_{2} \in S$ , because addition is inherited from $V$ , the sum ${\vec{s}}_{1} + {\vec{s}}_{2}$ in $S$ equals the sum ${\vec{s}}_{1} + {\vec{s}}_{2}$ in $V$ , and that equals the sum ${\vec{s}}_{2} + {\vec{s}}_{1}$ in $V$ (because $V$ is a vector space, its addition is commutative), and that in turn equals the sum ${\vec{s}}_{2} + {\vec{s}}_{1}$ in $S$ . The argument for the third condition is similar to that for the second. For the fourth, consider the zero vector of $V$ and note that closure of $S$ under linear combinations of pairs of vectors gives that $0 \cdot \vec{s} + 0 \cdot \vec{s} = \vec{0}$ is an element of $S$ (where $\vec{s}$ is any member of the nonempty set $S$ ); checking that $\vec{0}$ acts under the inherited operations as the additive identity of $S$ is easy. The fifth condition is satisfied because for any $\vec{s} \in S$ , closure under linear combinations of pairs of vectors shows that $0 \cdot \vec{0} + (- 1) \cdot \vec{s}$ is an element of $S$ , and it is obviously the additive inverse of $\vec{s}$ under the inherited operations. The verifications for the scalar multiplication conditions are similar; see Exercise 2.35.

QED

We will usually verify that a subset is a subspace by checking that it satisfies statement (2).

Remark 2.10 At the start of this chapter we introduced vector spaces as collections in which linear combinations “make sense.” Lemma 2.9’s statements (1)-(3) say that we can always make sense of an expression like $r_{1} {\vec{s}}_{1} + r_{2} {\vec{s}}_{2}$ in that the vector described is in the set $S$ .

As a contrast, consider the set $T$ of two-tall vectors whose entries add to a number greater than or equal to zero. Here we cannot just write any linear combination such as $2 {\vec{t}}_{1} - 3 {\vec{t}}_{2}$ and be confident the result is an element of $T$ .

Lemma 2.9 suggests that a good way to think of a vector space is as a collection of unrestricted linear combinations. The next two examples take some spaces and recasts their descriptions to be in that form.

Example 2.11 We can show that this plane through the origin subset of $ℝ^{3}$

$S = {(\begin{matrix} x \\ y \\ z \end{matrix}) ∣ x - 2 y + z = 0}$

is a subspace under the usual addition and scalar multiplication operations of column vectors by checking that it is nonempty and closed under linear combinations of two vectors. But there is another way. Think of $x - 2 y + z = 0$ as a one-equation linear system and parametrize it by expressing the leading variable in terms of the free variables $x = 2 y - z$ .

$\begin{matrix} S = {(\begin{matrix} 2 y - z \\ y \\ z \end{matrix}) ∣ y, z \in ℝ} = {y (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) + z (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) ∣ y, z \in ℝ} & (*) \end{matrix}$

Now, to show that this is a subspace consider $r_{1} {\vec{s}}_{1} + r_{2} {\vec{s}}_{2}$ . Each ${\vec{s}}_{i}$ is a linear combination of the two vectors in ( $*$ ) so this is a linear combination of linear combinations.

$r_{1} \cdot (y_{1} (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) + z_{1} (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix})) + r_{2} \cdot (y_{2} (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) + z_{2} (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}))$

The Linear Combination Lemma, Lemma One.III.2.3, shows that the total is a linear combination of the two vectors and so Lemma 2.9’s statement (2) is satisfied.

Example 2.12 This is a subspace of the $2 \times 2$ matrices $ℳ_{2 \times 2}$ .

$L = {(\begin{matrix} a & 0 \\ b & c \end{matrix}) ∣ a + b + c = 0}$

To parametrize, express the condition as $a = - b - c$ .

$L = {(\begin{matrix} - b - c & 0 \\ b & c \end{matrix}) ∣ b, c \in ℝ} = {b (\begin{matrix} - 1 & 0 \\ 1 & 0 \end{matrix}) + c (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) ∣ b, c \in ℝ}$

As above, we’ve described the subspace as a collection of unrestricted linear combinations. To show it is a subspace, note that a linear combination of vectors from $L$ is a linear combination of linear combinations and so statement (2) is true.

Definition 2.13 The span (or linear closure) of a nonempty subset $S$ of a vector space is the set of all linear combinations of vectors from $S$ .

$[S] = {c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} ∣ c_{1}, \dots, c_{n} \in ℝ and {\vec{s}}_{1}, \dots, {\vec{s}}_{n} \in S}$

The span of the empty subset of a vector space is its trivial subspace.

No notation for the span is completely standard. The square brackets used here are common but so are ‘ $span (S)$ ’ and ‘ $sp (S)$ ’.

Remark 2.14 In Chapter One, after we showed that we can write the solution set of a homogeneous linear system as ${c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k} ∣ c_{1}, \dots, c_{k} \in ℝ}$ , we described that as the set ‘generated’ by the $\vec{β}$ ’s. We now call that the span of ${{\vec{β}}_{1}, \dots, {\vec{β}}_{k}}$ .

Recall also from that proof that the span of the empty set is defined to be the set ${\vec{0}}$ because of the convention that a trivial linear combination, a combination of zero-many vectors, adds to $\vec{0}$ . Besides, defining the empty set’s span to be the trivial subspace is convenient because it keeps results like the next one from needing exceptions for the empty set.

Lemma 2.15 In a vector space, the span of any subset is a subspace.

Proof If the subset $S$ is empty then by definition its span is the trivial subspace. If $S$ is not empty then by Lemma 2.9 we need only check that the span $[S]$ is closed under linear combinations of pairs of elements. For a pair of vectors from that span, $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ and $\vec{w} = c_{n + 1} {\vec{s}}_{n + 1} + \dots + c_{m} {\vec{s}}_{m}$ , a linear combination

$\begin{matrix} p \cdot (c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}) + r \cdot (c_{n + 1} {\vec{s}}_{n + 1} + \dots + c_{m} {\vec{s}}_{m}) \\ = p c_{1} {\vec{s}}_{1} + \dots + p c_{n} {\vec{s}}_{n} + r c_{n + 1} {\vec{s}}_{n + 1} + \dots + r c_{m} {\vec{s}}_{m} \end{matrix}$

is a linear combination of elements of $S$ and so is an element of $[S]$ (possibly some of the ${\vec{s}}_{i}$ ’s from $\vec{v}$ equal some of the ${\vec{s}}_{j}$ ’s from $\vec{w}$ but that does not matter).

QED

The converse of the lemma holds: any subspace is the span of some set, because a subspace is obviously the span of itself, the set of all of its members. Thus a subset of a vector space is a subspace if and only if it is a span. This fits the intuition that a good way to think of a vector space is as a collection in which linear combinations are sensible.

Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset $S$ of a vector space is the smallest subspace containing all of the members of $S$ .

Example 2.16 In any vector space $V$ , for any vector $\vec{v} \in V$ , the set ${r \cdot \vec{v} ∣ r \in ℝ}$ is a subspace of $V$ . For instance, for any vector $\vec{v} \in ℝ^{3}$ the line through the origin containing that vector ${k \vec{v} ∣ k \in ℝ}$ is a subspace of $ℝ^{3}$ . This is true even if $\vec{v}$ is the zero vector, in which case it is the degenerate line, the trivial subspace.

Example 2.17 The span of this set is all of $ℝ^{2}$ .

${(\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \end{matrix})}$

We know that the span is some subspace of $ℝ^{2}$ . To check that it is all of $ℝ^{2}$ we must show that any member of $ℝ^{2}$ is a linear combination of these two vectors. So we ask: for which vectors with real components $x$ and $y$ are there scalars $c_{1}$ and $c_{2}$ such that this holds?

$\begin{matrix} c_{1} (\begin{matrix} 1 \\ 1 \end{matrix}) + c_{2} (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} x \\ y \end{matrix}) & (*) \end{matrix}$

Gauss’s Method

$\begin{array}{rcrcr} c_{1} & + & c_{2} & = & x \\ c_{1} & - & c_{2} & = & y \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcr} c_{1} & + & c_{2} & = & x \\ - 2 c_{2} & = & - x + y \end{array}$

with back substitution gives $c_{2} = (x - y) / 2$ and $c_{1} = (x + y) / 2$ . This shows that for any $x, y$ there are appropriate coefficients $c_{1}, c_{2}$ making ( $*$ ) true— we can write any element of $ℝ^{2}$ as a linear combination of the two given ones. For instance, for $x = 1$ and $y = 2$ the coefficients $c_{2} = - 1 / 2$ and $c_{1} = 3 / 2$ will do.

Since spans are subspaces, and we know that a good way to understand a subspace is to parametrize its description, we can try to understand a set’s span in that way.

Example 2.18 Consider, in the vector space of quadratic polynomials $𝒫_{2}$ , the span of the set $S = {3 x - x^{2}, 2 x}$ . By the definition of span, it is the set of unrestricted linear combinations of the two ${c_{1} (3 x - x^{2}) + c_{2} (2 x) ∣ c_{1}, c_{2} \in ℝ}$ . Clearly polynomials in this span must have a constant term of zero. Is that necessary condition also sufficient?

We are asking: for which members $a_{2} x^{2} + a_{1} x + a_{0}$ of $𝒫_{2}$ are there $c_{1}$ and $c_{2}$ such that $a_{2} x^{2} + a_{1} x + a_{0} = c_{1} (3 x - x^{2}) + c_{2} (2 x)$ ? Polynomials are equal when their coefficients are equal so we want conditions on $a_{2}$ , $a_{1}$ , and $a_{0}$ making that triple a solution of this system.

$\begin{array}{rcrcr} - c_{1} & = & a_{2} \\ 3 c_{1} & + & 2 c_{2} & = & a_{1} \\ 0 & = & a_{0} \end{array}$

Gauss’s Method and back-substitution gives $c_{1} = - a_{2}$ , and $c_{2} = (3 / 2) a_{2} + (1 / 2) a_{1}$ , and $0 = a_{0}$ . Thus as long as there is no constant term $a_{0} = 0$ we can give coefficients $c_{1}$ and $c_{2}$ to describe that polynomial as an element of the span. For instance, for the polynomial $0 - 4 x + 3 x^{2}$ , the coefficients $c_{1} = - 3$ and $c_{2} = 5 / 2$ will do. So the span of the given set is $[S] = {a_{1} x + a_{2} x^{2} ∣ a_{1}, a_{2} \in ℝ}$ .

Incidentally, this shows that the set ${x, x^{2}}$ spans the same subspace. A space can have more than one spanning set. Two other sets spanning this subspace are ${x, x^{2}, - x + 2 x^{2}}$ and ${x, x + x^{2}, x + 2 x^{2}, \dots}$ .

Example 2.19 The picture below shows the subspaces of $ℝ^{3}$ that we now know of: the trivial subspace, lines through the origin, planes through the origin, and the whole space. (Of course, the picture shows only a few of the infinitely many cases. Line segments connect subsets with their supersets.) In the next section we will prove that $ℝ^{3}$ has no other kind of subspace, so in fact this lists them all.

This describes each subspace as the span of a set with a minimal number of members. With this, the subspaces fall naturally into levels—planes on one level, lines on another, etc.

So far in this chapter we have seen that to study the properties of linear combinations, the right setting is a collection that is closed under these combinations. In the first subsection we introduced such collections, vector spaces, and we saw a great variety of examples. In this subsection we saw still more spaces, ones that are subspaces of others. In all of the variety there is a commonality. Example 2.19 above brings it out: vector spaces and subspaces are best understood as a span, and especially as a span of a small number of vectors. The next section studies spanning sets that are minimal.

Exercises

Exercise 2.20 Worked answer

Recommended. Which of these subsets of the vector space of $2 \times 2$ matrices are subspaces under the inherited operations? For each one that is a subspace, parametrize its description. For each that is not, give a condition that fails.
1. ${(\begin{matrix} a & 0 \\ 0 & b \end{matrix}) ∣ a, b \in ℝ}$
2. ${(\begin{matrix} a & 0 \\ 0 & b \end{matrix}) ∣ a + b = 0}$
3. ${(\begin{matrix} a & 0 \\ 0 & b \end{matrix}) ∣ a + b = 5}$
4. ${(\begin{matrix} a & c \\ 0 & b \end{matrix}) ∣ a + b = 0, c \in ℝ}$
Back to Exercise 2.20

Answer. By Lemma 2.9, to see if each subset of $ℳ_{2 \times 2}$ is a subspace, we need only check if it is nonempty and closed.
1. Yes, we can easily check that it is nonempty and closed. This is a parametrization.
  
  ${a (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) + b (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) ∣ a, b \in ℝ}$
  
  By the way, the parametrization also shows that it is a subspace, since it is given as the span of the two-matrix set, and any span is a subspace.
2. Yes; it is easily checked to be nonempty and closed. Alternatively, as mentioned in the prior answer, the existence of a parametrization shows that it is a subspace. For the parametrization, the condition $a + b = 0$ can be rewritten as $a = - b$ . Then we have this.
  
  ${(\begin{matrix} - b & 0 \\ 0 & b \end{matrix}) ∣ b \in ℝ} = {b (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) ∣ b \in ℝ}$
3. No. It is not closed under addition. For instance,
  
  $(\begin{matrix} 5 & 0 \\ 0 & 0 \end{matrix}) + (\begin{matrix} 5 & 0 \\ 0 & 0 \end{matrix}) = (\begin{matrix} 10 & 0 \\ 0 & 0 \end{matrix})$
  
  is not in the set. (This set is also not closed under scalar multiplication, for instance, it does not contain the zero matrix.)
4. Yes.
  
  ${b (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) + c (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) ∣ b, c \in ℝ}$
Exercise 2.21 Worked answer

Recommended. Is this a subspace of $𝒫_{2}$ : ${a_{0} + a_{1} x + a_{2} x^{2} ∣ a_{0} + 2 a_{1} + a_{2} = 4}$ ? If it is then parametrize its description.

Back to Exercise 2.21

Answer. No, it is not closed. In particular, it is not closed under scalar multiplication because it does not contain the zero polynomial.
Exercise 2.22 Worked answer

Is the vector in the span of the set?

$(\begin{matrix} 1 \\ 0 \\ 3 \end{matrix}) {(\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})}$

Back to Exercise 2.22

Answer. The equation

$(\begin{matrix} 1 \\ 0 \\ 3 \end{matrix}) = c_{1} (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix}) + c_{2} (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$

gives rise to a linear system

$(\begin{array}{ccc} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ - 1 & 1 & 3 \end{array}) ⟶_{\begin{array}{l} (1 / 2) ρ_{1} + ρ_{3} \end{array}}^{(- 1 / 2) ρ_{1} + ρ_{2}} (\begin{array}{ccc} 2 & 1 & 1 \\ 0 & - 3 / 2 & - 1 / 2 \\ 0 & 0 & 3 \end{array})$

that has no solution, so the vector is not in the span.
Exercise 2.23 Worked answer

Recommended. Decide if the vector lies in the span of the set, inside of the space.
1. $(\begin{matrix} 2 \\ 0 \\ 1 \end{matrix})$ , ${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}$ , in $ℝ^{3}$
2. $x - x^{3}$ , ${x^{2}, 2 x + x^{2}, x + x^{3}}$ , in $𝒫_{3}$
3. $(\begin{matrix} 0 & 1 \\ 4 & 2 \end{matrix})$ , ${(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}), (\begin{matrix} 2 & 0 \\ 2 & 3 \end{matrix})}$ , in $ℳ_{2 \times 2}$
Back to Exercise 2.23

Answer.
1. Yes, solving the linear system arising from
  
  $r_{1} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + r_{2} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix})$
  
  gives $r_{1} = 2$ and $r_{2} = 1$ .
2. Yes; the linear system arising from $r_{1} (x^{2}) + r_{2} (2 x + x^{2}) + r_{3} (x + x^{3}) = x - x^{3}$
  
  $\begin{array}{rcrcrcr} 2 r_{2} & + & r_{3} & = & 1 \\ r_{1} & + & r_{2} & = & 0 \\ r_{3} & = & - 1 \end{array}$
  
  gives that $- 1 (x^{2}) + 1 (2 x + x^{2}) - 1 (x + x^{3}) = x - x^{3}$ .
3. No; any combination of the two given matrices has a zero in the upper right.
Exercise 2.24 Worked answer

A superhero is at the origin of a two dimensional plane. The superhero has two devices, a hoverboard that moves any distance in the direction $(\binom{3}{1})$ and a magic carpet that moves any distance in the direction $(\binom{1}{2})$ .
1. An evil villain is hiding out in the plane at the point $(- 5, 7)$ . How many hoverboard units and magic carpet units does the superhero have to move to get to the villain?
2. Is there anywhere in the plane that the villain could safely hide and not be reached? If so, give one such location. If not, explain why not.
3. The superhero and the villain are transported to a three dimensional space where the superhero now has three devices.
  
  $hoverboard: (\begin{matrix} - 1 \\ 0 \\ 3 \end{matrix}) magic carpet: (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) scooter: (\begin{matrix} 5 \\ 4 \\ - 9 \end{matrix})$
  
  Is there anywhere that the villain could safely hide? If so, give one such location and if not, explain why not.
Back to Exercise 2.24

Answer.
1. This relationship
  
  $(\begin{matrix} - 5 \\ 7 \end{matrix}) = a (\begin{matrix} 3 \\ 1 \end{matrix}) + b (\begin{matrix} 1 \\ 2 \end{matrix})$
  
  gives this system of equations.
  
  $\begin{array}{rcrcr} 3 a & + & b & = & - 5 \\ a & + & 2 b & = & 7 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (1 / 3) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 3 a & + & b & = & - 5 \\ (5 / 3) b & = & 26 / 3 \end{array}$
  
  So $b = 26 / 5$ . Substituting $3 a + (26 / 5) = - 5$ gives $a = - 17 / 5$ . In the terms of the section, the vector $(\binom{- 5}{7})$ is in the span of the set of the other two.
2. There is no place to hide. This Gauss’s method reduction
  
  $\begin{array}{rcrcr} 3 a & + & b & = & x \\ a & + & 2 b & = & y \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (1 / 3) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 3 a & + & b & = & x \\ (5 / 3) b & = & - (1 / 3) x + y \end{array}$
  
  shows that for any $x, y \in ℝ$ there is a pair $a, b \in ℝ$ . In the terms of the section, the span of the two vectors is the entire plane.
3. No place to hide. Consider this system.
  
  $a (\begin{matrix} - 1 \\ 0 \\ 3 \end{matrix}) + b (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) + c (\begin{matrix} 5 \\ 4 \\ - 9 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  After reduction
  
  $\begin{aligned} \begin{array}{rcrcrcr} - a & + & 2 b & + & 5 c & = & x \\ b & + & 4 c & = & y \\ 3 a & - & 9 c & = & z \end{array} & ⟶_{\begin{matrix} \end{matrix}}^{3 ρ_{1} + ρ_{3}} \begin{array}{rcrcrcr} - a & + & 2 b & + & 5 c & = & x \\ b & + & 4 c & = & y \\ 6 b & + & 6 c & = & 3 x + z \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- 6 ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} - a & + & 2 b & + & 5 c & = & x \\ b & + & 4 c & = & y \\ - 18 c & = & 3 x + - 6 y + z \end{array} \end{aligned}$
  
  it shows that for any $x, y, z \in ℝ$ the required $a$ , $b$ , and $c$ exist. The superhero’s devices span the plane.
Exercise 2.25 Worked answer

Which of these are members of the span $[{\cos^{2} x, \sin^{2} x}]$ in the vector space of real-valued functions of one real variable?
1. $f (x) = 1$
2. $f (x) = 3 + x^{2}$
3. $f (x) = \sin x$
4. $f (x) = \cos (2 x)$
Back to Exercise 2.25

Answer.
1. Yes; it is in that span since $1 \cdot \cos^{2} x + 1 \cdot \sin^{2} x = f (x)$ .
2. No, since $r_{1} \cos^{2} x + r_{2} \sin^{2} x = 3 + x^{2}$ has no scalar solutions that work for all $x$ . For instance, setting $x$ to be $0$ and $π$ gives the two equations $r_{1} \cdot 1 + r_{2} \cdot 0 = 3$ and $r_{1} \cdot 1 + r_{2} \cdot 0 = 3 + π^{2}$ , which are not consistent with each other.
3. No; consider what happens on setting $x$ to be $π / 2$ and $3 π / 2$ .
4. Yes, $\cos (2 x) = 1 \cdot \cos^{2} (x) - 1 \cdot \sin^{2} (x)$ .
Exercise 2.26 Worked answer

Recommended. Which of these sets spans $ℝ^{3}$ ? That is, which of these sets has the property that any three-tall vector can be expressed as a suitable linear combination of the set’s elements?
1. ${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 3 \end{matrix})}$
2. ${(\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}$
3. ${(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 3 \\ 0 \\ 0 \end{matrix})}$
4. ${(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 3 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 1 \\ 5 \end{matrix})}$
5. ${(\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 3 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 5 \\ 1 \\ 2 \end{matrix}), (\begin{matrix} 6 \\ 0 \\ 2 \end{matrix})}$
Back to Exercise 2.26

Answer.
1. Yes, for any $x, y, z \in ℝ$ this equation
  
  $r_{1} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + r_{2} (\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}) + r_{3} (\begin{matrix} 0 \\ 0 \\ 3 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  has the solution $r_{1} = x$ , $r_{2} = y / 2$ , and $r_{3} = z / 3$ .
2. Yes, the equation
  
  $r_{1} (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}) + r_{2} (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + r_{3} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  gives rise to this
  
  $\begin{array}{rcrcrcr} 2 r_{1} & + & r_{2} & = & x \\ r_{2} & = & y \\ r_{1} & + & r_{3} & = & z \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{3}} ⟶_{\begin{array}{l} \end{array}}^{(1 / 2) ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} 2 r_{1} & + & r_{2} & = & x \\ r_{2} & = & y \\ r_{3} & = & - (1 / 2) x + (1 / 2) y + z \end{array}$
  
  so that, given any $x$ , $y$ , and $z$ , we can compute that $r_{3} = (- 1 / 2) x + (1 / 2) y + z$ , $r_{2} = y$ , and $r_{1} = (1 / 2) x - (1 / 2) y$ .
3. No. In particular, we cannot get the vector
  
  $(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$
  
  as a linear combination since the two given vectors both have a third component of zero.
4. Yes. The equation
  
  $r_{1} (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) + r_{2} (\begin{matrix} 3 \\ 1 \\ 0 \end{matrix}) + r_{3} (\begin{matrix} - 1 \\ 0 \\ 0 \end{matrix}) + r_{4} (\begin{matrix} 2 \\ 1 \\ 5 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  leads to this reduction.
  
  $(\begin{array}{ccccc} 1 & 3 & - 1 & 2 & x \\ 0 & 1 & 0 & 1 & y \\ 1 & 0 & 0 & 5 & z \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{3}} ⟶_{\begin{array}{l} \end{array}}^{3 ρ_{2} + ρ_{3}} (\begin{array}{ccccc} 1 & 3 & - 1 & 2 & x \\ 0 & 1 & 0 & 1 & y \\ 0 & 0 & 1 & 6 & - x + 3 y + z \end{array})$
  
  We have infinitely many solutions. We can, for example, set $r_{4}$ to be zero and solve for $r_{3}$ , $r_{2}$ , and $r_{1}$ in terms of $x$ , $y$ , and $z$ by the usual methods of back-substitution.
5. No. The equation
  
  $r_{1} (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}) + r_{2} (\begin{matrix} 3 \\ 0 \\ 1 \end{matrix}) + r_{3} (\begin{matrix} 5 \\ 1 \\ 2 \end{matrix}) + r_{4} (\begin{matrix} 6 \\ 0 \\ 2 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  leads to this reduction.
  
  $\begin{matrix} (\begin{array}{ccccc} 2 & 3 & 5 & 6 & x \\ 1 & 0 & 1 & 0 & y \\ 1 & 1 & 2 & 2 & z \end{array}) \\ ⟶_{\begin{array}{l} - (1 / 2) ρ_{1} + ρ_{3} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- (1 / 3) ρ_{2} + ρ_{3}} (\begin{array}{ccccc} 2 & 3 & 5 & 6 & x \\ 0 & - 3 / 2 & - 3 / 2 & - 3 & - (1 / 2) x + y \\ 0 & 0 & 0 & 0 & - (1 / 3) x - (1 / 3) y + z \end{array}) \end{matrix}$
  
  This shows that not every three-tall vector can be so expressed. Only the vectors satisfying the restriction that $- (1 / 3) x - (1 / 3) y + z = 0$ are in the span. (To see that any such vector is indeed expressible, take $r_{3}$ and $r_{4}$ to be zero and solve for $r_{1}$ and $r_{2}$ in terms of $x$ , $y$ , and $z$ by back-substitution.)
Exercise 2.27 Worked answer

Recommended. Parametrize each subspace’s description. Then express each subspace as a span.
1. The subset ${(\begin{matrix} a & b & c \end{matrix}) ∣ a - c = 0}$ of the three-wide row vectors
2. This subset of $ℳ_{2 \times 2}$
  
  ${(\begin{matrix} a & b \\ c & d \end{matrix}) ∣ a + d = 0}$
3. This subset of $ℳ_{2 \times 2}$
  
  ${(\begin{matrix} a & b \\ c & d \end{matrix}) ∣ 2 a - c - d = 0 and a + 3 b = 0}$
4. The subset ${a + b x + c x^{3} ∣ a - 2 b + c = 0}$ of $𝒫_{3}$
5. The subset of $𝒫_{2}$ of quadratic polynomials $p$ such that $p (7) = 0$
Back to Exercise 2.27

Answer.
1. ${(\begin{matrix} c & b & c \end{matrix}) ∣ b, c \in ℝ} = {b (\begin{matrix} 0 & 1 & 0 \end{matrix}) + c (\begin{matrix} 1 & 0 & 1 \end{matrix}) ∣ b, c \in ℝ}$ The obvious choice for the set that spans is ${(\begin{matrix} 0 & 1 & 0 \end{matrix}), (\begin{matrix} 1 & 0 & 1 \end{matrix})}$ .
2. ${(\begin{matrix} - d & b \\ c & d \end{matrix}) ∣ b, c, d \in ℝ} = {b (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) + c (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}) + d (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) ∣ b, c, d \in ℝ}$ One set that spans this space consists of those three matrices.
3. The system
  
  $\begin{array}{rcrcrcrcr} a & + & 3 b & = & 0 \\ 2 a & - c & - & d & = & 0 \end{array}$
  
  gives $b = - (c + d) / 6$ and $a = (c + d) / 2$ . So one description is this.
  
  ${c (\begin{matrix} 1 / 2 & - 1 / 6 \\ 1 & 0 \end{matrix}) + d (\begin{matrix} 1 / 2 & - 1 / 6 \\ 0 & 1 \end{matrix}) ∣ c, d \in ℝ}$
  
  That shows that a set spanning this subspace consists of those two matrices.
4. The $a = 2 b - c$ gives that the set ${(2 b - c) + b x + c x^{3} ∣ b, c \in ℝ}$ equals the set ${b (2 + x) + c (- 1 + x^{3}) ∣ b, c \in ℝ}$ . So the subspace is the span of the set ${2 + x, - 1 + x^{3}}$ .
5. The set ${a + b x + c x^{2} ∣ a + 7 b + 49 c = 0}$ can be parametrized as
  
  ${b (- 7 + x) + c (- 49 + x^{2}) ∣ b, c \in ℝ}$
  
  and so has the spanning set ${- 7 + x, - 49 + x^{2}}$ .
Exercise 2.28 Worked answer

Recommended. Find a set to span the given subspace of the given space. (Hint. Parametrize each.)
1. the $x z$ -plane in $ℝ^{3}$
2. ${(\begin{matrix} x \\ y \\ z \end{matrix}) ∣ 3 x + 2 y + z = 0}$ in $ℝ^{3}$
3. ${(\begin{matrix} x \\ y \\ z \\ w \end{matrix}) ∣ 2 x + y + w = 0 and y + 2 z = 0}$ in $ℝ^{4}$
4. ${a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} ∣ a_{0} + a_{1} = 0 and a_{2} - a_{3} = 0}$ in $𝒫_{3}$
5. The set $𝒫_{4}$ in the space $𝒫_{4}$
6. $ℳ_{2 \times 2}$ in $ℳ_{2 \times 2}$
Back to Exercise 2.28

Answer.
1. We can parametrize in this way
  
  ${(\begin{matrix} x \\ 0 \\ z \end{matrix}) ∣ x, z \in ℝ} = {x (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + z (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) ∣ x, z \in ℝ}$
  
  giving this for a spanning set.
  
  ${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}$
2. Here is a parametrization, and the associated spanning set.
  
  ${y (\begin{matrix} - 2 / 3 \\ 1 \\ 0 \end{matrix}) + z (\begin{matrix} - 1 / 3 \\ 0 \\ 1 \end{matrix}) ∣ y, z \in ℝ} {(\begin{matrix} - 2 / 3 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 / 3 \\ 0 \\ 1 \end{matrix})}$
3. ${(\begin{matrix} 1 \\ - 2 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 / 2 \\ 0 \\ 0 \\ 1 \end{matrix})}$
4. Parametrize the description as ${- a_{1} + a_{1} x + a_{3} x^{2} + a_{3} x^{3} ∣ a_{1}, a_{3} \in ℝ}$ to get ${- 1 + x, x^{2} + x^{3}} .$
5. ${1, x, x^{2}, x^{3}, x^{4}}$
6. ${(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})}$
Exercise 2.29 Worked answer

Is $ℝ^{2}$ a subspace of $ℝ^{3}$ ?

Back to Exercise 2.29

Answer. Technically, no. Subspaces of $ℝ^{3}$ are sets of three-tall vectors, while $ℝ^{2}$ is a set of two-tall vectors. Clearly though, $ℝ^{2}$ is “just like” this subspace of $ℝ^{3}$ .

${(\begin{matrix} x \\ y \\ 0 \end{matrix}) ∣ x, y \in ℝ}$
Exercise 2.30 Worked answer

Recommended. Decide if each is a subspace of the vector space of real-valued functions of one real variable.
1. The even functions ${f : ℝ \to ℝ ∣ f (- x) = f (x) for all x}$ . For example, two members of this set are $f_{1} (x) = x^{2}$ and $f_{2} (x) = \cos (x)$ .
2. The odd functions ${f : ℝ \to ℝ ∣ f (- x) = - f (x) for all x}$ . Two members are $f_{3} (x) = x^{3}$ and $f_{4} (x) = \sin (x)$ .
Back to Exercise 2.30

Answer. Of course, the addition and scalar multiplication operations are the ones inherited from the enclosing space.
1. This is a subspace. It is not empty as it contains at least the two example functions given. It is closed because if $f_{1}, f_{2}$ are even and $c_{1}, c_{2}$ are scalars then we have this.
  
  $(c_{1} f_{1} + c_{2} f_{2}) (- x) = c_{1} f_{1} (- x) + c_{2} f_{2} (- x) = c_{1} f_{1} (x) + c_{2} f_{2} (x) = (c_{1} f_{1} + c_{2} f_{2}) (x)$
2. This is also a subspace; the check is similar to the prior one.
Exercise 2.31 Worked answer

Example 2.16 says that for any vector $\vec{v}$ that is an element of a vector space $V$ , the set ${r \cdot \vec{v} ∣ r \in ℝ}$ is a subspace of $V$ . (This is simply the span of the singleton set ${\vec{v}}$ .) Must any such subspace be a proper subspace?

Back to Exercise 2.31

Answer. It can be improper. For instance, if the vector space is $ℝ^{1}$ and $\vec{v} \neq \vec{0}$ then the subspace ${r \cdot \vec{v} ∣ r \in ℝ}$ is all of $ℝ^{1}$ .
Exercise 2.32 Worked answer

An example following the definition of a vector space shows that the solution set of a homogeneous linear system is a vector space. In the terminology of this subsection, it is a subspace of $ℝ^{n}$ where the system has $n$ variables. What about a non-homogeneous linear system; do its solutions form a subspace (under the inherited operations)?

Back to Exercise 2.32

Answer. No, such a set is not closed. For one thing, it does not contain the zero vector.
Exercise 2.33 Worked answer

[Cleary] Give an example of each or explain why it would be impossible to do so.
1. A nonempty subset of $ℳ_{2 \times 2}$ that is not a subspace.
2. A set of two vectors in $ℝ^{2}$ that does not span the space.
Back to Exercise 2.33

Answer.
1. This nonempty subset of $ℳ_{2 \times 2}$ is not a subspace.
  
  $A = {(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}), (\begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix})}$
  
  One reason that it is not a subspace of $ℳ_{2 \times 2}$ is that it does not contain the zero matrix. (Another reason is that it is not closed under addition, since the sum of the two is not an element of $A$ . It is also not closed under scalar multiplication.)
2. This set of two vectors does not span $ℝ^{2}$ .
  
  ${(\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 3 \\ 3 \end{matrix})}$
  
  No linear combination of these two can give a vector whose second component is unequal to its first component.
Exercise 2.34 Worked answer

Example 2.19 shows that $ℝ^{3}$ has infinitely many subspaces. Does every nontrivial space have infinitely many subspaces?

Back to Exercise 2.34

Answer. No. The only subspaces of $ℝ^{1}$ are the space itself and its trivial subspace. Any subspace $S$ of $ℝ$ that contains a nonzero member $\vec{v}$ must contain the set of all of its scalar multiples ${r \cdot \vec{v} ∣ r \in ℝ}$ . But this set is all of $ℝ$ .
Exercise 2.35 Worked answer

Finish the proof of Lemma 2.9.

Back to Exercise 2.35

Answer. Item (1) is checked in the text.

Item (2) has five conditions. First, for closure, if $c \in ℝ$ and $\vec{s} \in S$ then $c \cdot \vec{s} \in S$ as $c \cdot \vec{s} = c \cdot \vec{s} + 0 \cdot \vec{0}$ . Second, because the operations in $S$ are inherited from $V$ , for $c, d \in ℝ$ and $\vec{s} \in S$ , the scalar product $(c + d) \cdot \vec{s}$ in $S$ equals the product $(c + d) \cdot \vec{s}$ in $V$ , and that equals $c \cdot \vec{s} + d \cdot \vec{s}$ in $V$ , which equals $c \cdot \vec{s} + d \cdot \vec{s}$ in $S$ .

The check for the third, fourth, and fifth conditions are similar to the second condition’s check just given.
Exercise 2.36 Worked answer

Show that each vector space has only one trivial subspace.

Back to Exercise 2.36

Answer. An exercise in the prior subsection shows that every vector space has only one zero vector (that is, there is only one vector that is the additive identity element of the space). But a trivial space has only one element and that element must be this (unique) zero vector.
Exercise 2.37 Worked answer

Show that for any subset $S$ of a vector space, the span of the span equals the span $[[S]] = [S]$ . (Hint. Members of $[S]$ are linear combinations of members of $S$ . Members of $[[S]]$ are linear combinations of linear combinations of members of $S$ .)

Back to Exercise 2.37

Answer. As the hint suggests, the basic reason is the Linear Combination Lemma from the first chapter. For the full proof, we will show mutual containment between the two sets.

The first containment $[[S]] \supseteq [S]$ is an instance of the more general, and obvious, fact that for any subset $T$ of a vector space, $[T] \supseteq T$ .

For the other containment, that $[[S]] \subseteq [S]$ , take $m$ vectors from $[S]$ , namely $c_{1, 1} {\vec{s}}_{1, 1} + \dots + c_{1, n_{1}} {\vec{s}}_{1, n_{1}}$ , …, $c_{1, m} {\vec{s}}_{1, m} + \dots + c_{1, n_{m}} {\vec{s}}_{1, n_{m}}$ , and note that any linear combination of those

$r_{1} (c_{1, 1} {\vec{s}}_{1, 1} + \dots + c_{1, n_{1}} {\vec{s}}_{1, n_{1}}) + \dots + r_{m} (c_{1, m} {\vec{s}}_{1, m} + \dots + c_{1, n_{m}} {\vec{s}}_{1, n_{m}})$

is a linear combination of elements of $S$

$= (r_{1} c_{1, 1}) {\vec{s}}_{1, 1} + \dots + (r_{1} c_{1, n_{1}}) {\vec{s}}_{1, n_{1}} + \dots + (r_{m} c_{1, m}) {\vec{s}}_{1, m} + \dots + (r_{m} c_{1, n_{m}}) {\vec{s}}_{1, n_{m}}$

and so is in $[S]$ . That is, simply recall that a linear combination of linear combinations (of members of $S$ ) is a linear combination (again of members of $S$ ).
Exercise 2.38 Worked answer

All of the subspaces that we’ve seen in some way use zero in their description. For example, the subspace in Example 2.3 consists of all the vectors from $ℝ^{2}$ with a second component of zero. In contrast, the collection of vectors from $ℝ^{2}$ with a second component of one does not form a subspace (it is not closed under scalar multiplication). Another example is Example 2.2, where the condition on the vectors is that the three components add to zero. If the condition there were that the three components add to one then it would not be a subspace (again, it would fail to be closed). However, a reliance on zero is not strictly necessary. Consider the set

${(\begin{matrix} x \\ y \\ z \end{matrix}) ∣ x + y + z = 1}$

under these operations.

$(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} - 1 \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix}) r (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r x - r + 1 \\ r y \\ r z \end{matrix})$
1. Show that it is not a subspace of $ℝ^{3}$ . (Hint. See Example 2.7).
2. Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply.
3. Show that any subspace of $ℝ^{3}$ must pass through the origin, and so any subspace of $ℝ^{3}$ must involve zero in its description. Does the converse hold? Does any subset of $ℝ^{3}$ that contains the origin become a subspace when given the inherited operations?
Back to Exercise 2.38

Answer.
1. It is not a subspace because these are not the inherited operations. For one thing, in this space,
  
  $0 \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})$
  
  while this does not, of course, hold in $ℝ^{3}$ .
2. We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If $r_{1}, r_{2}, x_{1}, x_{2}, y_{1}, y_{2}, z_{1}, z_{2}$ are in $ℝ$ then
  
  $\begin{matrix} r_{1} (\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + r_{2} (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} r_{1} x_{1} - r_{1} + 1 \\ r_{1} y_{1} \\ r_{1} z_{1} \end{matrix}) + (\begin{matrix} r_{2} x_{2} - r_{2} + 1 \\ r_{2} y_{2} \\ r_{2} z_{2} \end{matrix}) \\ = (\begin{matrix} r_{1} x_{1} - r_{1} + r_{2} x_{2} - r_{2} + 1 \\ r_{1} y_{1} + r_{2} y_{2} \\ r_{1} z_{1} + r_{2} z_{2} \end{matrix}) \end{matrix}$
  
  (note that the definition of addition in this space is that the first components combine as $(r_{1} x_{1} - r_{1} + 1) + (r_{2} x_{2} - r_{2} + 1) - 1$ , so the first component of the last vector does not say ‘ $+ 2$ ’). Adding the three components of the last vector gives $r_{1} (x_{1} - 1 + y_{1} + z_{1}) + r_{2} (x_{2} - 1 + y_{2} + z_{2}) + 1 = r_{1} \cdot 0 + r_{2} \cdot 0 + 1 = 1$ .
  
  Most of the other checks of the conditions are easy (although the oddness of the operations keeps them from being routine). Commutativity of addition goes like this.
  
  $(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} - 1 \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix}) = (\begin{matrix} x_{2} + x_{1} - 1 \\ y_{2} + y_{1} \\ z_{2} + z_{1} \end{matrix}) = (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) + (\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix})$
  
  Associativity of addition has
  
  $((\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix})) + (\begin{matrix} x_{3} \\ y_{3} \\ z_{3} \end{matrix}) = (\begin{matrix} (x_{1} + x_{2} - 1) + x_{3} - 1 \\ (y_{1} + y_{2}) + y_{3} \\ (z_{1} + z_{2}) + z_{3} \end{matrix})$
  
  while
  
  $(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + ((\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) + (\begin{matrix} x_{3} \\ y_{3} \\ z_{3} \end{matrix})) = (\begin{matrix} x_{1} + (x_{2} + x_{3} - 1) - 1 \\ y_{1} + (y_{2} + y_{3}) \\ z_{1} + (z_{2} + z_{3}) \end{matrix})$
  
  and they are equal. The identity element with respect to this addition operation works this way
  
  $(\begin{matrix} x \\ y \\ z \end{matrix}) + (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} x + 1 - 1 \\ y + 0 \\ z + 0 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  and the additive inverse is similar.
  
  $(\begin{matrix} x \\ y \\ z \end{matrix}) + (\begin{matrix} - x + 2 \\ - y \\ - z \end{matrix}) = (\begin{matrix} x + (- x + 2) - 1 \\ y - y \\ z - z \end{matrix}) = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})$
  
  The conditions on scalar multiplication are also easy. For the first condition,
  
  $(r + s) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} (r + s) x - (r + s) + 1 \\ (r + s) y \\ (r + s) z \end{matrix})$
  
  while
  
  $\begin{matrix} r (\begin{matrix} x \\ y \\ z \end{matrix}) + s (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r x - r + 1 \\ r y \\ r z \end{matrix}) + (\begin{matrix} s x - s + 1 \\ s y \\ s z \end{matrix}) \\ = (\begin{matrix} (r x - r + 1) + (s x - s + 1) - 1 \\ r y + s y \\ r z + s z \end{matrix}) \end{matrix}$
  
  and the two are equal. The second condition compares
  
  $r \cdot ((\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix})) = r \cdot (\begin{matrix} x_{1} + x_{2} - 1 \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix}) = (\begin{matrix} r (x_{1} + x_{2} - 1) - r + 1 \\ r (y_{1} + y_{2}) \\ r (z_{1} + z_{2}) \end{matrix})$
  
  with
  
  $\begin{matrix} r (\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) + r (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} r x_{1} - r + 1 \\ r y_{1} \\ r z_{1} \end{matrix}) + (\begin{matrix} r x_{2} - r + 1 \\ r y_{2} \\ r z_{2} \end{matrix}) \\ = (\begin{matrix} (r x_{1} - r + 1) + (r x_{2} - r + 1) - 1 \\ r y_{1} + r y_{2} \\ r z_{1} + r z_{2} \end{matrix}) \end{matrix}$
  
  and they are equal. For the third condition,
  
  $(r s) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r s x - r s + 1 \\ r s y \\ r s z \end{matrix})$
  
  while
  
  $r (s (\begin{matrix} x \\ y \\ z \end{matrix})) = r ((\begin{matrix} s x - s + 1 \\ s y \\ s z \end{matrix})) = (\begin{matrix} r (s x - s + 1) - r + 1 \\ r s y \\ r s z \end{matrix})$
  
  and the two are equal. For scalar multiplication by $1$ we have this.
  
  $1 \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 x - 1 + 1 \\ 1 y \\ 1 z \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})$
  
  Thus all the conditions on a vector space are met by these two operations.
  
  Remark. A way to understand this vector space is to think of it as the plane in $ℝ^{3}$
  
  $P = {(\begin{matrix} x \\ y \\ z \end{matrix}) ∣ x + y + z = 0}$
  
  displaced away from the origin by $1$ along the $x$ -axis. Then addition becomes: to add two members of this space,
  
  $(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}), (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix})$
  
  (such that $x_{1} + y_{1} + z_{1} = 1$ and $x_{2} + y_{2} + z_{2} = 1$ ) move them back by $1$ to place them in $P$ and add as usual,
  
  $(\begin{matrix} x_{1} - 1 \\ y_{1} \\ z_{1} \end{matrix}) + (\begin{matrix} x_{2} - 1 \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} - 2 \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix}) (in P)$
  
  and then move the result back out by $1$ along the $x$ -axis.
  
  $(\begin{matrix} x_{1} + x_{2} - 1 \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix}) .$
  
  Scalar multiplication is similar.
3. For the subspace to be closed under the inherited scalar multiplication, where $\vec{v}$ is a member of that subspace,
  
  $0 \cdot \vec{v} = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
  
  must also be a member.
  
  The converse does not hold. Here is a subset of $ℝ^{3}$ that contains the origin
  
  ${(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}$
  
  (this subset has only two elements) but is not a subspace.
Exercise 2.39 Worked answer

We can give a justification for the convention that the sum of zero-many vectors equals the zero vector. Consider this sum of three vectors ${\vec{v}}_{1} + {\vec{v}}_{2} + {\vec{v}}_{3}$ .
1. What is the difference between this sum of three vectors and the sum of the first two of these three?
2. What is the difference between the prior sum and the sum of just the first one vector?
3. What should be the difference between the prior sum of one vector and the sum of no vectors?
4. So what should be the definition of the sum of no vectors?
Back to Exercise 2.39

Answer.
1. $({\vec{v}}_{1} + {\vec{v}}_{2} + {\vec{v}}_{3}) - ({\vec{v}}_{1} + {\vec{v}}_{2}) = {\vec{v}}_{3}$
2. $({\vec{v}}_{1} + {\vec{v}}_{2}) - ({\vec{v}}_{1}) = {\vec{v}}_{2}$
3. Surely, ${\vec{v}}_{1}$ .
4. Taking the one-long sum and subtracting gives ( ${\vec{v}}_{1}) - {\vec{v}}_{1} = \vec{0}$ .
Exercise 2.40 Worked answer

Is a space determined by its subspaces? That is, if two vector spaces have the same subspaces, must the two be equal?

Back to Exercise 2.40

Answer. Yes; any space is a subspace of itself, so each space contains the other.
Exercise 2.41 Worked answer
1. Give a set that is closed under scalar multiplication but not addition.
2. Give a set closed under addition but not scalar multiplication.
3. Give a set closed under neither.
Back to Exercise 2.41

Answer.
1. The union of the $x$ -axis and the $y$ -axis in $ℝ^{2}$ is one.
2. The set of integers, as a subset of $ℝ^{1}$ , is one.
3. The subset ${\vec{v}}$ of $ℝ^{2}$ is one, where $\vec{v}$ is any nonzero vector.
Exercise 2.42 Worked answer

Show that the span of a set of vectors does not depend on the order in which the vectors are listed in that set.

Back to Exercise 2.42

Answer. Because vector space addition is commutative, a reordering of summands leaves a linear combination unchanged.
Exercise 2.43 Worked answer

Which trivial subspace is the span of the empty set? Is it

${(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})} \subseteq ℝ^{3}, or {0 + 0 x} \subseteq 𝒫_{1},$

or some other subspace?

Back to Exercise 2.43

Answer. We always consider that span in the context of an enclosing space.
Exercise 2.44 Worked answer

Show that if a vector is in the span of a set then adding that vector to the set won’t make the span any bigger. Is that also ‘only if’?

Back to Exercise 2.44

Answer. It is both ‘if’ and ‘only if’.

For ‘if’, let $S$ be a subset of a vector space $V$ and assume $\vec{v} \in S$ satisfies $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ where $c_{1}, \dots, c_{n}$ are scalars and ${\vec{s}}_{1}, \dots, {\vec{s}}_{n} \in S$ . We must show that $[S \cup {\vec{v}}] = [S]$ .

Containment one way, $[S] \subseteq [S \cup {\vec{v}}]$ is obvious. For the other direction, $[S \cup {\vec{v}}] \subseteq [S]$ , note that if a vector is in the set on the left then it has the form $d_{0} \vec{v} + d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m}$ where the $d$ ’s are scalars and the $\vec{t}$ ’s are in $S$ . Rewrite that as $d_{0} (c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}) + d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m}$ and note that the result is a member of the span of $S$ .

The ‘only if’ is clearly true—adding $\vec{v}$ enlarges the span to include at least $\vec{v}$ .
Exercise 2.45 Worked answer

Recommended. Subspaces are subsets and so we naturally consider how ‘is a subspace of’ interacts with the usual set operations.
1. If $A, B$ are subspaces of a vector space, must their intersection $A \cap B$ be a subspace? Always? Sometimes? Never?
2. Must the union $A \cup B$ be a subspace?
3. If $A$ is a subspace of some $V$ , must its set complement $V - A$ be a subspace?
(Hint. Try some test subspaces from Example 2.19.)
Back to Exercise 2.45

Answer.
1. Always.
  
  Assume that $A, B$ are subspaces of $V$ . Note that their intersection is not empty as both contain the zero vector. If $\vec{w}, \vec{s} \in A \cap B$ and $r, s$ are scalars then $r \vec{v} + s \vec{w} \in A$ because each vector is in $A$ and so a linear combination is in $A$ , and $r \vec{v} + s \vec{w} \in B$ for the same reason. Thus the intersection is closed. Now Lemma 2.9 applies.
2. Sometimes (more precisely, only if $A \subseteq B$ or $B \subseteq A$ ).
  
  To see the answer is not ‘always’, take $V$ to be $ℝ^{3}$ , take $A$ to be the $x$ -axis, and $B$ to be the $y$ -axis. Note that
  
  $(\begin{matrix} 1 \\ 0 \end{matrix}) \in A and (\begin{matrix} 0 \\ 1 \end{matrix}) \in B but (\begin{matrix} 1 \\ 0 \end{matrix}) + (\begin{matrix} 0 \\ 1 \end{matrix}) \notin A \cup B$
  
  as the sum is in neither $A$ nor $B$ .
  
  The answer is not ‘never’ because if $A \subseteq B$ or $B \subseteq A$ then clearly $A \cup B$ is a subspace.
  
  To show that $A \cup B$ is a subspace only if one subspace contains the other, we assume that $A ⊈ B$ and $B ⊈ A$ and prove that the union is not a subspace. The assumption that $A$ is not a subset of $B$ means that there is an $\vec{a} \in A$ with $\vec{a} \notin B$ . The other assumption gives a $\vec{b} \in B$ with $\vec{b} \notin A$ . Consider $\vec{a} + \vec{b}$ . Note that sum is not an element of $A$ or else $(\vec{a} + \vec{b}) - \vec{a}$ would be in $A$ , which it is not. Similarly the sum is not an element of $B$ . Hence the sum is not an element of $A \cup B$ , and so the union is not a subspace.
3. Never. As $A$ is a subspace it contains the zero vector and therefore the set that is $A$ ’s complement, $V - A$ , does not. Without the zero vector, the complement cannot be a vector space.
Exercise 2.46 Worked answer

Does the span of a set depend on the enclosing space? That is, if $W$ is a subspace of $V$ and $S$ is a subset of $W$ (and so also a subset of $V$ ), might the span of $S$ in $W$ differ from the span of $S$ in $V$ ?

Back to Exercise 2.46

Answer. The span of a set does not depend on the enclosing space. A linear combination of vectors from $S$ gives the same sum whether we regard the operations as those of $W$ or as those of $V$ , because the operations of $W$ are inherited from $V$ .
Exercise 2.47 Worked answer

Is the relation ‘is a subspace of’ transitive? That is, if $V$ is a subspace of $W$ and $W$ is a subspace of $X$ , must $V$ be a subspace of $X$ ?

Back to Exercise 2.47

Answer. It is; apply Lemma 2.9. (You must consider the following. Suppose $B$ is a subspace of a vector space $V$ and suppose $A \subseteq B \subseteq V$ is a subspace. From which space does $A$ inherit its operations? The answer is that it doesn’t matter— $A$ will inherit the same operations in either case.)
Exercise 2.48 Worked answer

Because ‘span of’ is an operation on sets we naturally consider how it interacts with the usual set operations.
1. If $S \subseteq T$ are subsets of a vector space, is $[S] \subseteq [T]$ ? Always? Sometimes? Never?
2. If $S, T$ are subsets of a vector space, is $[S \cup T] = [S] \cup [T]$ ?
3. If $S, T$ are subsets of a vector space, is $[S \cap T] = [S] \cap [T]$ ?
4. Is the span of the complement equal to the complement of the span?
Back to Exercise 2.48

Answer.
1. Always; if $S \subseteq T$ then a linear combination of elements of $S$ is also a linear combination of elements of $T$ .
2. Sometimes (more precisely, if and only if $S \subseteq T$ or $T \subseteq S$ ).
  
  The answer is not ‘always’ as is shown by this example from $ℝ^{3}$
  
  $S = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})}, T = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}$
  
  because of this.
  
  $(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \in [S \cup T] (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \notin [S] \cup [T]$
  
  The answer is not ‘never’ because if either set contains the other then equality is clear. We can characterize equality as happening only when either set contains the other by assuming $S ⊈ T$ (implying the existence of a vector $\vec{s} \in S$ with $\vec{s} \notin T$ ) and $T ⊈ S$ (giving a $\vec{t} \in T$ with $\vec{t} \notin S$ ), noting $\vec{s} + \vec{t} \in [S \cup T]$ , and showing that $\vec{s} + \vec{t} \notin [S] \cup [T]$ .
3. Sometimes.
  
  Clearly $[S \cap T] \subseteq [S] \cap [T]$ because any linear combination of vectors from $S \cap T$ is a combination of vectors from $S$ and also a combination of vectors from $T$ .
  
  Containment the other way does not always hold. For instance, in $ℝ^{2}$ , take
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}, T = {(\begin{matrix} 2 \\ 0 \end{matrix})}$
  
  so that $[S] \cap [T]$ is the $x$ -axis but $[S \cap T]$ is the trivial subspace.
  
  Characterizing exactly when equality holds is tough. Clearly equality holds if either set contains the other, but that is not ‘only if’ by this example in $ℝ^{3}$ .
  
  $S = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})}, T = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}$
4. Never, as the span of the complement is a subspace, while the complement of the span is not (it does not contain the zero vector).
Exercise 2.49 Worked answer

Find a structure that is closed under linear combinations, and yet is not a vector space.

Back to Exercise 2.49

Answer. For this to happen, one of the conditions giving the sensibleness of the addition and scalar multiplication operations must be violated. Consider $ℝ^{2}$ with these operations.

$(\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + (\begin{matrix} x_{2} \\ y_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) r (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$

The set $ℝ^{2}$ is closed under these operations. But it is not a vector space.

$1 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) \neq (\begin{matrix} 1 \\ 1 \end{matrix})$

References cited in this section

Cleary

R. Cleary, private communication, Nov. 2011.

More information on equivalence of statements is in the appendix.↩︎