Linear Independence

The prior section shows how to understand a vector space as a span, as an unrestricted linear combination of some of its elements. For example, the space of linear polynomials ${a + b x ∣ a, b \in ℝ}$ is spanned by the set ${1, x}$ . The prior section also showed that a space can have many sets that span it. Two more sets that span the space of linear polynomials are ${1, 2 x}$ and ${1, x, 2 x}$ .

At the end of that section we described some spanning sets as ‘minimal’ but we never precisely defined that word. We could mean that a spanning set is minimal if it contains the smallest number of members of any set with the same span, so that ${1, x, 2 x}$ is not minimal because it has three members while we can give two-element sets spanning the same space. Or we could mean that a spanning set is minimal when it has no elements that we can remove without changing the span. Under this meaning ${1, x, 2 x}$ is not minimal because removing the $2 x$ to get ${1, x}$ leaves the span unchanged.

The first sense of minimality appears to be a global requirement, in that to check if a spanning set is minimal we seemingly must look at all the sets that span and find one with the least number of elements. The second sense of minimality is local since we need to look only at the set and consider the span with and without various elements. For instance, using the second sense we could compare the span of ${1, x, 2 x}$ with the span of ${1, x}$ and note that $2 x$ is a “repeat” in that its removal doesn’t shrink the span.

In this section we will use the second sense of ‘minimal spanning set’ because of this technical convenience. However, the most important result of this book is that the two senses coincide. We will prove that in the next section.

Definition and Examples

We saw “repeats” in the first chapter. There, Gauss’s Method turned them into $0 = 0$ equations.

Example 1.1 Recall the Statics example from Chapter One’s opening. We got two balances with the pair of unknown-mass objects, one at $40$ cm and $15$ cm and another at $- 50$ cm and $25$ cm, and we then computed the value of those masses. Had we instead gotten the second balance at $20$ cm and $7.5$ cm then Gauss’s Method on the resulting two-equations, two-unknowns system would not have yielded a solution, it would have yielded a $0 = 0$ equation along with an equation containing a free variable. Intuitively, the problem is that $(\begin{matrix} 20 & 7.5 \end{matrix})$ is half of $(\begin{matrix} 40 & 15 \end{matrix})$ , that is, $(\begin{matrix} 20 & 7.5 \end{matrix})$ is in the span of the set ${(\begin{matrix} 40 & 15 \end{matrix})}$ and so is repeated data. We would have been trying to solve a two-unknowns problem with essentially only one piece of information.

We take $\vec{v}$ to be a “repeat” of the vectors in a set $S$ if $\vec{v} \in [S]$ so that it depends on, that is, is expressible in terms of, elements of the set $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ .

Lemma 1.2 Where $V$ is a vector space, $S$ is a subset of that space, and $\vec{v}$ is an element of that space, $[S \cup {\vec{v}}] = [S]$ if and only if $\vec{v} \in [S]$ .

Proof Half of the if and only if is immediate: if $\vec{v} \notin [S]$ then the sets are not equal because $\vec{v} \in [S \cup {\vec{v}}]$ .

For the other half assume that $\vec{v} \in [S]$ so that $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ for some scalars $c_{i}$ and vectors ${\vec{s}}_{i} \in S$ . We will use mutual containment to show that the sets $[S \cup {\vec{v}}]$ and $[S]$ are equal. The containment $[S \cup {\vec{v}}] \supseteq [S]$ is clear.

To show containment in the other direction let $\vec{w}$ be an element of $[S \cup {\vec{v}}]$ . Then $\vec{w}$ is a linear combination of elements of $S \cup {\vec{v}}$ , which we can write as $\vec{w} = c_{n + 1} {\vec{s}}_{n + 1} + \dots + c_{n + k} {\vec{s}}_{n + k} + c_{n + k + 1} \vec{v}$ . (Possibly some of the ${\vec{s}}_{i}$ ’s from $\vec{w}$ ’s equation are the same as some of those from $\vec{v}$ ’s equation but that does not matter.) Expand $\vec{v}$ .

$\vec{w} = c_{n + 1} {\vec{s}}_{n + 1} + \dots + c_{n + k} {\vec{s}}_{n + k} + c_{n + k + 1} \cdot (c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n})$

Recognize the right hand side as a linear combination of linear combinations of vectors from $S$ . Thus $\vec{w} \in [S]$ .

QED

The discussion at the section’s opening involved removing vectors, not adding them.

Corollary 1.3 For $\vec{v} \in S$ , omitting that vector does not shrink the span if and only if that vector is dependent on other vectors in the set. That is, $[S] = [S - {\vec{v}}]$ if and only if $\vec{v} \in [S - {\vec{v}}]$ .

Thus, to know whether removing a vector will decrease the span, we need to know whether the vector is a linear combination of others in the set.

Definition 1.4 In any vector space, a set of vectors is linearly independent if none of its elements is a linear combination of the others from the set.¹ Otherwise the set is linearly dependent.

Thus the set ${{\vec{s}}_{0}, \dots, {\vec{s}}_{n}}$ is independent if there is no equality ${\vec{s}}_{i} = c_{0} {\vec{s}}_{0} + \dots + c_{i - 1} {\vec{s}}_{i - 1} + c_{i + 1} {\vec{s}}_{i + 1} + \dots + c_{n} {\vec{s}}_{n}$ . The definition’s use of the word ‘others’ means that writing ${\vec{s}}_{i}$ as a linear combination via ${\vec{s}}_{i} = 1 \cdot {\vec{s}}_{i}$ does not count.

Observe that, although this way of writing one vector as a combination of the others

${\vec{s}}_{0} = c_{1} {\vec{s}}_{1} + c_{2} {\vec{s}}_{2} + \dots + c_{n} {\vec{s}}_{n}$

visually sets off ${\vec{s}}_{0}$ , algebraically there is nothing special about that vector in that equation. For any ${\vec{s}}_{i}$ with a coefficient $c_{i}$ that is non- $0$ , we can rewrite to isolate ${\vec{s}}_{i}$ .

${\vec{s}}_{i} = (1 / c_{i}) {\vec{s}}_{0} + \dots + (- c_{i - 1} / c_{i}) {\vec{s}}_{i - 1} + (- c_{i + 1} / c_{i}) {\vec{s}}_{i + 1} + \dots + (- c_{n} / c_{i}) {\vec{s}}_{n}$

When we don’t want to single out any vector we will instead say that ${\vec{s}}_{0}, {\vec{s}}_{1}, \dots, {\vec{s}}_{n}$ are in a linear relationship and put all of the vectors on the same side. The next result rephrases the linear independence definition in this style. It is how we usually compute whether a finite set is dependent or independent.

Lemma 1.5 A subset $S$ of a vector space is linearly independent if and only if among its elements the only linear relationship $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = \vec{0}$ is the trivial one, $c_{1} = 0, \dots, c_{n} = 0$ (where ${\vec{s}}_{i} \neq {\vec{s}}_{j}$ when $i \neq j$ ) .

Proof If $S$ is linearly independent then no vector ${\vec{s}}_{i}$ is a linear combination of other vectors from $S$ , so there is no linear relationship where some of the $\vec{s}$ ’s have nonzero coefficients.

If $S$ is not linearly independent then some ${\vec{s}}_{i}$ is a linear combination ${\vec{s}}_{i} = c_{1} {\vec{s}}_{1} + \dots + c_{i - 1} {\vec{s}}_{i - 1} + c_{i + 1} {\vec{s}}_{i + 1} + \dots + c_{n} {\vec{s}}_{n}$ of other vectors from $S$ . Subtracting ${\vec{s}}_{i}$ from both sides gives a relationship involving a nonzero coefficient, the $- 1$ in front of ${\vec{s}}_{i}$ .

QED

Example 1.6 In the vector space of two-wide row vectors, the two-element set ${(\begin{matrix} 40 & 15 \end{matrix}), (\begin{matrix} - 50 & 25 \end{matrix})}$ is linearly independent. To check this, take

$c_{1} \cdot (\begin{matrix} 40 & 15 \end{matrix}) + c_{2} \cdot (\begin{matrix} - 50 & 25 \end{matrix}) = (\begin{matrix} 0 & 0 \end{matrix})$

and solve the resulting system.

$\begin{array}{rcrcr} 40 c_{1} & - & 50 c_{2} & = & 0 \\ 15 c_{1} & + & 25 c_{2} & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (15 / 40) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 40 c_{1} & - & 50 c_{2} & = & 0 \\ (175 / 4) c_{2} & = & 0 \end{array}$

Both $c_{1}$ and $c_{2}$ are zero. So the only linear relationship between the two given row vectors is the trivial relationship.

In the same vector space, the set ${(\begin{matrix} 40 & 15 \end{matrix}), (\begin{matrix} 20 & 7.5 \end{matrix})}$ is linearly dependent since we can satisfy $c_{1} \cdot (\begin{matrix} 40 & 15 \end{matrix}) + c_{2} \cdot (\begin{matrix} 20 & 7.5 \end{matrix}) = (\begin{matrix} 0 & 0 \end{matrix})$ with $c_{1} = 1$ and $c_{2} = - 2$ .

Example 1.7 The set ${1 + x, 1 - x}$ is linearly independent in $𝒫_{2}$ , the space of quadratic polynomials with real coefficients, because

$0 + 0 x + 0 x^{2} = c_{1} (1 + x) + c_{2} (1 - x) = (c_{1} + c_{2}) + (c_{1} - c_{2}) x + 0 x^{2}$

gives

$\begin{array}{rcrcr} c_{1} & + & c_{2} & = & 0 \\ c_{1} & - & c_{2} & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcr} c_{1} & + & c_{2} & = & 0 \\ 2 c_{2} & = & 0 \end{array}$

since polynomials are equal only if their coefficients are equal. Thus, the only linear relationship between these two members of $𝒫_{2}$ is the trivial one.

Remark 1.8 The lemma specifies that ${\vec{s}}_{i} \neq {\vec{s}}_{j}$ when $i \neq j$ because of course if some vector $\vec{s}$ appears twice then we can get a nontrivial $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = \vec{0}$ , by taking the associated coefficients to be $1$ and $- 1$ . Besides, if some vector appears more than once in an expression then we can always combine the coefficients.

Note that the lemma allows the opposite of appearing more than once, that some vectors from $S$ don’t appear at all. For instance, if $S$ is infinite then because linear relationships involve only finitely many vectors, any such relationship leaves out many of $S$ ’s vectors. However, note also that if $S$ is finite then where convenient we can take a combination $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ to contain each of $S$ ’s vectors once and only once. If a vector is missing then we can add it by using a coefficient of $0$ .

Example 1.9 The rows of this matrix

$A = (\begin{matrix} 2 & 3 & 1 & 0 \\ 0 & - 1 & 0 & - 2 \\ 0 & 0 & 0 & 1 \end{matrix})$

form a linearly independent set. This is easy to check for this case but also recall that Lemma One.III.2.5 shows that the rows of any echelon form matrix make a linearly independent set.

Example 1.10 In $ℝ^{3}$ , where

${\vec{v}}_{1} = (\begin{matrix} 3 \\ 4 \\ 5 \end{matrix}) {\vec{v}}_{2} = (\begin{matrix} 2 \\ 9 \\ 2 \end{matrix}) {\vec{v}}_{3} = (\begin{matrix} 4 \\ 18 \\ 4 \end{matrix})$

the set $S = {{\vec{v}}_{1}, {\vec{v}}_{2}, {\vec{v}}_{3}}$ is linearly dependent because this is a relationship

$0 \cdot {\vec{v}}_{1} + 2 \cdot {\vec{v}}_{2} - 1 \cdot {\vec{v}}_{3} = \vec{0}$

where not all of the scalars are zero (the fact that some of the scalars are zero doesn’t matter).

That example illustrates why, although Definition 1.4 is a clearer statement of what independence means, Lemma 1.5 is better for computations. Working straight from the definition, someone trying to compute whether $S$ is linearly independent would start by setting ${\vec{v}}_{1} = c_{2} {\vec{v}}_{2} + c_{3} {\vec{v}}_{3}$ and concluding that there are no such $c_{2}$ and $c_{3}$ . But knowing that the first vector is not dependent on the other two is not enough. This person would have to go on to try ${\vec{v}}_{2} = c_{1} {\vec{v}}_{1} + c_{3} {\vec{v}}_{3}$ , in order to find the dependence $c_{1} = 0$ , $c_{3} = 1 / 2$ . Lemma 1.5 gets the same conclusion with only one computation.

Example 1.11 The empty subset of a vector space is linearly independent. There is no nontrivial linear relationship among its members as it has no members.

Example 1.12 In any vector space, any subset containing the zero vector is linearly dependent. One example is, in the space $𝒫_{2}$ of quadratic polynomials, the subset ${1 + x, x + x^{2}, 0}$ . It is linearly dependent because $0 \cdot {\vec{v}}_{1} + 0 \cdot {\vec{v}}_{2} + 1 \cdot \vec{0} = \vec{0}$ is a nontrivial relationship, since not all of the coefficients are zero.

There is a subtle point that we shall see a number of times and that bears on the prior example. It is about the trivial sum, the sum of the empty set. One way to see how to define the trivial sum is to consider the progression ${\vec{v}}_{1} + {\vec{v}}_{2} + {\vec{v}}_{3}$ , followed by ${\vec{v}}_{1} + {\vec{v}}_{2}$ , followed by ${\vec{v}}_{1}$ . The difference between the sum of three vectors and the sum of two is ${\vec{v}}_{3}$ . Then the difference between the sum of two and the sum of one is ${\vec{v}}_{2}$ . In next passing to the trivial sum, the sum of zero-many vectors, we can expect to subtract ${\vec{v}}_{1}$ . So we define the sum of zero-many vectors to be the zero vector.

The relation with the prior example is that if the zero vector is in a set then that set has an element that is a combination of a subset of other vectors from the set, specifically, the zero vector is a combination of the empty subset. Even the set $S = {\vec{0}}$ is linearly dependent, because $\vec{0}$ is the sum of the empty set and the empty set is a subset of $S$ .

Remark 1.13 The definition of linear independence, Definition 1.4, refers to a ‘set’ of vectors. Sets are the most familiar kind of collection and in practice everyone uses the word ‘set’ in this context. But to be complete, we will note that sets are not quite the right kind of collection for this purpose.

Recall that a set is a collection with two properties: (i) order does not matter, so that the set ${1, 2}$ equals the set ${2, 1}$ , and (ii) duplicates collapse, so that the set ${1, 1, 2}$ equals the set ${1, 2}$ .

Now consider this matrix reduction.

$(\begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 1 & 2 & 3 \end{matrix}) ⟶_{\begin{array}{l} \end{array}}^{(1 / 2) ρ_{2}} (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{matrix})$

On the left the set of matrix rows ${(\begin{matrix} 1 & 1 & 1 \end{matrix}), (\begin{matrix} 2 & 2 & 2 \end{matrix}), (\begin{matrix} 1 & 2 & 3 \end{matrix})}$ is linearly dependent. On the right the set of rows is ${(\begin{matrix} 1 & 1 & 1 \end{matrix}), (\begin{matrix} 1 & 1 & 1 \end{matrix}), (\begin{matrix} 1 & 2 & 3 \end{matrix})}$ . Because duplicates collapse, that equals the set ${(\begin{matrix} 1 & 1 & 1 \end{matrix}), (\begin{matrix} 1 & 2 & 3 \end{matrix})}$ , which is linearly independent. This is a problem because Gauss’s Method should preserve linear dependence.

That is, strictly speaking, we need a type of collection where duplicates do not collapse. A collection where order does not matter and duplicates don’t collapse is a multiset.

However, while insisting on being completely correct has advantages, departing from the standard terminology of ‘set’ would have pitfalls of its own, so we will continue to use that word. Later, we shall occasionally need to take combinations without letting duplicates collapse and we shall do that without further comment.

Corollary 1.14 A set $S$ is linearly independent if and only if for any $\vec{v} \in S$ , its removal shrinks the span $[S - {v}] ⊊ [S]$ .

Proof This follows from Corollary 1.3. If $S$ is linearly independent then none of its vectors is dependent on the other elements, so removal of any vector will shrink the span. If $S$ is not linearly independent then it contains a vector that is dependent on other elements of the set, and removal of that vector will not shrink the span.

QED

So a spanning set is minimal if and only if it is linearly independent.

The prior result addresses removing elements from a linearly independent set. The next one adds elements.

Lemma 1.15 Suppose that $S$ is linearly independent and that $\vec{v} \notin S$ . Then the set $S \cup {\vec{v}}$ is linearly independent if and only if $\vec{v} \notin [S]$ .

Proof We will show that $S \cup {\vec{v}}$ is not linearly independent if and only if $\vec{v} \in [S]$ .

Suppose first that $\vec{v} \in [S]$ . Express $\vec{v}$ as a combination $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ . Rewrite that $\vec{0} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} - 1 \cdot \vec{v}$ . Since $\vec{v} \notin S$ , it does not equal any of the ${\vec{s}}_{i}$ so this is a nontrivial linear dependence among the elements of $S \cup {\vec{v}}$ . Thus that set is not linearly independent.

Now suppose that $S \cup {\vec{v}}$ is not linearly independent and consider a nontrivial dependence among its members $\vec{0} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} + c_{n + 1} \cdot \vec{v}$ . If $c_{n + 1} = 0$ then that is a dependence among the elements of $S$ , but we are assuming that $S$ is independent, so $c_{n + 1} \neq 0$ . Rewrite the equation as $\vec{v} = (c_{1} / c_{n + 1}) {\vec{s}}_{1} + \dots + (c_{n} / c_{n + 1}) {\vec{s}}_{n}$ to get $\vec{v} \in [S]$

QED

Example 1.16 This subset of $ℝ^{3}$ is linearly independent.

$S = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}$

The span of $S$ is the $x$ -axis. Here are two supersets, one that is linearly dependent and the other independent.

dependent: ${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} - 3 \\ 0 \\ 0 \end{matrix})}$ independent: ${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})}$

We got the dependent superset by adding a vector from the $x$ -axis and so the span did not grow. We got the independent superset by adding a vector that isn’t in $[S]$ , because it has a nonzero $y$ component, causing the span to grow.

For the independent set

$S = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})}$

the span $[S]$ is the $x y$ -plane. Here are two supersets.

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

As above, the additional member of the dependent superset comes from $[S]$ , the $x y$ -plane, while the added member of the independent superset comes from outside of that span.

Finally, consider this independent set

$S = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}$

with $[S] = ℝ^{3}$ . We can get a linearly dependent superset.

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

But there is no linearly independent superset of $S$ . One way to see that is to note that for any vector that we would add to $S$ , the equation

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

has a solution $c_{1} = x$ , $c_{2} = y$ , and $c_{3} = z$ . Another way to see it is that we cannot add any vectors from outside of the span $[S]$ because that span is $ℝ^{3}$ .

Corollary 1.17 In a vector space, any finite set has a linearly independent subset with the same span.

Proof If $S = {{\vec{s}}_{1}, \dots, {\vec{s}}_{n}}$ is linearly independent then $S$ itself satisfies the statement, so assume that it is linearly dependent.

By the definition of dependent, $S$ contains a vector ${\vec{v}}_{1}$ that is a linear combination of the others. Define the set $S_{1} = S - {{\vec{v}}_{1}}$ . By Corollary 1.3 the span does not shrink $[S_{1}] = [S]$ .

If $S_{1}$ is linearly independent then we are done. Otherwise iterate: take a vector ${\vec{v}}_{2}$ that is a linear combination of other members of $S_{1}$ and discard it to derive $S_{2} = S_{1} - {{\vec{v}}_{2}}$ such that $[S_{2}] = [S_{1}]$ . Repeat this until a linearly independent set $S_{j}$ appears; one must appear eventually because $S$ is finite and the empty set is linearly independent. (Formally, this argument uses induction on the number of elements in $S$ . Exercise 1.42 asks for the details.)

QED

Thus if we have a set that is linearly dependent then we can, without changing the span, pare down by discarding what we have called “repeat” vectors.

Example 1.18 This set spans $ℝ^{3}$ (the check is routine) but is not linearly independent.

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

We will calculate which vectors to drop in order to get a subset that is independent but has the same span. This linear relationship

$\begin{matrix} c_{1} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}) + c_{3} (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}) + c_{4} (\begin{matrix} 0 \\ - 1 \\ 1 \end{matrix}) + c_{5} (\begin{matrix} 3 \\ 3 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) & (*) \end{matrix}$

gives a system

$\begin{array}{rcrcrcrcrcr} c_{1} & + & c_{3} & + & + & 3 c_{5} & = & 0 \\ 2 c_{2} & + & 2 c_{3} & - & c_{4} & + & 3 c_{5} & = & 0 \\ c_{4} & = & 0 \end{array}$

whose solution set has this parametrization.

${(\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \\ c_{4} \\ c_{5} \end{matrix}) = c_{3} (\begin{matrix} - 1 \\ - 1 \\ 1 \\ 0 \\ 0 \end{matrix}) + c_{5} (\begin{matrix} - 3 \\ - 3 / 2 \\ 0 \\ 0 \\ 1 \end{matrix}) ∣ c_{3}, c_{5} \in ℝ}$

Set $c_{5} = 1$ and $c_{3} = 0$ to get an instance of ( $*$ ).

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

This shows that the vector from $S$ that we’ve associated with $c_{5}$ is in the span of the set of $c_{1}$ ’s vector and $c_{2}$ ’s vector. We can discard $S$ ’s fifth vector without shrinking the span.

Similarly, set $c_{3} = 1$ , and $c_{5} = 0$ to get an instance of ( $*$ ) that shows we can discard $S$ ’s third vector without shrinking the span. Thus this set has the same span as $S$ .

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

The check that it is linearly independent is routine.

Corollary 1.19 A subset $S = {{\vec{s}}_{1}, \dots, {\vec{s}}_{n}}$ of a vector space is linearly dependent if and only if some $\vec{s_{i}}$ is a linear combination of the vectors ${\vec{s}}_{1}$ , …, ${\vec{s}}_{i - 1}$ listed before it.

Proof Consider $S_{0} = {}$ , $S_{1} = {\vec{s_{1}}}$ , $S_{2} = {{\vec{s}}_{1}, {\vec{s}}_{2}}$ , etc. Some index $i \geq 1$ is the first one with $S_{i - 1} \cup {{\vec{s}}_{i}}$ linearly dependent, and there ${\vec{s}}_{i} \in [S_{i - 1}]$ .

QED

The proof of Corollary 1.17 describes producing a linearly independent set by shrinking, by taking subsets. And the proof of Corollary 1.19 describes finding a linearly dependent set by taking supersets. We finish this subsection by considering how linear independence and dependence interact with the subset relation between sets.

Lemma 1.20 Any subset of a linearly independent set is also linearly independent. Any superset of a linearly dependent set is also linearly dependent.

Proof Both are clear.

QED

Restated, subset preserves independence and superset preserves dependence.

Those are two of the four possible cases. The third case, whether subset preserves linear dependence, is covered by Example 1.18, which gives a linearly dependent set $S$ with one subset that is linearly dependent and another that is independent. The fourth case, whether superset preserves linear independence, is covered by Example 1.16, which gives cases where a linearly independent set has both an independent and a dependent superset. This table summarizes.

	$\hat{S} \subset S$	$\hat{S} \supset S$
$S$ independent	$\hat{S}$ must be independent	$\hat{S}$ may be either
$S$ dependent	$\hat{S}$ may be either	$\hat{S}$ must be dependent

Example 1.16 has something else to say about the interaction between linear independence and superset. It names a linearly independent set that is maximal in that it has no supersets that are linearly independent. By Lemma 1.15 a linearly independent set is maximal if and only if it spans the entire space, because that is when all the vectors in the space are already in the span. This nicely complements Lemma 1.14, that a spanning set is minimal if and only if it is linearly independent.

Exercises

Exercise 1.21 Worked answer

Recommended. Decide whether each subset of $ℝ^{3}$ is linearly dependent or linearly independent.
1. ${(\begin{matrix} 1 \\ - 3 \\ 5 \end{matrix}), (\begin{matrix} 2 \\ 2 \\ 4 \end{matrix}), (\begin{matrix} 4 \\ - 4 \\ 14 \end{matrix})}$
2. ${(\begin{matrix} 1 \\ 7 \\ 7 \end{matrix}), (\begin{matrix} 2 \\ 7 \\ 7 \end{matrix}), (\begin{matrix} 3 \\ 7 \\ 7 \end{matrix})}$
3. ${(\begin{matrix} 0 \\ 0 \\ - 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 4 \end{matrix})}$
4. ${(\begin{matrix} 9 \\ 9 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 3 \\ 5 \\ - 4 \end{matrix}), (\begin{matrix} 12 \\ 12 \\ - 1 \end{matrix})}$
Back to Exercise 1.21

Answer. For each of these, when the subset is independent you must prove it, and when the subset is dependent you must give an example of a dependence.
1. It is dependent. Considering
  
  $c_{1} (\begin{matrix} 1 \\ - 3 \\ 5 \end{matrix}) + c_{2} (\begin{matrix} 2 \\ 2 \\ 4 \end{matrix}) + c_{3} (\begin{matrix} 4 \\ - 4 \\ 14 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
  
  gives this linear system.
  
  $\begin{array}{rcrcrcr} c_{1} & + & 2 c_{2} & + & 4 c_{3} & = & 0 \\ - 3 c_{1} & + & 2 c_{2} & - & 4 c_{3} & = & 0 \\ 5 c_{1} & + & 4 c_{2} & + & 14 c_{3} & = & 0 \end{array}$
  
  Gauss’s Method
  
  $(\begin{array}{cccr} 1 & 2 & 4 & 0 \\ - 3 & 2 & - 4 & 0 \\ 5 & 4 & 14 & 0 \end{array}) ⟶_{\begin{array}{l} - 5 ρ_{1} + ρ_{3} \end{array}}^{3 ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{(3 / 4) ρ_{2} + ρ_{3}} (\begin{array}{cccr} 1 & 2 & 4 & 0 \\ 0 & 8 & 8 & 0 \\ 0 & 0 & 0 & 0 \end{array})$
  
  yields a free variable, so there are infinitely many solutions. For an example of a particular dependence we can set $c_{3}$ to be, say, $1$ . Then we get $c_{2} = - 1$ and $c_{1} = - 2$ .
2. It is dependent. The linear system that arises here
  
  $(\begin{array}{cccr} 1 & 2 & 3 & 0 \\ 7 & 7 & 7 & 0 \\ 7 & 7 & 7 & 0 \end{array}) ⟶_{\begin{array}{l} - 7 ρ_{1} + ρ_{3} \end{array}}^{- 7 ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{2} + ρ_{3}} (\begin{array}{cccr} 1 & 2 & 3 & 0 \\ 0 & - 7 & - 14 & 0 \\ 0 & 0 & 0 & 0 \end{array})$
  
  has infinitely many solutions. We can get a particular solution by taking $c_{3}$ to be, say, $1$ , and back-substituting to get the resulting $c_{2}$ and $c_{1}$ .
3. It is linearly independent. The system
  
  $(\begin{array}{ccr} 0 & 1 & 0 \\ 0 & 0 & 0 \\ - 1 & 4 & 0 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{ρ_{1} \leftrightarrow ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{3} \leftrightarrow ρ_{1}} (\begin{array}{ccc} - 1 & 4 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array})$
  
  has only the solution $c_{1} = 0$ and $c_{2} = 0$ . (We could also have gotten the answer by inspection—the second vector is obviously not a multiple of the first, and vice versa.)
4. It is linearly dependent. The linear system
  
  $(\begin{array}{ccccr} 9 & 2 & 3 & 12 & 0 \\ 9 & 0 & 5 & 12 & 0 \\ 0 & 1 & - 4 & - 1 & 0 \end{array})$
  
  has more unknowns than equations, and so Gauss’s Method must end with at least one variable free (there can’t be a contradictory equation because the system is homogeneous, and so has at least the solution of all zeroes). To exhibit a combination, we can do the reduction
  
  $⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{(1 / 2) ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 9 & 2 & 3 & 12 & 0 \\ 0 & - 2 & 2 & 0 & 0 \\ 0 & 0 & - 3 & - 1 & 0 \end{array})$
  
  and take, say, $c_{4} = 1$ . Then we have that $c_{3} = - 1 / 3$ , $c_{2} = - 1 / 3$ , and $c_{1} = - 31 / 27$ .
Exercise 1.22 Worked answer

Recommended. Which of these subsets of $𝒫_{3}$ are linearly dependent and which are independent?
1. ${3 - x + 9 x^{2}, 5 - 6 x + 3 x^{2}, 1 + 1 x - 5 x^{2}}$
2. ${- x^{2}, 1 + 4 x^{2}}$
3. ${2 + x + 7 x^{2}, 3 - x + 2 x^{2}, 4 - 3 x^{2}}$
4. ${8 + 3 x + 3 x^{2}, x + 2 x^{2}, 2 + 2 x + 2 x^{2}, 8 - 2 x + 5 x^{2}}$
Back to Exercise 1.22

Answer. In the cases of independence, you must prove that it is independent. Otherwise, you must exhibit a dependence. (Here we give a specific dependence but others are possible.)
1. This set is independent. Setting up the relation $c_{1} (3 - x + 9 x^{2}) + c_{2} (5 - 6 x + 3 x^{2}) + c_{3} (1 + 1 x - 5 x^{2}) = 0 + 0 x + 0 x^{2}$ gives a linear system
  
  $(\begin{array}{cccr} 3 & 5 & 1 & 0 \\ - 1 & - 6 & 1 & 0 \\ 9 & 3 & - 5 & 0 \end{array}) ⟶_{\begin{array}{l} - 3 ρ_{1} + ρ_{3} \end{array}}^{(1 / 3) ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{3 ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- (12 / 13) ρ_{2} + ρ_{3}} (\begin{array}{cccr} 3 & 5 & 1 & 0 \\ 0 & - 13 & 4 & 0 \\ 0 & 0 & - 128 / 13 & 0 \end{array})$
  
  with only one solution: $c_{1} = 0$ , $c_{2} = 0$ , and $c_{3} = 0$ .
2. This set is independent. We can see this by inspection, straight from the definition of linear independence. Obviously neither is a multiple of the other.
3. This set is linearly independent. The linear system reduces in this way
  
  $(\begin{array}{cccr} 2 & 3 & 4 & 0 \\ 1 & - 1 & 0 & 0 \\ 7 & 2 & - 3 & 0 \end{array}) ⟶_{\begin{array}{l} - (7 / 2) ρ_{1} + ρ_{3} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{- (17 / 5) ρ_{2} + ρ_{3}} (\begin{array}{cccr} 2 & 3 & 4 & 0 \\ 0 & - 5 / 2 & - 2 & 0 \\ 0 & 0 & - 51 / 5 & 0 \end{array})$
  
  to show that there is only the solution $c_{1} = 0$ , $c_{2} = 0$ , and $c_{3} = 0$ .
4. This set is linearly dependent. The linear system
  
  $(\begin{array}{ccccr} 8 & 0 & 2 & 8 & 0 \\ 3 & 1 & 2 & - 2 & 0 \\ 3 & 2 & 2 & 5 & 0 \end{array})$
  
  must, after reduction, end with at least one variable free (there are more variables than equations, and there is no possibility of a contradictory equation because the system is homogeneous). We can take the free variables as parameters to describe the solution set. We can then set the parameter to a nonzero value to get a nontrivial linear relation.
Exercise 1.23 Worked answer

Determine if each set is linearly independent in the natural space.
1. ${(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix})}$
2. ${(\begin{matrix} 1 & 3 & 1 \end{matrix}), (\begin{matrix} - 1 & 4 & 3 \end{matrix}), (\begin{matrix} - 1 & 11 & 7 \end{matrix})}$
3. ${(\begin{matrix} 5 & 4 \\ 1 & 2 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 1 & 0 \\ - 1 & 4 \end{matrix})}$
Back to Exercise 1.23

Answer.
1. The natural vector space is $ℝ^{3}$ . Set up the equation
  
  $(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) = c_{1} (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix})$
  
  and consider the resulting homogeneous system.
  
  $(\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{ccc} 1 & - 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{array})$
  
  This has the unique solution, that $c_{1} = 0$ , $c_{2} = 0$ . So it is linearly independent.
2. The natural vector space is the set of three-wide row vectors. The equation
  
  $(\begin{matrix} 0 & 0 & 0 \end{matrix}) = c_{1} (\begin{matrix} 1 & 3 & 1 \end{matrix}) + c_{2} (\begin{matrix} - 1 & 4 & 3 \end{matrix}) + c_{3} (\begin{matrix} - 1 & 11 & 7 \end{matrix})$
  
  gives rise to a linear system
  
  $\begin{aligned} (\begin{array}{cccc} 1 & - 1 & - 1 & 0 \\ 3 & 4 & 11 & 0 \\ 1 & 3 & 7 & 0 \end{array}) & ⟶_{\begin{matrix} - ρ_{1} + ρ_{3} \end{matrix}}^{- 3 ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & - 1 & - 1 & 0 \\ 0 & 7 & 14 & 0 \\ 0 & 4 & 8 & 0 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{(4 / 7) ρ_{2} + ρ_{3}} (\begin{array}{cccc} 1 & - 1 & - 1 & 0 \\ 0 & 7 & 14 & 0 \\ 0 & 0 & 0 & 0 \end{array}) \end{aligned}$
  
  with infinitely many solutions, that is, more than just the trivial solution.
  
  ${(\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix}) = (\begin{matrix} - 1 \\ - 2 \\ 1 \end{matrix}) c_{3} ∣ c_{3} \in ℝ}$
  
  So the set is linearly dependent. One dependence comes from setting $c_{3} = 2$ , giving $c_{1} = - 2$ and $c_{2} = - 4$ .
3. Without having to set up a system we can see that the second element of the set is a multiple of the first (namely, $0$ times the first).
Exercise 1.24 Worked answer

Recommended. Prove that each set ${f, g}$ is linearly independent in the vector space of all functions from $ℝ^{+}$ to $ℝ$ .
1. $f (x) = x$ and $g (x) = 1 / x$
2. $f (x) = \cos (x)$ and $g (x) = \sin (x)$
3. $f (x) = e^{x}$ and $g (x) = \ln (x)$
Back to Exercise 1.24

Answer. Let $Z : ℝ^{+} \to ℝ$ be the zero function $Z (x) = 0$ , which is the additive identity in the vector space under discussion.
1. This set is linearly independent. Consider $c_{1} \cdot f (x) + c_{2} \cdot g (x) = Z (x)$ . Plugging in $x = 1$ and $x = 2$ gives a linear system
  
  $\begin{array}{rcrcr} c_{1} \cdot 1 & + & c_{2} \cdot 1 & = & 0 \\ c_{1} \cdot 2 & + & c_{2} \cdot (1 / 2) & = & 0 \end{array}$
  
  with the unique solution $c_{1} = 0$ , $c_{2} = 0$ .
2. This set is linearly independent. Consider $c_{1} \cdot f (x) + c_{2} \cdot g (x) = Z (x)$ and plug in $x = π$ and $x = π / 2$ to get
  
  $\begin{array}{rcrcr} c_{1} \cdot (- 1) & + & c_{2} \cdot 0 & = & 0 \\ c_{1} \cdot 0 & + & c_{2} \cdot 1 & = & 0 \end{array}$
  
  which obviously gives $c_{1} = 0$ , $c_{2} = 0$ .
3. This set is also linearly independent. Considering $c_{1} \cdot f (x) + c_{2} \cdot g (x) = Z (x)$ and plugging in $x = 1$ and $x = e$
  
  $\begin{array}{rcrcr} c_{1} \cdot e & + & c_{2} \cdot 0 & = & 0 \\ c_{1} \cdot e^{e} & + & c_{2} \cdot 1 & = & 0 \end{array}$
  
  gives that $c_{1} = 0$ and $c_{2} = 0$ .
Exercise 1.25 Worked answer

Recommended. Which of these subsets of the space of real-valued functions of one real variable is linearly dependent and which is linearly independent? (We have abbreviated some constant functions; e.g., in the first item, the ‘ $2$ ’ stands for the constant function $f (x) = 2$ .)
1. ${2, 4 \sin^{2} (x), \cos^{2} (x)}$
2. ${1, \sin (x), \sin (2 x)}$
3. ${x, \cos (x)}$
4. ${(1 + x)^{2}, x^{2} + 2 x, 3}$
5. ${\cos (2 x), \sin^{2} (x), \cos^{2} (x)}$
6. ${0, x, x^{2}}$
Back to Exercise 1.25

Answer. In each case, if the set is independent then you must prove that and if it is dependent then you must exhibit a dependence.
1. This set is dependent. The familiar relation $\sin^{2} (x) + \cos^{2} (x) = 1$ shows that $2 = c_{1} \cdot (4 \sin^{2} (x)) + c_{2} \cdot (\cos^{2} (x))$ is satisfied by $c_{1} = 1 / 2$ and $c_{2} = 2$ .
2. This set is independent. Consider the relationship $c_{1} \cdot 1 + c_{2} \cdot \sin (x) + c_{3} \cdot \sin (2 x) = 0$ (that ‘ $0$ ’ is the zero function). Taking three suitable points such as $x = π$ , $x = π / 2$ , $x = π / 4$ gives a system
  
  $\begin{array}{rcrcrcr} c_{1} & = & 0 \\ c_{1} & + & c_{2} & = & 0 \\ c_{1} & + & (\sqrt{2} / 2) c_{2} & + & c_{3} & = & 0 \end{array}$
  
  whose only solution is $c_{1} = 0$ , $c_{2} = 0$ , and $c_{3} = 0$ .
3. By inspection, this set is independent. Any dependence $\cos (x) = c \cdot x$ is not possible since the cosine function is not a multiple of the identity function (we are applying Corollary 1.19).
4. By inspection, we spot that there is a dependence. Because $(1 + x)^{2} = x^{2} + 2 x + 1$ , we get that $c_{1} \cdot (1 + x)^{2} + c_{2} \cdot (x^{2} + 2 x) = 3$ is satisfied by $c_{1} = 3$ and $c_{2} = - 3$ .
5. This set is dependent. The easiest way to see that is to recall the trigonometric relationship $\cos^{2} (x) - \sin^{2} (x) = \cos (2 x)$ . (Remark. A person who doesn’t recall this, and tries some $x$ ’s, simply never gets a system leading to a unique solution, and never gets to conclude that the set is independent. Of course, this person might wonder if they simply never tried the right set of $x$ ’s, but a few tries will lead most people to look instead for a dependence.)
6. This set is dependent, because it contains the zero object in the vector space, the zero polynomial.
Exercise 1.26 Worked answer

Does the equation $\sin^{2} (x) / \cos^{2} (x) = \tan^{2} (x)$ show that this set of functions ${\sin^{2} (x), \cos^{2} (x), \tan^{2} (x)}$ is a linearly dependent subset of the set of all real-valued functions with domain the interval $(- π / 2. . π / 2)$ of real numbers between $- π / 2$ and $π / 2)$ ?

Back to Exercise 1.26

Answer. No, that equation is not a linear relationship. In fact this set is independent, as the system arising from taking $x$ to be $0$ , $π / 6$ and $π / 4$ shows.
Exercise 1.27 Worked answer

Is the $x y$ -plane subset of the vector space $ℝ^{3}$ linearly independent?

Back to Exercise 1.27

Answer. No. Here are two members of the plane where the second is a multiple of the first.

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

(Another reason that the answer is “no” is the the zero vector is a member of the plane and no set containing the zero vector is linearly independent.)
Exercise 1.28 Worked answer

Recommended. Show that the nonzero rows of an echelon form matrix form a linearly independent set.

Back to Exercise 1.28

Answer. We have already showed this: the Linear Combination Lemma and its corollary state that in an echelon form matrix, no nonzero row is a linear combination of the others.
Exercise 1.29 Worked answer
1. Show that if the set ${\vec{u}, \vec{v}, \vec{w}}$ is linearly independent then so is the set ${\vec{u}, \vec{u} + \vec{v}, \vec{u} + \vec{v} + \vec{w}}$ .
2. What is the relationship between the linear independence or dependence of ${\vec{u}, \vec{v}, \vec{w}}$ and the independence or dependence of ${\vec{u} - \vec{v}, \vec{v} - \vec{w}, \vec{w} - \vec{u}}$ ?
Back to Exercise 1.29

Answer.
1. Assume that ${\vec{u}, \vec{v}, \vec{w}}$ is linearly independent, so that any relationship $d_{0} \vec{u} + d_{1} \vec{v} + d_{2} \vec{w} = \vec{0}$ leads to the conclusion that $d_{0} = 0$ , $d_{1} = 0$ , and $d_{2} = 0$ .
  
  Consider the relationship $c_{1} (\vec{u}) + c_{2} (\vec{u} + \vec{v}) + c_{3} (\vec{u} + \vec{v} + \vec{w}) = \vec{0}$ . Rewrite it to get $(c_{1} + c_{2} + c_{3}) \vec{u} + (c_{2} + c_{3}) \vec{v} + (c_{3}) \vec{w} = \vec{0}$ . Taking $d_{0}$ to be $c_{1} + c_{2} + c_{3}$ , taking $d_{1}$ to be $c_{2} + c_{3}$ , and taking $d_{2}$ to be $c_{3}$ we have this system.
  
  $\begin{array}{rcrcrcr} c_{1} & + & c_{2} & + & c_{3} & = & 0 \\ c_{2} & + & c_{3} & = & 0 \\ c_{3} & = & 0 \end{array}$
  
  Conclusion: the $c$ ’s are all zero, and so the set is linearly independent.
2. The second set is dependent
  
  $1 \cdot (\vec{u} - \vec{v}) + 1 \cdot (\vec{v} - \vec{w}) + 1 \cdot (\vec{w} - \vec{u}) = \vec{0}$
  
  whether or not the first set is independent.
Exercise 1.30 Worked answer

Example 1.11 shows that the empty set is linearly independent.
1. When is a one-element set linearly independent?
2. How about a set with two elements?
Back to Exercise 1.30

Answer.
1. A singleton set ${\vec{v}}$ is linearly independent if and only if $\vec{v} \neq \vec{0}$ . For the ‘if’ direction, with $\vec{v} \neq \vec{0}$ , we can apply Lemma 1.5 by considering the relationship $c \cdot \vec{v} = \vec{0}$ and noting that the only solution is the trivial one: $c = 0$ . For the ‘only if’ direction, just recall that Example 1.12 shows that ${\vec{0}}$ is linearly dependent, and so if the set ${\vec{v}}$ is linearly independent then $\vec{v} \neq \vec{0}$ .
  
  (Remark. Another answer is to say that this is the special case of Lemma 1.15 where $S = \emptyset$ .)
2. A set with two elements is linearly independent if and only if neither member is a multiple of the other (note that if one is the zero vector then it is a multiple of the other). This is an equivalent statement: a set is linearly dependent if and only if one element is a multiple of the other.
  
  The proof is easy. A set ${{\vec{v}}_{1}, {\vec{v}}_{2}}$ is linearly dependent if and only if there is a relationship $c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} = \vec{0}$ with either $c_{1} \neq 0$ or $c_{2} \neq 0$ (or both). That holds if and only if ${\vec{v}}_{1} = (- c_{2} / c_{1}) {\vec{v}}_{2}$ or ${\vec{v}}_{2} = (- c_{1} / c_{2}) {\vec{v}}_{1}$ (or both).
Exercise 1.31 Worked answer

In any vector space $V$ , the empty set is linearly independent. What about all of $V$ ?

Back to Exercise 1.31

Answer. This set is linearly dependent set because it contains the zero vector.
Exercise 1.32 Worked answer

Show that if ${\vec{x}, \vec{y}, \vec{z}}$ is linearly independent then so are all of its proper subsets: ${\vec{x}, \vec{y}}$ , ${\vec{x}, \vec{z}}$ , ${\vec{y}, \vec{z}}$ , ${\vec{x}}$ , ${\vec{y}}$ , ${\vec{z}}$ , and ${}$ . Is that ‘only if’ also?

Back to Exercise 1.32

Answer. Lemma 1.20 gives the ‘if’ half. The converse (the ‘only if’ statement) does not hold. An example is to consider the vector space $ℝ^{2}$ and these vectors.

$\vec{x} = (\begin{matrix} 1 \\ 0 \end{matrix}), \vec{y} = (\begin{matrix} 0 \\ 1 \end{matrix}), \vec{z} = (\begin{matrix} 1 \\ 1 \end{matrix})$
Exercise 1.33 Worked answer
1. Show that this
  
  $S = {(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 \\ 2 \\ 0 \end{matrix})}$
  
  is a linearly independent subset of $ℝ^{3}$ .
2. Show that
  
  $(\begin{matrix} 3 \\ 2 \\ 0 \end{matrix})$
  
  is in the span of $S$ by finding $c_{1}$ and $c_{2}$ giving a linear relationship.
  
  $c_{1} (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} - 1 \\ 2 \\ 0 \end{matrix}) = (\begin{matrix} 3 \\ 2 \\ 0 \end{matrix})$
  
  Show that the pair $c_{1}, c_{2}$ is unique.
3. Assume that $S$ is a subset of a vector space and that $\vec{v}$ is in $[S]$ , so that $\vec{v}$ is a linear combination of vectors from $S$ . Prove that if $S$ is linearly independent then a linear combination of vectors from $S$ adding to $\vec{v}$ is unique (that is, unique up to reordering and adding or taking away terms of the form $0 \cdot \vec{s}$ ). Thus $S$ as a spanning set is minimal in this strong sense: each vector in $[S]$ is a combination of elements of $S$ a minimum number of times—only once.
4. Prove that it can happen when $S$ is not linearly independent that distinct linear combinations sum to the same vector.
Back to Exercise 1.33

Answer.
1. The linear system arising from
  
  $c_{1} (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} - 1 \\ 2 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
  
  has the unique solution $c_{1} = 0$ and $c_{2} = 0$ .
2. The linear system arising from
  
  $c_{1} (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} - 1 \\ 2 \\ 0 \end{matrix}) = (\begin{matrix} 3 \\ 2 \\ 0 \end{matrix})$
  
  has the unique solution $c_{1} = 8 / 3$ and $c_{2} = - 1 / 3$ .
3. Suppose that $S$ is linearly independent. Suppose that we have both $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n}$ and $\vec{v} = d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m}$ (where the vectors are members of $S$ ). Now,
  
  $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = \vec{v} = d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m}$
  
  can be rewritten in this way.
  
  $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} - d_{1} {\vec{t}}_{1} - \dots - d_{m} {\vec{t}}_{m} = \vec{0}$
  
  Possibly some of the $\vec{s}$ ’s equal some of the $\vec{t}$ ’s; we can combine the associated coefficients (i.e., if ${\vec{s}}_{i} = {\vec{t}}_{j}$ then $\dots + c_{i} {\vec{s}}_{i} + \dots - d_{j} {\vec{t}}_{j} - \dots$ can be rewritten as $\dots + (c_{i} - d_{j}) {\vec{s}}_{i} + \dots$ ). That equation is a linear relationship among distinct (after the combining is done) members of the set $S$ . We’ve assumed that $S$ is linearly independent, so all of the coefficients are zero. If $i$ is such that ${\vec{s}}_{i}$ does not equal any ${\vec{t}}_{j}$ then $c_{i}$ is zero. If $j$ is such that ${\vec{t}}_{j}$ does not equal any ${\vec{s}}_{i}$ then $d_{j}$ is zero. In the final case, we have that $c_{i} - d_{j} = 0$ and so $c_{i} = d_{j}$ .
  
  Therefore, the original two sums are the same, except perhaps for some $0 \cdot {\vec{s}}_{i}$ or $0 \cdot {\vec{t}}_{j}$ terms that we can neglect.
4. This set is not linearly independent:
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 0 \end{matrix})} \subset ℝ^{2}$
  
  and these two linear combinations give the same result
  
  $(\begin{matrix} 0 \\ 0 \end{matrix}) = 2 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) - 1 \cdot (\begin{matrix} 2 \\ 0 \end{matrix}) = 4 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) - 2 \cdot (\begin{matrix} 2 \\ 0 \end{matrix})$
  
  Thus, a linearly dependent set might have indistinct sums.
  
  In fact, this stronger statement holds: if a set is linearly dependent then it must have the property that there are two distinct linear combinations that sum to the same vector. Briefly, where $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = \vec{0}$ then multiplying both sides of the relationship by two gives another relationship. If the first relationship is nontrivial then the second is also.
Exercise 1.34 Worked answer

Prove that a polynomial gives rise to the zero function if and only if it is the zero polynomial. (Comment. This question is not a Linear Algebra matter but we often use the result. A polynomial gives rise to a function in the natural way: $x \mapsto c_{n} x^{n} + \dots + c_{1} x + c_{0}$ .)

Back to Exercise 1.34

Answer. In this ‘if and only if’ statement, the ‘if’ half is clear—if the polynomial is the zero polynomial then the function that arises from the action of the polynomial must be the zero function $x \mapsto 0$ . For ‘only if’ we write $p (x) = c_{n} x^{n} + \dots + c_{0}$ . Plugging in zero $p (0) = 0$ gives that $c_{0} = 0$ . Taking the derivative and plugging in zero $p^{'} (0) = 0$ gives that $c_{1} = 0$ . Similarly we get that each $c_{i}$ is zero, and $p$ is the zero polynomial.
Exercise 1.35 Worked answer

Return to Section 1.2 and redefine point, line, plane, and other linear surfaces to avoid degenerate cases.

Back to Exercise 1.35

Answer. The work in this section suggests that we should define an $n$ -dimensional non-degenerate linear surface as the span of a linearly independent set of $n$ vectors.
Exercise 1.36 Worked answer
1. Show that any set of four vectors in $ℝ^{2}$ is linearly dependent.
2. Is this true for any set of five? Any set of three?
3. What is the most number of elements that a linearly independent subset of $ℝ^{2}$ can have?
Back to Exercise 1.36

Answer.
1. For any $a_{1, 1}$ , …, $a_{2, 4}$ ,
  
  $c_{1} (\begin{matrix} a_{1, 1} \\ a_{2, 1} \end{matrix}) + c_{2} (\begin{matrix} a_{1, 2} \\ a_{2, 2} \end{matrix}) + c_{3} (\begin{matrix} a_{1, 3} \\ a_{2, 3} \end{matrix}) + c_{4} (\begin{matrix} a_{1, 4} \\ a_{2, 4} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$
  
  yields a linear system
  
  $\begin{array}{rcrcrcrcr} a_{1, 1} c_{1} & + & a_{1, 2} c_{2} & + & a_{1, 3} c_{3} & + & a_{1, 4} c_{4} & = & 0 \\ a_{2, 1} c_{1} & + & a_{2, 2} c_{2} & + & a_{2, 3} c_{3} & + & a_{2, 4} c_{4} & = & 0 \end{array}$
  
  that has infinitely many solutions (Gauss’s Method leaves at least two variables free). Hence there are nontrivial linear relationships among the given members of $ℝ^{2}$ .
2. Any set five vectors is a superset of a set of four vectors, and so is linearly dependent.
  
  With three vectors from $ℝ^{2}$ , the argument from the prior item still applies, with the slight change that Gauss’s Method now only leaves at least one variable free (but that still gives infinitely many solutions).
3. The prior item shows that no three-element subset of $ℝ^{2}$ is independent. We know that there are two-element subsets of $ℝ^{2}$ that are independent—one is
  
  ${(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}$
  
  and so the answer is two.
Exercise 1.37 Worked answer

Is there a set of four vectors in $ℝ^{3}$ such that any three form a linearly independent set?

Back to Exercise 1.37

Answer. Yes; here is one.

${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})}$
Exercise 1.38 Worked answer

Must every linearly dependent set have a subset that is dependent and a subset that is independent?

Back to Exercise 1.38

Answer. Yes. Fix any vector space $V$ , and let $S$ be a linearly dependent subset of $V$ . For a subset of $S$ that is dependent we can take $S$ itself. For a subset of $S$ that is independent we can take the empty set.
Exercise 1.39 Worked answer

In $ℝ^{4}$ what is the biggest linearly independent set you can find? The smallest? The biggest linearly dependent set? The smallest? (‘Biggest’ and ‘smallest’ mean that there are no supersets or subsets with the same property.)

Back to Exercise 1.39

Answer. In $ℝ^{4}$ the biggest linearly independent set has four vectors. There are many examples of such sets, this is one.

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

To see that no set with five or more vectors can be independent, set up

$c_{1} (\begin{matrix} a_{1, 1} \\ a_{2, 1} \\ a_{3, 1} \\ a_{4, 1} \end{matrix}) + c_{2} (\begin{matrix} a_{1, 2} \\ a_{2, 2} \\ a_{3, 2} \\ a_{4, 2} \end{matrix}) + c_{3} (\begin{matrix} a_{1, 3} \\ a_{2, 3} \\ a_{3, 3} \\ a_{4, 3} \end{matrix}) + c_{4} (\begin{matrix} a_{1, 4} \\ a_{2, 4} \\ a_{3, 4} \\ a_{4, 4} \end{matrix}) + c_{5} (\begin{matrix} a_{1, 5} \\ a_{2, 5} \\ a_{3, 5} \\ a_{4, 5} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix})$

and note that the resulting linear system

$\begin{array}{rcrcrcrcrcr} a_{1, 1} c_{1} & + & a_{1, 2} c_{2} & + & a_{1, 3} c_{3} & + & a_{1, 4} c_{4} & + & a_{1, 5} c_{5} & = & 0 \\ a_{2, 1} c_{1} & + & a_{2, 2} c_{2} & + & a_{2, 3} c_{3} & + & a_{2, 4} c_{4} & + & a_{2, 5} c_{5} & = & 0 \\ a_{3, 1} c_{1} & + & a_{3, 2} c_{2} & + & a_{3, 3} c_{3} & + & a_{3, 4} c_{4} & + & a_{3, 5} c_{5} & = & 0 \\ a_{4, 1} c_{1} & + & a_{4, 2} c_{2} & + & a_{4, 3} c_{3} & + & a_{4, 4} c_{4} & + & a_{4, 5} c_{5} & = & 0 \end{array}$

has four equations and five unknowns, so Gauss’s Method must end with at least one $c$ variable free, so there are infinitely many solutions, and so the above linear relationship among the four-tall vectors has more solutions than just the trivial solution.

The smallest linearly independent set is the empty set.

The biggest linearly dependent set is $ℝ^{4}$ . The smallest is ${\vec{0}}$ .
Exercise 1.40 Worked answer

Recommended. Linear independence and linear dependence are properties of sets. We can thus naturally ask how the properties of linear independence and dependence act with respect to the familiar elementary set relations and operations. In this body of this subsection we have covered the subset and superset relations. We can also consider the operations of intersection, complementation, and union.
1. How does linear independence relate to intersection: can an intersection of linearly independent sets be independent? Must it be?
2. How does linear independence relate to complementation?
3. Show that the union of two linearly independent sets can be linearly independent.
4. Show that the union of two linearly independent sets need not be linearly independent.
Back to Exercise 1.40

Answer.
1. The intersection of two linearly independent sets $S \cap T$ must be linearly independent as it is a subset of the linearly independent set $S$ (as well as the linearly independent set $T$ also, of course).
2. The complement of a linearly independent set is linearly dependent as it contains the zero vector.
3. A simple example in $ℝ^{2}$ is these two sets.
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix})} T = {(\begin{matrix} 0 \\ 1 \end{matrix})}$
  
  A somewhat subtler example, again in $ℝ^{2}$ , is these two.
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix})} T = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}$
4. We must produce an example. One, in $ℝ^{2}$ , is
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix})} T = {(\begin{matrix} 2 \\ 0 \end{matrix})}$
  
  since the linear dependence of $S_{1} \cup S_{2}$ is easy to see.
Exercise 1.41 Worked answer

Continued from prior exercise. What is the interaction between the property of linear independence and the operation of union?
1. We might conjecture that the union $S \cup T$ of linearly independent sets is linearly independent if and only if their spans have a trivial intersection $[S] \cap [T] = {\vec{0}}$ . What is wrong with this argument for the ‘if’ direction of that conjecture? “If the union $S \cup T$ is linearly independent then the only solution to $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} + d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m} = \vec{0}$ is the trivial one $c_{1} = 0$ , …, $d_{m} = 0$ . So any member of the intersection of the spans must be the zero vector because in $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m}$ each scalar is zero.”
2. Give an example showing that the conjecture is false.
3. Find linearly independent sets $S$ and $T$ so that the union of $S - (S \cap T)$ and $T - (S \cap T)$ is linearly independent, but the union $S \cup T$ is not linearly independent.
4. Characterize when the union of two linearly independent sets is linearly independent, in terms of the intersection of spans.
Back to Exercise 1.41

Answer.
1. Lemma 1.5 requires that the vectors ${\vec{s}}_{1}, \dots, {\vec{s}}_{n}, {\vec{t}}_{1}, \dots, {\vec{t}}_{m}$ be distinct. But we could have that the union $S \cup T$ is linearly independent with some ${\vec{s}}_{i}$ equal to some ${\vec{t}}_{j}$ .
2. One example in $ℝ^{2}$ is these two.
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix})} T = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}$
3. An example from $ℝ^{2}$ is these sets.
  
  $S = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})} T = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix})}$
4. The union of two linearly independent sets $S \cup T$ is linearly independent if and only if their spans of $S$ and $T - (S \cap T)$ have a trivial intersection $[S] \cap [T - (S \cap T)] = {\vec{0}}$ . To prove that, assume that $S$ and $T$ are linearly independent subsets of some vector space.
  
  For the ‘only if’ direction, assume that the intersection of the spans is trivial $[S] \cap [T - (S \cap T)] = {\vec{0}}$ . Consider the set $S \cup (T - (S \cap T)) = S \cup T$ and consider the linear relationship $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} + d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m} = \vec{0}$ . Subtracting gives $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = - d_{1} {\vec{t}}_{1} - \dots - d_{m} {\vec{t}}_{m}$ . The left side of that equation sums to a vector in $[S]$ , and the right side is a vector in $[T - (S \cap T)]$ . Therefore, since the intersection of the spans is trivial, both sides equal the zero vector. Because $S$ is linearly independent, all of the $c$ ’s are zero. Because $T$ is linearly independent so also is $T - (S \cap T)$ linearly independent, and therefore all of the $d$ ’s are zero. Thus, the original linear relationship among members of $S \cup T$ only holds if all of the coefficients are zero. Hence, $S \cup T$ is linearly independent.
  
  For the ‘if’ half we can make the same argument in reverse. Suppose that the union $S \cup T$ is linearly independent. Consider a linear relationship among members of $S$ and $T - (S \cap T)$ . $c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} + d_{1} {\vec{t}}_{1} + \dots + d_{m} {\vec{t}}_{m} = \vec{0}$ Note that no ${\vec{s}}_{i}$ is equal to a ${\vec{t}}_{j}$ so that is a combination of distinct vectors, as required by Lemma 1.5. So the only solution is the trivial one $c_{1} = 0$ , …, $d_{m} = 0$ . Since any vector $\vec{v}$ in the intersection of the spans $[S] \cap [T - (S \cap T)]$ we can write $\vec{v} = c_{1} {\vec{s}}_{1} + \dots + c_{n} {\vec{s}}_{n} = - d_{1} {\vec{t}}_{1} - \dots - d_{m} {\vec{t}}_{m}$ , and it must be the zero vector because each scalar is zero.
Exercise 1.42 Worked answer

For Corollary 1.17,
1. fill in the induction for the proof;
2. give an alternate proof that starts with the empty set and builds a sequence of linearly independent subsets of the given finite set until one appears with the same span as the given set.
Back to Exercise 1.42

Answer.
1. We do induction on the number of vectors in the finite set $S$ .
  
  The base case is that $S$ has no elements. In this case $S$ is linearly independent and there is nothing to check—a subset of $S$ that has the same span as $S$ is $S$ itself.
  
  For the inductive step assume that the theorem is true for all sets of size $n = 0$ , $n = 1$ , …, $n = k$ in order to prove that it holds when $S$ has $n = k + 1$ elements. If the $k + 1$ -element set $S = {{\vec{s}}_{0}, \dots, {\vec{s}}_{k}}$ is linearly independent then the theorem is trivial, so assume that it is dependent. By Corollary 1.19 there is an ${\vec{s}}_{i}$ that is a linear combination of other vectors in $S$ . Define $S_{1} = S - {{\vec{s}}_{i}}$ and note that $S_{1}$ has the same span as $S$ by Corollary 1.3. The set $S_{1}$ has $k$ elements and so the inductive hypothesis applies to give that it has a linearly independent subset with the same span. That subset of $S_{1}$ is the desired subset of $S$ .
2. Here is a sketch of the argument. We have left out the induction argument details.
  
  If the finite set $S$ is empty then there is nothing to prove. If $S = {\vec{0}}$ then the empty subset will do.
  
  Otherwise, take some nonzero vector ${\vec{s}}_{1} \in S$ and define $S_{1} = {{\vec{s}}_{1}}$ . If $[S_{1}] = [S]$ then we are finished with this proof by noting that $S_{1}$ is linearly independent.
  
  If not, then there is a nonzero vector ${\vec{s}}_{2} \in S - [S_{1}]$ (if every $\vec{s} \in S$ is in $[S_{1}]$ then $[S_{1}] = [S]$ ). Define $S_{2} = S_{1} \cup {{\vec{s}}_{2}}$ . If $[S_{2}] = [S]$ then we are finished by using Theorem 1.19 to show that $S_{2}$ is linearly independent.
  
  Repeat the last paragraph until a set with a big enough span appears. That must eventually happen because $S$ is finite, and $[S]$ will be reached at worst when we have used every vector from $S$ .
Exercise 1.43 Worked answer

With a some calculation we can get formulas to determine whether or not a set of vectors is linearly independent.
1. Show that this subset of $ℝ^{2}$
  
  ${(\begin{matrix} a \\ c \end{matrix}), (\begin{matrix} b \\ d \end{matrix})}$
  
  is linearly independent if and only if $a d - b c \neq 0$ .
2. Show that this subset of $ℝ^{3}$
  
  ${(\begin{matrix} a \\ d \\ g \end{matrix}), (\begin{matrix} b \\ e \\ h \end{matrix}), (\begin{matrix} c \\ f \\ i \end{matrix})}$
  
  is linearly independent iff $a e i + b f g + c d h - h f a - i d b - g e c \neq 0$ .
3. When is this subset of $ℝ^{3}$
  
  ${(\begin{matrix} a \\ d \\ g \end{matrix}), (\begin{matrix} b \\ e \\ h \end{matrix})}$
  
  linearly independent?
4. This is an opinion question: for a set of four vectors from $ℝ^{4}$ , must there be a formula involving the sixteen entries that determines independence of the set? (You needn’t produce such a formula, just decide if one exists.)
Back to Exercise 1.43

Answer.
1. Assuming first that $a \neq 0$ ,
  
  $x (\begin{matrix} a \\ c \end{matrix}) + y (\begin{matrix} b \\ d \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$
  
  gives
  
  $\begin{array}{rcrcr} a x & + & b y & = & 0 \\ c x & + & d y & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (c / a) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} a x & + & b y & = & 0 \\ (- (c / a) b + d) y & = & 0 \end{array}$
  
  which has a solution if and only if $0 \neq - (c / a) b + d = (- c b + a d) / d$ (we’ve assumed in this case that $a \neq 0$ , and so back substitution yields a unique solution).
  
  The $a = 0$ case is also not hard—break it into the $c \neq 0$ and $c = 0$ subcases and note that in these cases $a d - b c = 0 \cdot d - b c$ .
  
  Comment. An earlier exercise showed that a two-vector set is linearly dependent if and only if either vector is a scalar multiple of the other. We could also use that to make the calculation.
2. The equation
  
  $c_{1} (\begin{matrix} a \\ d \\ g \end{matrix}) + c_{2} (\begin{matrix} b \\ e \\ h \end{matrix}) + c_{3} (\begin{matrix} c \\ f \\ i \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$
  
  expresses a homogeneous linear system. We proceed by writing it in matrix form and applying Gauss’s Method.
  
  We first reduce the matrix to upper-triangular. Assume that $a \neq 0$ . With that, we can clear down the first column.
  
  $⟶_{\begin{array}{l} \end{array}}^{(1 / a) ρ_{1}} (\begin{array}{cccc} 1 & b / a & c / a & 0 \\ d & e & f & 0 \\ g & h & i & 0 \end{array}) ⟶_{\begin{array}{l} - g ρ_{1} + ρ_{3} \end{array}}^{- d ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & b / a & c / a & 0 \\ 0 & (a e - b d) / a & (a f - c d) / a & 0 \\ 0 & (a h - b g) / a & (a i - c g) / a & 0 \end{array})$
  
  Then we get a $1$ in the second row, second column entry. (Assuming for the moment that $a e - b d \neq 0$ , in order to do the row reduction step.)
  
  $⟶_{\begin{array}{l} \end{array}}^{(a / (a e - b d)) ρ_{2}} (\begin{array}{cccc} 1 & b / a & c / a & 0 \\ 0 & 1 & (a f - c d) / (a e - b d) & 0 \\ 0 & (a h - b g) / a & (a i - c g) / a & 0 \end{array})$
  
  Then, under the assumptions, we perform the row operation $((a h - b g) / a) ρ_{2} + ρ_{3}$ to get this.
  
  $(\begin{array}{cccc} 1 & b / a & c / a & 0 \\ 0 & 1 & (a f - c d) / (a e - b d) & 0 \\ 0 & 0 & (a e i + b g f + c d h - h f a - i d b - g e c) / (a e - b d) & 0 \end{array})$
  
  Therefore, the original system is nonsingular if and only if the above $3, 3$ entry is nonzero (this fraction is defined because of the $a e - b d \neq 0$ assumption). It equals zero if and only if the numerator is zero.
  
  We next worry about the assumptions. First, if $a \neq 0$ but $a e - b d = 0$ then we swap
  
  $\begin{matrix} (\begin{array}{cccc} 1 & b / a & c / a & 0 \\ 0 & 0 & (a f - c d) / a & 0 \\ 0 & (a h - b g) / a & (a i - c g) / a & 0 \end{array}) \\ ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} \leftrightarrow ρ_{3}} (\begin{array}{cccc} 1 & b / a & c / a & 0 \\ 0 & (a h - b g) / a & (a i - c g) / a & 0 \\ 0 & 0 & (a f - c d) / a & 0 \end{array}) \end{matrix}$
  
  and conclude that the system is nonsingular if and only if either $a h - b g = 0$ or $a f - c d = 0$ . That’s the same as asking that their product be zero:
  
  $\begin{aligned} a h a f - a h c d - b g a f + b g c d & = 0 \\ a h a f - a h c d - b g a f + a e g c & = 0 \\ a (h a f - h c d - b g f + e g c) & = 0 \end{aligned}$
  
  (in going from the first line to the second we’ve applied the case assumption that $a e - b d = 0$ by substituting $a e$ for $b d$ ). Since we are assuming that $a \neq 0$ , we have that $h a f - h c d - b g f + e g c = 0$ . With $a e - b d = 0$ we can rewrite this to fit the form we need: in this $a \neq 0$ and $a e - b d = 0$ case, the given system is nonsingular when $h a f - h c d - b g f + e g c - i (a e - b d) = 0$ , as required.
  
  The remaining cases have the same character. Do the $a = 0$ but $d \neq 0$ case and the $a = 0$ and $d = 0$ but $g \neq 0$ case by first swapping rows and then going on as above. The $a = 0$ , $d = 0$ , and $g = 0$ case is easy—a set with a zero vector is linearly dependent, and the formula comes out to equal zero.
3. It is linearly dependent if and only if either vector is a multiple of the other. That is, it is not independent iff
  
  $(\begin{matrix} a \\ d \\ g \end{matrix}) = r \cdot (\begin{matrix} b \\ e \\ h \end{matrix}) or (\begin{matrix} b \\ e \\ h \end{matrix}) = s \cdot (\begin{matrix} a \\ d \\ g \end{matrix})$
  
  (or both) for some scalars $r$ and $s$ . Eliminating $r$ and $s$ in order to restate this condition only in terms of the given letters $a$ , $b$ , $d$ , $e$ , $g$ , $h$ , we have that it is not independent—it is dependent—iff $a e - b d = a h - g b = d h - g e$ .
4. Dependence or independence is a function of the indices, so there is indeed a formula (although at first glance a person might think the formula involves cases: “if the first component of the first vector is zero then …”, this guess turns out not to be correct).
Exercise 1.44 Worked answer

Recommended.
1. Prove that a set of two perpendicular nonzero vectors from $ℝ^{n}$ is linearly independent when $n > 1$ .
2. What if $n = 1$ ? $n = 0$ ?
3. Generalize to more than two vectors.
Back to Exercise 1.44

Answer. Recall that two vectors from $ℝ^{n}$ are perpendicular if and only if their dot product is zero.
1. Assume that $\vec{v}$ and $\vec{w}$ are perpendicular nonzero vectors in $ℝ^{n}$ , with $n > 1$ . With the linear relationship $c \vec{v} + d \vec{w} = \vec{0}$ , apply $\vec{v}$ to both sides to conclude that $c \cdot ‖ \vec{v} ‖^{2} + d \cdot 0 = 0$ . Because $\vec{v} \neq \vec{0}$ we have that $c = 0$ . A similar application of $\vec{w}$ shows that $d = 0$ .
2. Two vectors in $ℝ^{1}$ are perpendicular if and only if at least one of them is zero.
  
  We define $ℝ^{0}$ to be a trivial space, and so both $\vec{v}$ and $\vec{w}$ are the zero vector.
3. The right generalization is to look at a set ${{\vec{v}}_{1}, \dots, {\vec{v}}_{n}} \subseteq ℝ^{k}$ of vectors that are mutually orthogonal (also called pairwise perpendicular): if $i \neq j$ then ${\vec{v}}_{i}$ is perpendicular to ${\vec{v}}_{j}$ . Mimicking the proof of the first item above shows that such a set of nonzero vectors is linearly independent.
Exercise 1.45 Worked answer

Consider the set of functions from the interval $(- 1 \dots 1) \subseteq ℝ$ to $ℝ$ .
1. Show that this set is a vector space under the usual operations.
2. Recall the formula for the sum of an infinite geometric series: $1 + x + x^{2} + \dots = 1 / (1 - x)$ for all $x \in (- 1. .1)$ . Why does this not express a dependence inside of the set ${g (x) = 1 / (1 - x), f_{0} (x) = 1, f_{1} (x) = x, f_{2} (x) = x^{2}, \dots}$ (in the vector space that we are considering)? (Hint. Review the definition of linear combination.)
3. Show that the set in the prior item is linearly independent.
This shows that some vector spaces exist with linearly independent subsets that are infinite.
Back to Exercise 1.45

Answer.
1. This check is routine.
2. The summation is infinite (has infinitely many summands). The definition of linear combination involves only finite sums.
3. No nontrivial finite sum of members of ${g, f_{0}, f_{1}, \dots}$ adds to the zero object: assume that
  
  $c_{0} \cdot (1 / (1 - x)) + c_{1} \cdot 1 + \dots + c_{n} \cdot x^{n} = 0$
  
  (any finite sum uses a highest power, here $n$ ). Multiply both sides by $1 - x$ to conclude that each coefficient is zero, because a polynomial describes the zero function only when it is the zero polynomial.
Exercise 1.46 Worked answer

Show that, where $S$ is a subspace of $V$ , if a subset $T$ of $S$ is linearly independent in $S$ then $T$ is also linearly independent in $V$ . Is that ‘only if’?

Back to Exercise 1.46

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

Let $T$ be a subset of the subspace $S$ of the vector space $V$ . The assertion that any linear relationship $c_{1} {\vec{t}}_{1} + \dots + c_{n} {\vec{t}}_{n} = \vec{0}$ among members of $T$ must be the trivial relationship $c_{1} = 0$ , …, $c_{n} = 0$ is a statement that holds in $S$ if and only if it holds in $V$ , because the subspace $S$ inherits its addition and scalar multiplication operations from $V$ .

References cited in this section

See also Remark 1.13.↩︎