Maps Between Spaces

Isomorphisms

In the examples after the definition of a vector space we expressed the intuition that some spaces are essentially the same as others. For instance, we may perceive that the space of two-by-two matrices and the space of four tall vectors

$ℳ_{2 \times 2} = {(\begin{matrix} a & b \\ c & d \end{matrix}) ∣ a, b, c, d \in ℝ} ℝ^{4} = {(\begin{matrix} a \\ b \\ c \\ d \end{matrix}) ∣ a, b, c, d \in ℝ}$

are not equal because their elements are not the same, but that they differ only cosmetically. We will now make this precise.

This illustrates a common phase of a mathematical investigation. With the help of some examples we’ve gotten an idea. We will next give a formal definition and then produce some results backing our contention that the definition captures the idea. We’ve seen this happen already, for instance in the first section of the Vector Space chapter. There, the study of linear systems led us to consider collections closed under linear combinations. We defined such a collection as a vector space and we followed it with some supporting results.

That wasn’t an end point, instead it led to new insights such as the idea of a basis. Here also, after producing a definition and supporting it, we will get two pleasant surprises. First, we will find that the definition applies to some unforeseen, and interesting, cases. Second, the study of the definition will lead to new ideas. In this way, our investigation will build momentum.

Definition and Examples

We start with two examples that suggest the right definition.

Example 1.1 The space of two-wide row vectors and the space of two-tall column vectors are “the same” in that if we associate the vectors that have the same components, e.g.,

$(\begin{matrix} 1 & 2 \end{matrix}) ⟷ (\begin{matrix} 1 \\ 2 \end{matrix})$

(read the double arrow as “corresponds to”) then this association respects the operations. For instance these corresponding vectors add to corresponding totals

$(\begin{matrix} 1 & 2 \end{matrix}) + (\begin{matrix} 3 & 4 \end{matrix}) = (\begin{matrix} 4 & 6 \end{matrix}) ⟷ (\begin{matrix} 1 \\ 2 \end{matrix}) + (\begin{matrix} 3 \\ 4 \end{matrix}) = (\begin{matrix} 4 \\ 6 \end{matrix})$

and here is an example of the correspondence respecting scalar multiplication.

$5 \cdot (\begin{matrix} 1 & 2 \end{matrix}) = (\begin{matrix} 5 & 10 \end{matrix}) ⟷ 5 \cdot (\begin{matrix} 1 \\ 2 \end{matrix}) = (\begin{matrix} 5 \\ 10 \end{matrix})$

Stated generally, under the correspondence

$(\begin{matrix} a_{0} & a_{1} \end{matrix}) ⟷ (\begin{matrix} a_{0} \\ a_{1} \end{matrix})$

both operations are preserved:

$(\begin{matrix} a_{0} & a_{1} \end{matrix}) + (\begin{matrix} b_{0} & b_{1} \end{matrix}) = (\begin{matrix} a_{0} + b_{0} & a_{1} + b_{1} \end{matrix}) ⟷ (\begin{matrix} a_{0} \\ a_{1} \end{matrix}) + (\begin{matrix} b_{0} \\ b_{1} \end{matrix}) = (\begin{matrix} a_{0} + b_{0} \\ a_{1} + b_{1} \end{matrix})$

and

$r \cdot (\begin{matrix} a_{0} & a_{1} \end{matrix}) = (\begin{matrix} r a_{0} & r a_{1} \end{matrix}) ⟷ r \cdot (\begin{matrix} a_{0} \\ a_{1} \end{matrix}) = (\begin{matrix} r a_{0} \\ r a_{1} \end{matrix})$

(all of the variables are scalars).

Example 1.2 Another two spaces that we can think of as “the same” are $𝒫_{2}$ , the space of quadratic polynomials, and $ℝ^{3}$ . A natural correspondence is this.

$\begin{matrix} a_{0} + a_{1} x + a_{2} x^{2} ⟷ (\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \end{matrix}) & (e.g., 1 + 2 x + 3 x^{2} ⟷ (\begin{array}{c} 1 \\ 2 \\ 3 \end{array})) \end{matrix}$

This preserves structure: corresponding elements add in a corresponding way

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

and scalar multiplication corresponds also.

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

Definition 1.3 An isomorphism between two vector spaces $V$ and $W$ is a map $f : V \to W$ that

is a correspondence: $f$ is one-to-one and onto; ¹
preserves structure: if ${\vec{v}}_{1}, {\vec{v}}_{2} \in V$ then

$f ({\vec{v}}_{1} + {\vec{v}}_{2}) = f ({\vec{v}}_{1}) + f ({\vec{v}}_{2})$

and if $\vec{v} \in V$ and $r \in ℝ$ then

$f (r \vec{v}) = r f (\vec{v})$

(we write $V ≅ W$ , read “ $V$ is isomorphic to $W$ ”, when such a map exists).

“Morphism” means map, so “isomorphism” means a map expressing sameness.

Example 1.4 The vector space $G = {c_{1} \cos θ + c_{2} \sin θ ∣ c_{1}, c_{2} \in ℝ}$ of functions of $θ$ is isomorphic to $ℝ^{2}$ under this map.

$c_{1} \cos θ + c_{2} \sin θ \overset{f}{⟼} (\begin{matrix} c_{1} \\ c_{2} \end{matrix})$

We will check this by going through the conditions in the definition. We will first verify condition (1), that the map is a correspondence between the sets underlying the spaces.

To establish that $f$ is one-to-one we must prove that $f (\vec{a}) = f (\vec{b})$ only when $\vec{a} = \vec{b}$ . If

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

then by the definition of $f$

$(\begin{matrix} a_{1} \\ a_{2} \end{matrix}) = (\begin{matrix} b_{1} \\ b_{2} \end{matrix})$

from which we conclude that $a_{1} = b_{1}$ and $a_{2} = b_{2}$ , because column vectors are equal only when they have equal components. Thus $a_{1} \cos θ + a_{2} \sin θ = b_{1} \cos θ + b_{2} \sin θ$ , and as required we’ve verified that $f (\vec{a}) = f (\vec{b})$ implies that $\vec{a} = \vec{b}$ .

To prove that $f$ is onto we must check that any member of the codomain $ℝ^{2}$ is the image of some member of the domain $G$ . So, consider a member of the codomain

$(\begin{matrix} x \\ y \end{matrix})$

and note that it is the image under $f$ of $x \cos θ + y \sin θ$ .

Next we will verify condition (2), that $f$ preserves structure. This computation shows that $f$ preserves addition.

$\begin{matrix} f ((a_{1} \cos θ + a_{2} \sin θ) + (b_{1} \cos θ + b_{2} \sin θ)) \\ \begin{aligned} = f ((a_{1} + b_{1}) \cos θ + (a_{2} + b_{2}) \sin θ) \\ = (\begin{array}{c} a_{1} + b_{1} \\ a_{2} + b_{2} \end{array}) \\ = (\begin{array}{c} a_{1} \\ a_{2} \end{array}) + (\begin{array}{c} b_{1} \\ b_{2} \end{array}) \\ = f (a_{1} \cos θ + a_{2} \sin θ) + f (b_{1} \cos θ + b_{2} \sin θ) \end{aligned} \end{matrix}$

The computation showing that $f$ preserves scalar multiplication is similar.

$\begin{aligned} f (r \cdot (a_{1} \cos θ + a_{2} \sin θ)) & = f (r a_{1} \cos θ + r a_{2} \sin θ) \\ = (\begin{array}{c} r a_{1} \\ r a_{2} \end{array}) \\ = r \cdot (\begin{array}{c} a_{1} \\ a_{2} \end{array}) \\ = r \cdot f (a_{1} \cos θ + a_{2} \sin θ) \end{aligned}$

With both (1) and (2) verified, we know that $f$ is an isomorphism and we can say that the spaces are isomorphic $G ≅ ℝ^{2}$ .

Example 1.5 Let $V$ be the space ${c_{1} x + c_{2} y + c_{3} z ∣ c_{1}, c_{2}, c_{3} \in ℝ}$ of linear combinations of the three variables under the natural addition and scalar multiplication operations. Then $V$ is isomorphic to $𝒫_{2}$ , the space of quadratic polynomials.

To show this we must produce an isomorphism map. There is more than one possibility; for instance, here are four to choose among.

$c_{1} x + c_{2} y + c_{3} z$

$\overset{f_{1}}{⟼}$	$c_{1} + c_{2} x + c_{3} x^{2}$
$\overset{f_{2}}{⟼}$	$c_{2} + c_{3} x + c_{1} x^{2}$
$\overset{f_{3}}{⟼}$	$- c_{1} - c_{2} x - c_{3} x^{2}$
$\overset{f_{4}}{⟼}$	$c_{1} + (c_{1} + c_{2}) x + (c_{1} + c_{3}) x^{2}$

The first map is the more natural correspondence in that it just carries the coefficients over. However we shall do $f_{2}$ to underline that there are isomorphisms other than the obvious one. (Checking that $f_{1}$ is an isomorphism is Exercise 1.14.)

To show that $f_{2}$ is one-to-one we will prove that if $f_{2} (c_{1} x + c_{2} y + c_{3} z) = f_{2} (d_{1} x + d_{2} y + d_{3} z)$ then $c_{1} x + c_{2} y + c_{3} z = d_{1} x + d_{2} y + d_{3} z$ . The assumption that $f_{2} (c_{1} x + c_{2} y + c_{3} z) = f_{2} (d_{1} x + d_{2} y + d_{3} z)$ gives, by the definition of $f_{2}$ , that $c_{2} + c_{3} x + c_{1} x^{2} = d_{2} + d_{3} x + d_{1} x^{2}$ . Equal polynomials have equal coefficients so $c_{2} = d_{2}$ , $c_{3} = d_{3}$ , and $c_{1} = d_{1}$ . Hence $f_{2} (c_{1} x + c_{2} y + c_{3} z) = f_{2} (d_{1} x + d_{2} y + d_{3} z)$ implies that $c_{1} x + c_{2} y + c_{3} z = d_{1} x + d_{2} y + d_{3} z$ , and $f_{2}$ is one-to-one.

The map $f_{2}$ is onto because a member $a + b x + c x^{2}$ of the codomain is the image of a member of the domain, namely it is $f_{2} (c x + a y + b z)$ . For instance, $2 + 3 x - 4 x^{2}$ is $f_{2} (- 4 x + 2 y + 3 z)$ .

The computations for structure preservation are like those in the prior example. The map $f_{2}$ preserves addition

$\begin{matrix} f_{2} ((c_{1} x + c_{2} y + c_{3} z) + (d_{1} x + d_{2} y + d_{3} z)) \\ \begin{aligned} = f_{2} ((c_{1} + d_{1}) x + (c_{2} + d_{2}) y + (c_{3} + d_{3}) z) \\ = (c_{2} + d_{2}) + (c_{3} + d_{3}) x + (c_{1} + d_{1}) x^{2} \\ = (c_{2} + c_{3} x + c_{1} x^{2}) + (d_{2} + d_{3} x + d_{1} x^{2}) \\ = f_{2} (c_{1} x + c_{2} y + c_{3} z) + f_{2} (d_{1} x + d_{2} y + d_{3} z) \end{aligned} \end{matrix}$

and scalar multiplication.

$\begin{aligned} f_{2} (r \cdot (c_{1} x + c_{2} y + c_{3} z)) & = f_{2} (r c_{1} x + r c_{2} y + r c_{3} z) \\ = r c_{2} + r c_{3} x + r c_{1} x^{2} \\ = r \cdot (c_{2} + c_{3} x + c_{1} x^{2}) \\ = r \cdot f_{2} (c_{1} x + c_{2} y + c_{3} z) \end{aligned}$

Thus $f_{2}$ is an isomorphism. We write $V ≅ 𝒫_{2}$ .

Example 1.6 Every space is isomorphic to itself under the identity map. The check is easy.

Definition 1.7 An automorphism is an isomorphism of a space with itself.

Example 1.8 A dilation map $d_{s} : ℝ^{2} \to ℝ^{2}$ that multiplies all vectors by a nonzero scalar $s$ is an automorphism of $ℝ^{2}$ .

Dilation by 1.5: vectors u=(2,2) and v=(2,1) in the left axes correspond to (3,3) and (3,1.5) in the right axes. The map and both image vectors are labelled.

Another automorphism is a rotation or turning map, $t_{θ} : ℝ^{2} \to ℝ^{2}$ that rotates all vectors through an angle $θ$ .

Counterclockwise rotation through pi/6: the vector u=(3,1) in the left axes maps to the rotated vector in the right axes. Both the arrow between the axes and the image vector carry the rotation label.

A third type of automorphism of $ℝ^{2}$ is a map $f_{ℓ} : ℝ^{2} \to ℝ^{2}$ that flips or reflects all vectors over a line $ℓ$ through the origin.

Reflection across the dotted line y=x: the vector u=(3,2) in the left axes corresponds to (2,3) in the right axes. The line is labelled ell and the map is labelled f_ell.

Checking that these are automorphisms is Exercise 1.33.

Example 1.9 Consider the space $𝒫_{5}$ of polynomials of degree $5$ or less and the map $f$ that sends a polynomial $p (x)$ to $p (x - 1)$ . For instance, under this map $x^{2} \mapsto (x - 1)^{2} = x^{2} - 2 x + 1$ and $x^{3} + 2 x \mapsto (x - 1)^{3} + 2 (x - 1) = x^{3} - 3 x^{2} + 5 x - 3$ . This map is an automorphism of this space; the check is Exercise 1.25.

This isomorphism of $𝒫_{5}$ with itself does more than just tell us that the space is “the same” as itself. It gives us some insight into the space’s structure. Below is a family of parabolas, graphs of members of $𝒫_{5}$ . Each has a vertex at $y = - 1$ , and the left-most one has zeroes at $- 2.25$ and $- 1.75$ , the next one has zeroes at $- 1.25$ and $- 0.75$ , etc.

Five parabolas with vertices at height minus one and successive integer horizontal positions. The middle and next parabola are labelled p0 and p1; replacing x by x minus one shifts each graph right by one.

Substitution of $x - 1$ for $x$ in any function’s argument shifts its graph to the right by one. Thus, $f (p_{0}) = p_{1}$ , and $f$ ’s action is to shift all of the parabolas to the right by one. Notice that the picture before $f$ is applied is the same as the picture after $f$ is applied because while each parabola moves to the right, another one comes in from the left to take its place. This also holds true for cubics, etc. So the automorphism $f$ expresses the idea that $P_{5}$ has a certain horizontal-homogeneity: if we draw two pictures showing all members of $𝒫_{5}$ , one picture centered at $x = 0$ and the other centered at $x = 1$ , then the two pictures would be indistinguishable.

As described in the opening to this section, having given the definition of isomorphism, we next look to support the thesis that it captures our intuition of vector spaces being the same. First, the definition itself is persuasive: a vector space consists of a set and some structure and the definition simply requires that the sets correspond and that the structures correspond also. Also persuasive are the examples above, such as Example 1.1, which dramatize that isomorphic spaces are the same in all relevant respects. Sometimes people say, where $V ≅ W$ , that “ $W$ is just $V$ painted green”—differences are merely cosmetic.

The results below further support our contention that under an isomorphism all the things of interest in the two vector spaces correspond. Because we introduced vector spaces to study linear combinations, “of interest” means “pertaining to linear combinations.” Not of interest is the way that the vectors are presented typographically (or their color!).

Lemma 1.10 An isomorphism maps a zero vector to a zero vector.

Proof Where $f : V \to W$ is an isomorphism, fix some $\vec{v} \in V$ . Then $f ({\vec{0}}_{V}) = f (0 \cdot \vec{v}) = 0 \cdot f (\vec{v}) = {\vec{0}}_{W}$ .

QED

Lemma 1.11 For any map $f : V \to W$ between vector spaces these statements are equivalent.

$f$ preserves structure

$f ({\vec{v}}_{1} + {\vec{v}}_{2}) = f ({\vec{v}}_{1}) + f ({\vec{v}}_{2}) and f (c \vec{v}) = c f (\vec{v})$
$f$ preserves linear combinations of two vectors

$f (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) = c_{1} f ({\vec{v}}_{1}) + c_{2} f ({\vec{v}}_{2})$
$f$ preserves linear combinations of any finite number of vectors

$f (c_{1} {\vec{v}}_{1} + \dots + c_{n} {\vec{v}}_{n}) = c_{1} f ({\vec{v}}_{1}) + \dots + c_{n} f ({\vec{v}}_{n})$

Proof Since the implications $(3) ⟹ (2)$ and $(2) ⟹ (1)$ are clear, we need only show that $(1) ⟹ (3)$ . So assume statement (1). We will prove (3) by induction on the number of summands $n$ .

The one-summand base case, that $f (c {\vec{v}}_{1}) = c f ({\vec{v}}_{1})$ , is covered by the second clause of statement (1).

For the inductive step assume that statement (3) holds whenever there are $k$ or fewer summands. Consider the $k + 1$ -summand case. Use the first half of (1) to break the sum along the final ‘ $+$ ’.

$f (c_{1} {\vec{v}}_{1} + \dots + c_{k} {\vec{v}}_{k} + c_{k + 1} {\vec{v}}_{k + 1}) = f (c_{1} {\vec{v}}_{1} + \dots + c_{k} {\vec{v}}_{k}) + f (c_{k + 1} {\vec{v}}_{k + 1})$

Use the inductive hypothesis to break up the $k$ -term sum on the left.

$= f (c_{1} {\vec{v}}_{1}) + \dots + f (c_{k} {\vec{v}}_{k}) + f (c_{k + 1} {\vec{v}}_{k + 1})$

Now the second half of (1) gives

$= c_{1} f ({\vec{v}}_{1}) + \dots + c_{k} f ({\vec{v}}_{k}) + c_{k + 1} f ({\vec{v}}_{k + 1})$

when applied $k + 1$ times.

QED

We often use item (2) to simplify the verification that a map preserves structure.

Finally, a summary. In the prior chapter, after giving the definition of a vector space, we looked at examples and noted that some spaces seemed to be essentially the same as others. Here we have defined the relation ‘ $≅$ ’ and have argued that it is the right way to precisely say what we mean by “the same” because it preserves the features of interest in a vector space—in particular, it preserves linear combinations. In the next section we will show that isomorphism is an equivalence relation and so partitions the collection of vector spaces.

Exercises

Exercise 1.12 Worked answer

Recommended. Verify, using Example 1.4 as a model, that the two correspondences given before the definition are isomorphisms.
1. Example 1.1
2. Example 1.2
Back to Exercise 1.12

Answer.
1. Call the map $f$ .
  
  $(\begin{matrix} a & b \end{matrix}) \overset{f}{⟼} (\begin{matrix} a \\ b \end{matrix})$
  
  It is one-to-one because if $f$ sends two members of the domain to the same image, that is, if $f ((\begin{matrix} a & b \end{matrix})) = f ((\begin{matrix} c & d \end{matrix}))$ , then the definition of $f$ gives that
  
  $(\begin{matrix} a \\ b \end{matrix}) = (\begin{matrix} c \\ d \end{matrix})$
  
  and since column vectors are equal only if they have equal components, we have that $a = c$ and that $b = d$ . Thus, if $f$ maps two row vectors from the domain to the same column vector then the two row vectors are equal: $(\begin{matrix} a & b \end{matrix}) = (\begin{matrix} c & d \end{matrix})$ .
  
  To show that $f$ is onto we must show that any member of the codomain $ℝ^{2}$ is the image under $f$ of some row vector. That’s easy;
  
  $(\begin{matrix} x \\ y \end{matrix})$
  
  is $f ((\begin{matrix} x & y \end{matrix}))$ .
  
  The computation for preservation of addition is this.
  
  $\begin{matrix} f ((\begin{matrix} a & b \end{matrix}) + (\begin{matrix} c & d \end{matrix})) = f ((\begin{matrix} a + c & b + d \end{matrix})) = (\begin{matrix} a + c \\ b + d \end{matrix}) \\ = (\begin{matrix} a \\ b \end{matrix}) + (\begin{matrix} c \\ d \end{matrix}) = f ((\begin{matrix} a & b \end{matrix})) + f ((\begin{matrix} c & d \end{matrix})) \end{matrix}$
  
  The computation for preservation of scalar multiplication is similar.
  
  $f (r \cdot (\begin{matrix} a & b \end{matrix})) = f ((\begin{matrix} r a & r b \end{matrix})) = (\begin{matrix} r a \\ r b \end{matrix}) = r \cdot (\begin{matrix} a \\ b \end{matrix}) = r \cdot f ((\begin{matrix} a & b \end{matrix}))$
2. Denote the map from Example 1.2 by $f$ . To show that it is one-to-one, assume that $f (a_{0} + a_{1} x + a_{2} x^{2}) = f (b_{0} + b_{1} x + b_{2} x^{2})$ . Then by the definition of the function,
  
  $(\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \end{matrix}) = (\begin{matrix} b_{0} \\ b_{1} \\ b_{2} \end{matrix})$
  
  and so $a_{0} = b_{0}$ and $a_{1} = b_{1}$ and $a_{2} = b_{2}$ . Thus $a_{0} + a_{1} x + a_{2} x^{2} = b_{0} + b_{1} x + b_{2} x^{2}$ , and consequently $f$ is one-to-one.
  
  The function $f$ is onto because there is a polynomial sent to
  
  $(\begin{matrix} a \\ b \\ c \end{matrix})$
  
  by $f$ , namely, $a + b x + c x^{2}$ .
  
  As for structure, this shows that $f$ preserves addition
  
  $\begin{matrix} f ((a_{0} + a_{1} x + a_{2} x^{2}) + (b_{0} + b_{1} x + b_{2} x^{2})) \\ \begin{aligned} = f ((a_{0} + b_{0}) + (a_{1} + b_{1}) x + (a_{2} + b_{2}) x^{2}) \\ = (\begin{array}{c} a_{0} + b_{0} \\ a_{1} + b_{1} \\ a_{2} + b_{2} \end{array}) \\ = (\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}) + (\begin{array}{c} b_{0} \\ b_{1} \\ b_{2} \end{array}) \\ = f (a_{0} + a_{1} x + a_{2} x^{2}) + f (b_{0} + b_{1} x + b_{2} x^{2}) \end{aligned} \end{matrix}$
  
  and this shows
  
  $\begin{aligned} f (r (a_{0} + a_{1} x + a_{2} x^{2})) & = f ((r a_{0}) + (r a_{1}) x + (r a_{2}) x^{2}) \\ = (\begin{array}{c} r a_{0} \\ r a_{1} \\ r a_{2} \end{array}) \\ = r \cdot (\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}) \\ = r f (a_{0} + a_{1} x + a_{2} x^{2}) \end{aligned}$
  
  that it preserves scalar multiplication.
Exercise 1.13 Worked answer

Recommended. For the map $f : 𝒫_{1} \to ℝ^{2}$ given by

$a + b x \overset{f}{⟼} (\begin{matrix} a - b \\ b \end{matrix})$

Find the image of each of these elements of the domain.
1. $3 - 2 x$
2. $2 + 2 x$
3. $x$
Show that this map is an isomorphism.
Back to Exercise 1.13

Answer. These are the images.
1. $(\begin{matrix} 5 \\ - 2 \end{matrix})$
2. $(\begin{matrix} 0 \\ 2 \end{matrix})$
3. $(\begin{matrix} - 1 \\ 1 \end{matrix})$
To prove that $f$ is one-to-one, assume that it maps two linear polynomials to the same image $f (a_{1} + b_{1} x) = f (a_{2} + b_{2} x)$ . Then

$(\begin{matrix} a_{1} - b_{1} \\ b_{1} \end{matrix}) = (\begin{matrix} a_{2} - b_{2} \\ b_{2} \end{matrix})$

and so, since column vectors are equal only when their components are equal, $b_{1} = b_{2}$ and $a_{1} = a_{2}$ . That shows that the two linear polynomials are equal, and so $f$ is one-to-one.

To show that $f$ is onto, note that this member of the codomain

$(\begin{matrix} s \\ t \end{matrix})$

is the image of this member of the domain $(s + t) + t x$ .

To check that $f$ preserves structure, we can use item (2) of Lemma 1.11.

$\begin{aligned} f (c_{1} \cdot (a_{1} + b_{1} x) + c_{2} \cdot (a_{2} + b_{2} x)) & = f ((c_{1} a_{1} + c_{2} a_{2}) + (c_{1} b_{1} + c_{2} b_{2}) x) \\ = (\begin{array}{c} (c_{1} a_{1} + c_{2} a_{2}) - (c_{1} b_{1} + c_{2} b_{2}) \\ c_{1} b_{1} + c_{2} b_{2} \end{array}) \\ = c_{1} \cdot (\begin{array}{c} a_{1} - b_{1} \\ b_{1} \end{array}) + c_{2} \cdot (\begin{array}{c} a_{2} - b_{2} \\ b_{2} \end{array}) \\ = c_{1} \cdot f (a_{1} + b_{1} x) + c_{2} \cdot f (a_{2} + b_{2} x) \end{aligned}$
Exercise 1.14 Worked answer

Show that the natural map $f_{1}$ from Example 1.5 is an isomorphism.

Back to Exercise 1.14

Answer. To verify it is one-to-one, assume that $f_{1} (c_{1} x + c_{2} y + c_{3} z) = f_{1} (d_{1} x + d_{2} y + d_{3} z)$ . Then $c_{1} + c_{2} x + c_{3} x^{2} = d_{1} + d_{2} x + d_{3} x^{2}$ by the definition of $f_{1}$ . Members of $𝒫_{2}$ are equal only when they have the same coefficients, so this implies that $c_{1} = d_{1}$ and $c_{2} = d_{2}$ and $c_{3} = d_{3}$ . Therefore $f_{1} (c_{1} x + c_{2} y + c_{3} z) = f_{1} (d_{1} x + d_{2} y + d_{3} z)$ implies that $c_{1} x + c_{2} y + c_{3} z = d_{1} x + d_{2} y + d_{3} z$ , and so $f_{1}$ is one-to-one.

To verify that it is onto, consider an arbitrary member of the codomain $a_{1} + a_{2} x + a_{3} x^{2}$ and observe that it is indeed the image of a member of the domain, namely, it is $f_{1} (a_{1} x + a_{2} y + a_{3} z)$ . (For instance, $0 + 3 x + 6 x^{2} = f_{1} (0 x + 3 y + 6 z)$ .)

The computation checking that $f_{1}$ preserves addition is this.

$\begin{matrix} f_{1} ((c_{1} x + c_{2} y + c_{3} z) + (d_{1} x + d_{2} y + d_{3} z)) \\ \begin{aligned} = f_{1} ((c_{1} + d_{1}) x + (c_{2} + d_{2}) y + (c_{3} + d_{3}) z) \\ = (c_{1} + d_{1}) + (c_{2} + d_{2}) x + (c_{3} + d_{3}) x^{2} \\ = (c_{1} + c_{2} x + c_{3} x^{2}) + (d_{1} + d_{2} x + d_{3} x^{2}) \\ = f_{1} (c_{1} x + c_{2} y + c_{3} z) + f_{1} (d_{1} x + d_{2} y + d_{3} z) \end{aligned} \end{matrix}$

The check that $f_{1}$ preserves scalar multiplication is this.

$\begin{aligned} f_{1} (r \cdot (c_{1} x + c_{2} y + c_{3} z)) & = f_{1} ((r c_{1}) x + (r c_{2}) y + (r c_{3}) z) \\ = (r c_{1}) + (r c_{2}) x + (r c_{3}) x^{2} \\ = r \cdot (c_{1} + c_{2} x + c_{3} x^{2}) \\ = r \cdot f_{1} (c_{1} x + c_{2} y + c_{3} z) \end{aligned}$
Exercise 1.15 Worked answer

Show that the map $t : 𝒫_{2} \to 𝒫_{2}$ given by $t (a x^{2} + b x + c) = b x^{2} - (a + c) x + a$ is an isomorphism.

Back to Exercise 1.15

Answer. To see that the map is one-to-one suppose that $t ({\vec{v}}_{1}) = t ({\vec{v}}_{2})$ , aiming to conclude that ${\vec{v}}_{1} = {\vec{v}}_{2}$ . That is, $t (a_{1} x^{2} + b_{1} x + c_{1}) = t (a_{2} x^{2} + b_{2} x + c_{2})$ . Then $b_{1} x^{2} - (a_{1} + c_{1}) x + a_{1} = b_{2} x^{2} - (a_{2} + c_{2}) x + a_{2}$ and because quadratic polynomials are equal only if they have have the same quadratic terms, the same constant terms, and the same linear terms we conclude that $b_{1} = b_{2}$ , that $a_{1} = a_{2}$ , and from that, $c_{1} = c_{2}$ . Therefore $a_{1} x^{2} + b_{1} x + c_{1} = a_{2} x^{2} + b_{2} x + c_{2}$ and the function is one-to-one.

To see that the map is onto, we suppose that we are given a member $\vec{w}$ of the codomain and we find a member $\vec{v}$ of the domain that maps to it. Let the member of the codomain be $\vec{w} = p x^{2} + q x + r$ . Observe that where $\vec{v} = r x^{2} + p x + (- q - r)$ then $t (\vec{v}) = \vec{w}$ . Thus $t$ is onto.

To see that the map is a homomorphism we show that it respects linear combinations of two elements. By Lemma 1.11 this will show that the map preserves the operations.

$\begin{matrix} t (r_{1} (a_{1} x^{2} + b_{1} x + c_{1}) + r_{2} (a_{2} x^{2} + b_{2} x + c_{2})) \\ \begin{aligned} = t ((r_{1} a_{1} + r_{2} a_{2}) x^{2} + (r_{1} b_{1} + r_{2} b_{2}) x + (r_{1} c_{1} + r_{2} c_{2})) \\ = (r_{1} b_{1} + r_{2} b_{2}) x^{2} - ((r_{1} a_{1} + r_{2} a_{2}) + (r_{1} c_{1} + r_{2} c_{2})) x + (r_{1} a_{1} + r_{2} a_{2}) \\ = (r_{1} b_{1}) x^{2} - (r_{1} a_{1} + r_{1} c_{1}) x + r_{1} a_{1} + (r_{2} b_{2}) x^{2} - (r_{2} a_{2} + r_{2} c_{2}) x + r_{2} a_{2} \\ = r_{1} t (a_{1} x^{2} + b_{1} x + c_{1}) + r_{2} t (a_{2} x^{2} + b_{2} x + c_{2}) \end{aligned} \end{matrix}$
Exercise 1.16 Worked answer

Recommended. Verify that this map is an isomorphism: $h : ℝ^{4} \to ℳ_{2 \times 2}$ given by

$(\begin{matrix} a \\ b \\ c \\ d \end{matrix}) \mapsto (\begin{matrix} c & a + d \\ b & d \end{matrix})$

Back to Exercise 1.16

Answer. We first verify that $h$ is one-to-one. To do this we will show that $h ({\vec{v}}_{1}) = h ({\vec{v}}_{2})$ implies that ${\vec{v}}_{1} = {\vec{v}}_{2}$ . So assume that

$h ({\vec{v}}_{1}) = h ((\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \\ d_{1} \end{matrix})) = h ((\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \\ d_{2} \end{matrix})) = h ({\vec{v}}_{2})$

which gives

$(\begin{matrix} c_{1} & a_{1} + d_{1} \\ b_{1} & d_{1} \end{matrix}) = (\begin{matrix} c_{2} & a_{2} + d_{2} \\ b_{2} & d_{2} \end{matrix})$

from which we conclude that $c_{1} = c_{2}$ (by the upper-left entries), $b_{1} = b_{2}$ (by the lower-left entries), $d_{1} = d_{2}$ (by the lower-right entries), and with this last we get $a_{1} = a_{2}$ (by the upper right). Therefore ${\vec{v}}_{1} = {\vec{v}}_{2}$ .

Next we will show that the map is onto, that every member of the codomain $ℳ_{2 \times 2}$ is the image of some four-tall member of the domain. So, given

$\vec{w} = (\begin{matrix} m & n \\ p & q \end{matrix}) \in ℳ_{2 \times 2}$

observe that it is the image of this domain vector.

$\vec{v} = (\begin{matrix} n - q \\ p \\ m \\ q \end{matrix})$

To finish we verify that the map preserves linear combinations. By Lemma 1.11 this will show that the map preserves the operations.

$\begin{aligned} h (r_{1} \cdot (\begin{array}{c} a_{1} \\ b_{1} \\ c_{1} \\ d_{1} \end{array}) + r_{2} \cdot (\begin{array}{c} a_{2} \\ b_{2} \\ c_{2} \\ d_{2} \end{array})) & = h ((\begin{array}{c} r_{1} a_{1} + r_{2} a_{2} \\ r_{1} b_{1} + r_{2} b_{2} \\ r_{1} c_{1} + r_{2} c_{2} \\ r_{1} d_{1} + r_{2} d_{2} \end{array})) \\ = (\begin{array}{c} r_{1} c_{1} + r_{2} c_{2} & (r_{1} a_{1} + r_{2} a_{2}) + (r_{1} d_{1} + r_{2} d_{2}) \\ r_{1} b_{1} + r_{2} b_{2} & r_{1} d_{1} + r_{2} d_{2} \end{array}) \\ = r_{1} (\begin{array}{c} c_{1} & a_{1} + d_{1} \\ b_{1} & d_{1} \end{array}) + r_{2} (\begin{array}{c} c_{2} & a_{2} + d_{2} \\ b_{2} & d_{2} \end{array}) \\ = r_{1} \cdot h ((\begin{array}{c} a_{1} \\ b_{1} \\ c_{1} \\ d_{1} \end{array})) + r_{2} \cdot h ((\begin{array}{c} a_{2} \\ b_{2} \\ c_{2} \\ d_{2} \end{array})) \end{aligned}$
Exercise 1.17 Worked answer

Recommended. Decide whether each map is an isomorphism. If it is an isomorphism then prove it and if it isn’t then state a condition that it fails to satisfy.
1. $f : ℳ_{2 \times 2} \to ℝ$ given by
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto a d - b c$
2. $f : ℳ_{2 \times 2} \to ℝ^{4}$ given by
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto (\begin{matrix} a + b + c + d \\ a + b + c \\ a + b \\ a \end{matrix})$
3. $f : ℳ_{2 \times 2} \to 𝒫_{3}$ given by
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto c + (d + c) x + (b + a) x^{2} + a x^{3}$
4. $f : ℳ_{2 \times 2} \to 𝒫_{3}$ given by
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) \mapsto c + (d + c) x + (b + a + 1) x^{2} + a x^{3}$
Back to Exercise 1.17

Answer.
1. No; this map is not one-to-one. In particular, the matrix of all zeroes is mapped to the same image as the matrix of all ones.
2. Yes, this is an isomorphism.
  
  It is one-to-one:
  
  $\begin{aligned} if f ((\begin{array}{c} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array})) = f ((\begin{array}{c} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array})) \\ then (\begin{array}{c} a_{1} + b_{1} + c_{1} + d_{1} \\ a_{1} + b_{1} + c_{1} \\ a_{1} + b_{1} \\ a_{1} \end{array}) = (\begin{array}{c} a_{2} + b_{2} + c_{2} + d_{2} \\ a_{2} + b_{2} + c_{2} \\ a_{2} + b_{2} \\ a_{2} \end{array}) \end{aligned}$
  
  gives that $a_{1} = a_{2}$ , and that $b_{1} = b_{2}$ , and that $c_{1} = c_{2}$ , and that $d_{1} = d_{2}$ .
  
  It is onto, since this shows
  
  $(\begin{matrix} x \\ y \\ z \\ w \end{matrix}) = f ((\begin{matrix} w & z - w \\ y - z & x - y \end{matrix}))$
  
  that any four-tall vector is the image of a $2 \times 2$ matrix.
  
  Finally, it preserves combinations
  
  $\begin{matrix} f (r_{1} \cdot (\begin{matrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) + r_{2} \cdot (\begin{matrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix})) \\ \begin{aligned} = f ((\begin{array}{c} r_{1} a_{1} + r_{2} a_{2} & r_{1} b_{1} + r_{2} b_{2} \\ r_{1} c_{1} + r_{2} c_{2} & r_{1} d_{1} + r_{2} d_{2} \end{array})) \\ = (\begin{array}{c} r_{1} a_{1} + \dots + r_{2} d_{2} \\ r_{1} a_{1} + \dots + r_{2} c_{2} \\ r_{1} a_{1} + \dots + r_{2} b_{2} \\ r_{1} a_{1} + r_{2} a_{2} \end{array}) \\ = r_{1} \cdot (\begin{array}{c} a_{1} + \dots + d_{1} \\ a_{1} + \dots + c_{1} \\ a_{1} + b_{1} \\ a_{1} \end{array}) + r_{2} \cdot (\begin{array}{c} a_{2} + \dots + d_{2} \\ a_{2} + \dots + c_{2} \\ a_{2} + b_{2} \\ a_{2} \end{array}) \\ = r_{1} \cdot f ((\begin{array}{c} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array})) + r_{2} \cdot f ((\begin{array}{c} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array})) \end{aligned} \end{matrix}$
  
  and so item (2) of Lemma 1.11 shows that it preserves structure.
3. Yes, it is an isomorphism.
  
  To show that it is one-to-one, we suppose that two members of the domain have the same image under $f$ .
  
  $f ((\begin{matrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix})) = f ((\begin{matrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix}))$
  
  This gives, by the definition of $f$ , that $c_{1} + (d_{1} + c_{1}) x + (b_{1} + a_{1}) x^{2} + a_{1} x^{3} = c_{2} + (d_{2} + c_{2}) x + (b_{2} + a_{2}) x^{2} + a_{2} x^{3}$ and then the fact that polynomials are equal only when their coefficients are equal gives a set of linear equations
  
  $\begin{aligned} c_{1} & = c_{2} \\ d_{1} + c_{1} & = d_{2} + c_{2} \\ b_{1} + a_{1} & = b_{2} + a_{2} \\ a_{1} & = a_{2} \end{aligned}$
  
  that has only the solution $a_{1} = a_{2}$ , $b_{1} = b_{2}$ , $c_{1} = c_{2}$ , and $d_{1} = d_{2}$ .
  
  To show that $f$ is onto, we note that $p + q x + r x^{2} + s x^{3}$ is the image under $f$ of this matrix.
  
  $(\begin{matrix} s & r - s \\ p & q - p \end{matrix})$
  
  We can check that $f$ preserves structure by using item (2) of Lemma 1.11.
  
  $\begin{matrix} f (r_{1} \cdot (\begin{matrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) + r_{2} \cdot (\begin{matrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix})) \\ \begin{aligned} = f ((\begin{array}{c} r_{1} a_{1} + r_{2} a_{2} & r_{1} b_{1} + r_{2} b_{2} \\ r_{1} c_{1} + r_{2} c_{2} & r_{1} d_{1} + r_{2} d_{2} \end{array})) \\ = \begin{aligned} (r_{1} c_{1} + r_{2} c_{2}) + (r_{1} d_{1} + r_{2} d_{2} + r_{1} c_{1} + r_{2} c_{2}) x \\ + (r_{1} b_{1} + r_{2} b_{2} + r_{1} a_{1} + r_{2} a_{2}) x^{2} + (r_{1} a_{1} + r_{2} a_{2}) x^{3} \end{aligned} \\ = \begin{aligned} r_{1} \cdot (c_{1} + (d_{1} + c_{1}) x + (b_{1} + a_{1}) x^{2} + a_{1} x^{3}) \\ + r_{2} \cdot (c_{2} + (d_{2} + c_{2}) x + (b_{2} + a_{2}) x^{2} + a_{2} x^{3}) \end{aligned} \\ = r_{1} \cdot f ((\begin{array}{c} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array})) + r_{2} \cdot f ((\begin{array}{c} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array})) \end{aligned} \end{matrix}$
4. No, this map does not preserve structure. For instance, it does not send the matrix of all zeroes to the zero polynomial.
Exercise 1.18 Worked answer

Show that the map $f : ℝ^{1} \to ℝ^{1}$ given by $f (x) = x^{3}$ is one-to-one and onto. Is it an isomorphism?

Back to Exercise 1.18

Answer. It is one-to-one and onto, a correspondence, because it has an inverse (namely, $f^{- 1} (x) = \sqrt[3]{x}$ ). However, it is not an isomorphism. For instance, $f (1) + f (1) \neq f (1 + 1)$ .
Exercise 1.19 Worked answer

Recommended. Refer to Example 1.1. Produce two more isomorphisms (of course, you must also verify that they satisfy the conditions in the definition of isomorphism).

Back to Exercise 1.19

Answer. Many maps are possible. Here are two.

$(\begin{matrix} a & b \end{matrix}) \mapsto (\begin{matrix} b \\ a \end{matrix}) and (\begin{matrix} a & b \end{matrix}) \mapsto (\begin{matrix} 2 a \\ b \end{matrix})$

The verifications are straightforward adaptations of the others above.
Exercise 1.20 Worked answer

Refer to Example 1.2. Produce two more isomorphisms (and verify that they satisfy the conditions).

Back to Exercise 1.20

Answer. Here are two.

$a_{0} + a_{1} x + a_{2} x^{2} \mapsto (\begin{matrix} a_{1} \\ a_{0} \\ a_{2} \end{matrix}) and a_{0} + a_{1} x + a_{2} x^{2} \mapsto (\begin{matrix} a_{0} + a_{1} \\ a_{1} \\ a_{2} \end{matrix})$

Verification is straightforward (for the second, to show that it is onto, note that

$(\begin{matrix} s \\ t \\ u \end{matrix})$

is the image of $(s - t) + t x + u x^{2}$ ).
Exercise 1.21 Worked answer

Recommended. Show that, although $ℝ^{2}$ is not itself a subspace of $ℝ^{3}$ , it is isomorphic to the $x y$ -plane subspace of $ℝ^{3}$ .

Back to Exercise 1.21

Answer. The space $ℝ^{2}$ is not a subspace of $ℝ^{3}$ because it is not a subset of $ℝ^{3}$ . The two-tall vectors in $ℝ^{2}$ are not members of $ℝ^{3}$ .

The natural isomorphism $ι : ℝ^{2} \to ℝ^{3}$ (called the injection map) is this.

$(\begin{matrix} x \\ y \end{matrix}) \overset{ι}{⟼} (\begin{matrix} x \\ y \\ 0 \end{matrix})$

This map is one-to-one because

$f ((\begin{matrix} x_{1} \\ y_{1} \end{matrix})) = f ((\begin{matrix} x_{2} \\ y_{2} \end{matrix})) implies (\begin{matrix} x_{1} \\ y_{1} \\ 0 \end{matrix}) = (\begin{matrix} x_{2} \\ y_{2} \\ 0 \end{matrix})$

which in turn implies that $x_{1} = x_{2}$ and $y_{1} = y_{2}$ , and therefore the initial two two-tall vectors are equal.

Because

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

this map is onto the $x y$ -plane.

To show that this map preserves structure, we will use item (2) of Lemma 1.11 and show

$\begin{matrix} f (c_{1} \cdot (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + c_{2} \cdot (\begin{matrix} x_{2} \\ y_{2} \end{matrix})) = f ((\begin{matrix} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} y_{1} + c_{2} y_{2} \end{matrix})) = (\begin{matrix} c_{1} x_{1} + c_{2} x_{2} \\ c_{1} y_{1} + c_{2} y_{2} \\ 0 \end{matrix}) \\ = c_{1} \cdot (\begin{matrix} x_{1} \\ y_{1} \\ 0 \end{matrix}) + c_{2} \cdot (\begin{matrix} x_{2} \\ y_{2} \\ 0 \end{matrix}) = c_{1} \cdot f ((\begin{matrix} x_{1} \\ y_{1} \end{matrix})) + c_{2} \cdot f ((\begin{matrix} x_{2} \\ y_{2} \end{matrix})) \end{matrix}$

that it preserves combinations of two vectors.
Exercise 1.22 Worked answer

Find two isomorphisms between $ℝ^{16}$ and $ℳ_{4 \times 4}$ .

Back to Exercise 1.22

Answer. Here are two:

$(\begin{matrix} r_{1} \\ r_{2} \\ ⋮ \\ r_{16} \end{matrix}) \mapsto (\begin{matrix} r_{1} & r_{2} & \dots \\ \dots & r_{16} \end{matrix}) and (\begin{matrix} r_{1} \\ r_{2} \\ ⋮ \\ r_{16} \end{matrix}) \mapsto (\begin{matrix} r_{1} \\ r_{2} \\ ⋮ & ⋮ \\ r_{16} \end{matrix})$

Verification that each is an isomorphism is easy.
Exercise 1.23 Worked answer

Recommended. For what $k$ is $ℳ_{m \times n}$ isomorphic to $ℝ^{k}$ ?

Back to Exercise 1.23

Answer. When $k$ is the product $k = m n$ , here is an isomorphism.

$(\begin{matrix} r_{1} & r_{2} & \dots \\ ⋮ \\ \dots & r_{m \cdot n} \end{matrix}) \mapsto (\begin{matrix} r_{1} \\ r_{2} \\ ⋮ \\ r_{m \cdot n} \end{matrix})$

Checking that this is an isomorphism is easy.
Exercise 1.24 Worked answer

For what $k$ is $𝒫_{k}$ isomorphic to $ℝ^{n}$ ?

Back to Exercise 1.24

Answer. If $n \geq 1$ then $𝒫_{n - 1} ≅ ℝ^{n}$ . (If we take $𝒫_{- 1}$ and $ℝ^{0}$ to be trivial vector spaces, then the relationship extends one dimension lower.) The natural isomorphism between them is this.

$a_{0} + a_{1} x + \dots + a_{n - 1} x^{n - 1} \mapsto (\begin{matrix} a_{0} \\ a_{1} \\ ⋮ \\ a_{n - 1} \end{matrix})$

Checking that it is an isomorphism is straightforward.
Exercise 1.25 Worked answer

Prove that the map in Example 1.9, from $𝒫_{5}$ to $𝒫_{5}$ given by $p (x) \mapsto p (x - 1)$ , is a vector space isomorphism.

Back to Exercise 1.25

Answer. This is the map, expanded.

$\begin{matrix} f (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + a_{5} x^{5}) \\ \begin{aligned} = a_{0} + a_{1} (x - 1) + a_{2} (x - 1)^{2} + a_{3} (x - 1)^{3} \\ + a_{4} (x - 1)^{4} + a_{5} (x - 1)^{5} \\ = a_{0} + a_{1} (x - 1) + a_{2} (x^{2} - 2 x + 1) \\ + a_{3} (x^{3} - 3 x^{2} + 3 x - 1) \\ + a_{4} (x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1) \\ + a_{5} (x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x - 1) \\ = (a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - a_{5}) \\ + (a_{1} - 2 a_{2} + 3 a_{3} - 4 a_{4} + 5 a_{5}) x \\ + (a_{2} - 3 a_{3} + 6 a_{4} - 10 a_{5}) x^{2} \\ + (a_{3} - 4 a_{4} + 10 a_{5}) x^{3} \\ + (a_{4} - 5 a_{5}) x^{4} + a_{5} x^{5} \end{aligned} \end{matrix}$

This map is a correspondence because it has an inverse, the map $p (x) \mapsto p (x + 1)$ .

To finish checking that it is an isomorphism we apply item (2) of Lemma 1.11 and show that it preserves linear combinations of two polynomials. Briefly, $f (c \cdot (a_{0} + a_{1} x + \dots + a_{5} x^{5}) + d \cdot (b_{0} + b_{1} x + \dots + b_{5} x^{5}))$ equals this

$\begin{matrix} (c a_{0} - c a_{1} + c a_{2} - c a_{3} + c a_{4} - c a_{5} + d b_{0} - d b_{1} + d b_{2} - d b_{3} + d b_{4} - d b_{5}) \\ + \dots + (c a_{5} + d b_{5}) x^{5} \end{matrix}$

which equals $c \cdot f (a_{0} + a_{1} x + \dots + a_{5} x^{5}) + d \cdot f (b_{0} + b_{1} x + \dots + b_{5} x^{5})$ .
Exercise 1.26 Worked answer

Why, in Lemma 1.10, must there be a $\vec{v} \in V$ ? That is, why must $V$ be nonempty?

Back to Exercise 1.26

Answer. No vector space has the empty set underlying it. We can take $\vec{v}$ to be the zero vector.
Exercise 1.27 Worked answer

Are any two trivial spaces isomorphic?

Back to Exercise 1.27

Answer. Yes; where the two spaces are ${\vec{a}}$ and ${\vec{b}}$ , the map sending $\vec{a}$ to $\vec{b}$ is clearly one-to-one and onto, and also preserves what little structure there is.
Exercise 1.28 Worked answer

In the proof of Lemma 1.11, what about the zero-summands case (that is, if $n$ is zero)?

Back to Exercise 1.28

Answer. A linear combination of $n = 0$ vectors adds to the zero vector and so Lemma 1.10 shows that the three statements are equivalent in this case.
Exercise 1.29 Worked answer

Show that any isomorphism $f : 𝒫_{0} \to ℝ^{1}$ has the form $a \mapsto k a$ for some nonzero real number $k$ .

Back to Exercise 1.29

Answer. Consider the basis $⟨ 1 ⟩$ for $𝒫_{0}$ and let $f (1) \in ℝ$ be $k$ . For any $a \in 𝒫_{0}$ we have that $f (a) = f (a \cdot 1) = a f (1) = a k$ and so $f$ ’s action is multiplication by $k$ . Note that $k \neq 0$ or else the map is not one-to-one. (Incidentally, any such map $a \mapsto k a$ is an isomorphism, as is easy to check.)
Exercise 1.30 Worked answer

These prove that isomorphism is an equivalence relation.
1. Show that the identity map $id : V \to V$ is an isomorphism. Thus, any vector space is isomorphic to itself.
2. Show that if $f : V \to W$ is an isomorphism then so is its inverse $f^{- 1} : W \to V$ . Thus, if $V$ is isomorphic to $W$ then also $W$ is isomorphic to $V$ .
3. Show that a composition of isomorphisms is an isomorphism: if $f : V \to W$ is an isomorphism and $g : W \to U$ is an isomorphism then so also is $g \circ f : V \to U$ . Thus, if $V$ is isomorphic to $W$ and $W$ is isomorphic to $U$ , then also $V$ is isomorphic to $U$ .
Back to Exercise 1.30

Answer. In each item, following item (2) of Lemma 1.11, we show that the map preserves structure by showing that the it preserves linear combinations of two members of the domain.
1. The identity map is clearly one-to-one and onto. For linear combinations the check is easy.
  
  $id (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2}) = c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} = c_{1} \cdot id ({\vec{v}}_{1}) + c_{2} \cdot id ({\vec{v}}_{2})$
2. The inverse of a correspondence is also a correspondence (as stated in the appendix), so we need only check that the inverse preserves linear combinations. Assume that ${\vec{w}}_{1} = f ({\vec{v}}_{1})$ (so $f^{- 1} ({\vec{w}}_{1}) = {\vec{v}}_{1}$ ) and assume that ${\vec{w}}_{2} = f ({\vec{v}}_{2})$ .
  
  $\begin{aligned} f^{- 1} (c_{1} \cdot {\vec{w}}_{1} + c_{2} \cdot {\vec{w}}_{2}) & = f^{- 1} (c_{1} \cdot f ({\vec{v}}_{1}) + c_{2} \cdot f ({\vec{v}}_{2})) \\ = f^{- 1} (f (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2})) \\ = c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} \\ = c_{1} \cdot f^{- 1} ({\vec{w}}_{1}) + c_{2} \cdot f^{- 1} ({\vec{w}}_{2}) \end{aligned}$
3. The composition of two correspondences is a correspondence (as stated in the appendix), so we need only check that the composition map preserves linear combinations.
  
  $\begin{aligned} g \circ f (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2}) & = g (f (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2})) \\ = g (c_{1} \cdot f ({\vec{v}}_{1}) + c_{2} \cdot f ({\vec{v}}_{2})) \\ = c_{1} \cdot g (f ({\vec{v}}_{1})) + c_{2} \cdot g (f ({\vec{v}}_{2})) \\ = c_{1} \cdot g \circ f ({\vec{v}}_{1}) + c_{2} \cdot g \circ f ({\vec{v}}_{2}) \end{aligned}$
Exercise 1.31 Worked answer

Suppose that $f : V \to W$ preserves structure. Show that $f$ is one-to-one if and only if the unique member of $V$ mapped by $f$ to ${\vec{0}}_{W}$ is ${\vec{0}}_{V}$ .

Back to Exercise 1.31

Answer. One direction is easy: by definition, if $f$ is one-to-one then for any $\vec{w} \in W$ at most one $\vec{v} \in V$ has $f (\vec{v}) = \vec{w}$ , and so in particular, at most one member of $V$ is mapped to ${\vec{0}}_{W}$ . The proof of Lemma 1.10 does not use the fact that the map is a correspondence and therefore shows that any structure-preserving map $f$ sends ${\vec{0}}_{V}$ to ${\vec{0}}_{W}$ .

For the other direction, assume that the only member of $V$ that is mapped to ${\vec{0}}_{W}$ is ${\vec{0}}_{V}$ . To show that $f$ is one-to-one assume that $f ({\vec{v}}_{1}) = f ({\vec{v}}_{2})$ . Then $f ({\vec{v}}_{1}) - f ({\vec{v}}_{2}) = {\vec{0}}_{W}$ and so $f ({\vec{v}}_{1} - {\vec{v}}_{2}) = {\vec{0}}_{W}$ . Consequently ${\vec{v}}_{1} - {\vec{v}}_{2} = {\vec{0}}_{V}$ , so ${\vec{v}}_{1} = {\vec{v}}_{2}$ , and so $f$ is one-to-one.
Exercise 1.32 Worked answer

Suppose that $f : V \to W$ is an isomorphism. Prove that the set ${{\vec{v}}_{1}, \dots, {\vec{v}}_{k}} \subseteq V$ is linearly dependent if and only if the set of images ${f ({\vec{v}}_{1}), \dots, f ({\vec{v}}_{k})} \subseteq W$ is linearly dependent.

Back to Exercise 1.32

Answer. We will prove something stronger—not only is the existence of a dependence preserved by isomorphism, but each instance of a dependence is preserved, that is,

$\begin{matrix} {\vec{v}}_{i} = c_{1} {\vec{v}}_{1} + \dots + c_{i - 1} {\vec{v}}_{i - 1} + c_{i + 1} {\vec{v}}_{i + 1} + \dots + c_{k} {\vec{v}}_{k} \\ ⟺ f ({\vec{v}}_{i}) = c_{1} f ({\vec{v}}_{1}) + \dots + c_{i - 1} f ({\vec{v}}_{i - 1}) + c_{i + 1} f ({\vec{v}}_{i + 1}) + \dots + c_{k} f ({\vec{v}}_{k}) . \end{matrix}$

The $⟹$ direction of this statement holds by item (3) of Lemma 1.11. The $⟸$ direction holds by regrouping

$\begin{aligned} f ({\vec{v}}_{i}) & = c_{1} f ({\vec{v}}_{1}) + \dots + c_{i - 1} f ({\vec{v}}_{i - 1}) + c_{i + 1} f ({\vec{v}}_{i + 1}) + \dots + c_{k} f ({\vec{v}}_{k}) \\ = f (c_{1} {\vec{v}}_{1} + \dots + c_{i - 1} {\vec{v}}_{i - 1} + c_{i + 1} {\vec{v}}_{i + 1} + \dots + c_{k} {\vec{v}}_{k}) \end{aligned}$

and applying the fact that $f$ is one-to-one, and so for the two vectors ${\vec{v}}_{i}$ and $c_{1} {\vec{v}}_{1} + \dots + c_{i - 1} {\vec{v}}_{i - 1} + c_{i + 1} f {\vec{v}}_{i + 1} + \dots + c_{k} f ({\vec{v}}_{k}$ to be mapped to the same image by $f$ , they must be equal.
Exercise 1.33 Worked answer

Recommended. Show that each type of map from Example 1.8 is an automorphism.
1. Dilation $d_{s}$ by a nonzero scalar $s$ .
2. Rotation $t_{θ}$ through an angle $θ$ .
3. Reflection $f_{ℓ}$ over a line through the origin.
Hint. For the second and third items, polar coordinates are useful.
Back to Exercise 1.33

Answer.
1. This map is one-to-one because if $d_{s} ({\vec{v}}_{1}) = d_{s} ({\vec{v}}_{2})$ then by definition of the map, $s \cdot {\vec{v}}_{1} = s \cdot {\vec{v}}_{2}$ and so ${\vec{v}}_{1} = {\vec{v}}_{2}$ , as $s$ is nonzero. This map is onto as any $\vec{w} \in ℝ^{2}$ is the image of $\vec{v} = (1 / s) \cdot \vec{w}$ (again, note that $s$ is nonzero). (Another way to see that this map is a correspondence is to observe that it has an inverse: the inverse of $d_{s}$ is $d_{1 / s}$ .)
  
  To finish, note that this map preserves linear combinations
  
  $d_{s} (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2}) = s (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2}) = c_{1} s {\vec{v}}_{1} + c_{2} s {\vec{v}}_{2} = c_{1} \cdot d_{s} ({\vec{v}}_{1}) + c_{2} \cdot d_{s} ({\vec{v}}_{2})$
  
  and therefore is an isomorphism.
2. As in the prior item, we can show that the map $t_{θ}$ is a correspondence by noting that it has an inverse, $t_{- θ}$ .
  
  That the map preserves structure is geometrically easy to see. For instance, adding two vectors and then rotating them has the same effect as rotating first and then adding. For an algebraic argument, consider polar coordinates: the map $t_{θ}$ sends the vector with endpoint $(r, ϕ)$ to the vector with endpoint $(r, ϕ + θ)$ . Then the familiar trigonometric formulas $\cos (ϕ + θ) = \cos ϕ \cos θ - \sin ϕ \sin θ$ and $\sin (ϕ + θ) = \sin ϕ \cos θ + \cos ϕ \sin θ$ show how to express the map’s action in the usual rectangular coordinate system.
  
  $(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} r \cos ϕ \\ r \sin ϕ \end{matrix}) \overset{t_{θ}}{⟼} (\begin{matrix} r \cos (ϕ + θ) \\ r \sin (ϕ + θ) \end{matrix}) = (\begin{matrix} x \cos θ - y \sin θ \\ x \sin θ + y \cos θ \end{matrix})$
  
  Now the calculation for preservation of addition is routine.
  
  $\begin{matrix} (\begin{matrix} x_{1} + x_{2} \\ y_{1} + y_{2} \end{matrix}) \overset{t_{θ}}{⟼} (\begin{matrix} (x_{1} + x_{2}) \cos θ - (y_{1} + y_{2}) \sin θ \\ (x_{1} + x_{2}) \sin θ + (y_{1} + y_{2}) \cos θ \end{matrix}) \\ = (\begin{matrix} x_{1} \cos θ - y_{1} \sin θ \\ x_{1} \sin θ + y_{1} \cos θ \end{matrix}) + (\begin{matrix} x_{2} \cos θ - y_{2} \sin θ \\ x_{2} \sin θ + y_{2} \cos θ \end{matrix}) \end{matrix}$
  
  The calculation for preservation of scalar multiplication is similar.
3. This map is a correspondence because it has an inverse (namely, itself).
  
  As in the last item, that the reflection map preserves structure is geometrically easy to see: adding vectors and then reflecting gives the same result as reflecting first and then adding, for instance. For an algebraic proof, suppose that the line $ℓ$ has slope $k$ (the case of a line with undefined slope can be done as a separate, but easy, case). We can follow the hint and use polar coordinates: where the line $ℓ$ forms an angle of $ϕ$ with the $x$ -axis, the action of $f_{ℓ}$ is to send the vector with endpoint $(r \cos θ, r \sin θ)$ to the one with endpoint $(r \cos (2 ϕ - θ), r \sin (2 ϕ - θ))$ .
  
  To convert to rectangular coordinates, we will use some trigonometric formulas, as we did in the prior item. First observe that $\cos ϕ$ and $\sin ϕ$ can be determined from the slope $k$ of the line. This picture
  
  gives that $\cos ϕ = 1 / \sqrt{1 + k^{2}}$ and $\sin ϕ = k / \sqrt{1 + k^{2}}$ . Now,
  
  $\begin{aligned} \cos (2 ϕ - θ) & = \cos (2 ϕ) \cos θ + \sin (2 ϕ) \sin θ \\ = (\cos^{2} ϕ - \sin^{2} ϕ) \cos θ + (2 \sin ϕ \cos ϕ) \sin θ \\ = ((\frac{1}{\sqrt{1 + k^{2}}})^{2} - (\frac{k}{\sqrt{1 + k^{2}}})^{2}) \cos θ + (2 \frac{k}{\sqrt{1 + k^{2}}} \frac{1}{\sqrt{1 + k^{2}}}) \sin θ \\ = (\frac{1 - k^{2}}{1 + k^{2}}) \cos θ + (\frac{2 k}{1 + k^{2}}) \sin θ \end{aligned}$
  
  and thus the first component of the image vector is this.
  
  $r \cdot \cos (2 ϕ - θ) = \frac{1 - k^{2}}{1 + k^{2}} \cdot x + \frac{2 k}{1 + k^{2}} \cdot y$
  
  A similar calculation shows that the second component of the image vector is this.
  
  $r \cdot \sin (2 ϕ - θ) = \frac{2 k}{1 + k^{2}} \cdot x - \frac{1 - k^{2}}{1 + k^{2}} \cdot y$
  
  With this algebraic description of the action of $f_{ℓ}$
  
  $(\begin{matrix} x \\ y \end{matrix}) \overset{f_{ℓ}}{⟼} (\begin{matrix} (1 - k^{2} / 1 + k^{2}) \cdot x + (2 k / 1 + k^{2}) \cdot y \\ (2 k / 1 + k^{2}) \cdot x - (1 - k^{2} / 1 + k^{2}) \cdot y \end{matrix})$
  
  checking that it preserves structure is routine.
Exercise 1.34 Worked answer

Produce an automorphism of $𝒫_{2}$ other than the identity map, and other than a shift map $p (x) \mapsto p (x - k)$ .

Back to Exercise 1.34

Answer. First, the map $p (x) \mapsto p (x + k)$ doesn’t count because it is a version of $p (x) \mapsto p (x - k)$ . Here is a correct answer (many others are also correct): $a_{0} + a_{1} x + a_{2} x^{2} \mapsto a_{2} + a_{0} x + a_{1} x^{2}$ . Verification that this is an isomorphism is straightforward.
Exercise 1.35 Worked answer
1. Show that a function $f : ℝ^{1} \to ℝ^{1}$ is an automorphism if and only if it has the form $x \mapsto k x$ for some $k \neq 0$ .
2. Let $f$ be an automorphism of $ℝ^{1}$ such that $f (3) = 7$ . Find $f (- 2)$ .
3. Show that a function $f : ℝ^{2} \to ℝ^{2}$ is an automorphism if and only if it has the form
  
  $(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} a x + b y \\ c x + d y \end{matrix})$
  
  for some $a, b, c, d \in ℝ$ with $a d - b c \neq 0$ . Hint. Exercises in prior subsections have shown that
  
  $(\begin{matrix} b \\ d \end{matrix}) is not a multiple of (\begin{matrix} a \\ c \end{matrix})$
  
  if and only if $a d - b c \neq 0$ .
4. Let $f$ be an automorphism of $ℝ^{2}$ with
  
  $f ((\begin{matrix} 1 \\ 3 \end{matrix})) = (\begin{matrix} 2 \\ - 1 \end{matrix}) and f ((\begin{matrix} 1 \\ 4 \end{matrix})) = (\begin{matrix} 0 \\ 1 \end{matrix}) .$
  
  Find
  
  $f ((\begin{matrix} 0 \\ - 1 \end{matrix})) .$
Back to Exercise 1.35

Answer.
1. For the ‘only if’ half, let $f : ℝ^{1} \to ℝ^{1}$ to be an isomorphism. Consider the basis $⟨ 1 ⟩ \subseteq ℝ^{1}$ . Designate $f (1)$ by $k$ . Then for any $x$ we have that $f (x) = f (x \cdot 1) = x \cdot f (1) = x k$ , and so $f$ ’s action is multiplication by $k$ . To finish this half, just note that $k \neq 0$ or else $f$ would not be one-to-one.
  
  For the ‘if’ half we only have to check that such a map is an isomorphism when $k \neq 0$ . To check that it is one-to-one, assume that $f (x_{1}) = f (x_{2})$ so that $k x_{1} = k x_{2}$ and divide by the nonzero factor $k$ to conclude that $x_{1} = x_{2}$ . To check that it is onto, note that any $y \in ℝ^{1}$ is the image of $x = y / k$ (again, $k \neq 0$ ). Finally, to check that such a map preserves combinations of two members of the domain, we have this.
  
  $f (c_{1} x_{1} + c_{2} x_{2}) = k (c_{1} x_{1} + c_{2} x_{2}) = c_{1} k x_{1} + c_{2} k x_{2} = c_{1} f (x_{1}) + c_{2} f (x_{2})$
2. By the prior item, $f$ ’s action is $x \mapsto (7 / 3) x$ . Thus $f (- 2) = - 14 / 3$ .
3. For the ‘only if’ half, assume that $f : ℝ^{2} \to ℝ^{2}$ is an automorphism. Consider the standard basis $ℰ_{2}$ for $ℝ^{2}$ . Let
  
  $f ({\vec{e}}_{1}) = (\begin{matrix} a \\ c \end{matrix}) and f ({\vec{e}}_{2}) = (\begin{matrix} b \\ d \end{matrix}) .$
  
  Then the action of $f$ on any vector is determined by by its action on the two basis vectors.
  
  $f ((\begin{matrix} x \\ y \end{matrix})) = f (x \cdot {\vec{e}}_{1} + y \cdot {\vec{e}}_{2}) = x \cdot f ({\vec{e}}_{1}) + y \cdot f ({\vec{e}}_{2}) = x \cdot (\begin{matrix} a \\ c \end{matrix}) + y \cdot (\begin{matrix} b \\ d \end{matrix}) = (\begin{matrix} a x + b y \\ c x + d y \end{matrix})$
  
  To finish this half, note that if $a d - b c = 0$ , that is, if $f ({\vec{e}}_{2})$ is a multiple of $f ({\vec{e}}_{1})$ , then $f$ is not one-to-one.
  
  For ‘if’ we must check that the map is an isomorphism, under the condition that $a d - b c \neq 0$ . The structure-preservation check is easy; we will here show that $f$ is a correspondence. For the argument that the map is one-to-one, assume this.
  
  $f ((\begin{matrix} x_{1} \\ y_{1} \end{matrix})) = f ((\begin{matrix} x_{2} \\ y_{2} \end{matrix})) and so (\begin{matrix} a x_{1} + b y_{1} \\ c x_{1} + d y_{1} \end{matrix}) = (\begin{matrix} a x_{2} + b y_{2} \\ c x_{2} + d y_{2} \end{matrix})$
  
  Then, because $a d - b c \neq 0$ , the resulting system
  
  $\begin{array}{rcrcr} a (x_{1} - x_{2}) & + & b (y_{1} - y_{2}) & = & 0 \\ c (x_{1} - x_{2}) & + & d (y_{1} - y_{2}) & = & 0 \end{array}$
  
  has a unique solution, namely the trivial one $x_{1} - x_{2} = 0$ and $y_{1} - y_{2} = 0$ (this follows from the hint).
  
  The argument that this map is onto is closely related—this system
  
  $\begin{array}{rcrcr} a x_{1} & + & b y_{1} & = & x \\ c x_{1} & + & d y_{1} & = & y \end{array}$
  
  has a solution for any $x$ and $y$ if and only if this set
  
  ${(\begin{matrix} a \\ c \end{matrix}), (\begin{matrix} b \\ d \end{matrix})}$
  
  spans $ℝ^{2}$ , i.e., if and only if this set is a basis (because it is a two-element subset of $ℝ^{2}$ ), i.e., if and only if $a d - b c \neq 0$ .
4. $f ((\begin{matrix} 0 \\ - 1 \end{matrix})) = f ((\begin{matrix} 1 \\ 3 \end{matrix}) - (\begin{matrix} 1 \\ 4 \end{matrix})) = f ((\begin{matrix} 1 \\ 3 \end{matrix})) - f ((\begin{matrix} 1 \\ 4 \end{matrix})) = (\begin{matrix} 2 \\ - 1 \end{matrix}) - (\begin{matrix} 0 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ - 2 \end{matrix})$
Exercise 1.36 Worked answer

Refer to Lemma 1.10 and Lemma 1.11. Find two more things preserved by isomorphism.

Back to Exercise 1.36

Answer. There are many answers; two are linear independence and subspaces.

First we show that if a set ${{\vec{v}}_{1}, \dots, {\vec{v}}_{n}}$ is linearly independent then its image ${f ({\vec{v}}_{1}), \dots, f ({\vec{v}}_{n})}$ is also linearly independent. Consider a linear relationship among members of the image set.

$0 = c_{1} f ({\vec{v}}_{1}) + \dots + c_{n} f (\vec{v_{n}}) = f (c_{1} {\vec{v}}_{1}) + \dots + f (c_{n} \vec{v_{n}}) = f (c_{1} {\vec{v}}_{1} + \dots + c_{n} \vec{v_{n}})$

Because this map is an isomorphism, it is one-to-one. So $f$ maps only one vector from the domain to the zero vector in the range, that is, $c_{1} {\vec{v}}_{1} + \dots + c_{n} {\vec{v}}_{n}$ equals the zero vector (in the domain, of course). But, if ${{\vec{v}}_{1}, \dots, {\vec{v}}_{n}}$ is linearly independent then all of the $c$ ’s are zero, and so ${f ({\vec{v}}_{1}), \dots, f ({\vec{v}}_{n})}$ is linearly independent also. (Remark. There is a small point about this argument that is worth mention. In a set, repeats collapse, that is, strictly speaking, this is a one-element set: ${\vec{v}, \vec{v}}$ , because the things listed as in it are the same thing. Observe, however, the use of the subscript $n$ in the above argument. In moving from the domain set ${{\vec{v}}_{1}, \dots, {\vec{v}}_{n}}$ to the image set ${f ({\vec{v}}_{1}), \dots, f ({\vec{v}}_{n})}$ , there is no collapsing, because the image set does not have repeats, because the isomorphism $f$ is one-to-one.)

To show that if $f : V \to W$ is an isomorphism and if $U$ is a subspace of the domain $V$ then the set of image vectors $f (U) = {\vec{w} \in W ∣ \vec{w} = f (\vec{u}) for some \vec{u} \in U}$ is a subspace of $W$ , we need only show that it is closed under linear combinations of two of its members (it is nonempty because it contains the image of the zero vector). We have

$c_{1} \cdot f ({\vec{u}}_{1}) + c_{2} \cdot f ({\vec{u}}_{2}) = f (c_{1} {\vec{u}}_{1}) + f (c_{2} {\vec{u}}_{2}) = f (c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2})$

and $c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2}$ is a member of $U$ because of the closure of a subspace under combinations. Hence the combination of $f ({\vec{u}}_{1})$ and $f ({\vec{u}}_{2})$ is a member of $f (U)$ .
Exercise 1.37 Worked answer

We show that isomorphisms can be tailored to fit in that, sometimes, given vectors in the domain and in the range we can produce an isomorphism associating those vectors.
1. Let $B = ⟨ {\vec{β}}_{1}, {\vec{β}}_{2}, {\vec{β}}_{3} ⟩$ be a basis for $𝒫_{2}$ so that any $\vec{p} \in 𝒫_{2}$ has a unique representation as $\vec{p} = c_{1} {\vec{β}}_{1} + c_{2} {\vec{β}}_{2} + c_{3} {\vec{β}}_{3}$ , which we denote in this way.
  
  ${Rep}_{B} (\vec{p}) = (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix})$
  
  Show that the ${Rep}_{B} (\cdot)$ operation is a function from $𝒫_{2}$ to $ℝ^{3}$ (this entails showing that with every domain vector $\vec{v} \in 𝒫_{2}$ there is an associated image vector in $ℝ^{3}$ , and further, that with every domain vector $\vec{v} \in 𝒫_{2}$ there is at most one associated image vector).
2. Show that this ${Rep}_{B} (\cdot)$ function is one-to-one and onto.
3. Show that it preserves structure.
4. Produce an isomorphism from $𝒫_{2}$ to $ℝ^{3}$ that fits these specifications.
  
  $x + x^{2} \mapsto (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) and 1 - x \mapsto (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})$
Back to Exercise 1.37

Answer.
1. The association
  
  $\vec{p} = c_{1} {\vec{β}}_{1} + c_{2} {\vec{β}}_{2} + c_{3} {\vec{β}}_{3} \overset{{Rep}_{B} (\cdot)}{⟼} (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix})$
  
  is a function if every member $\vec{p}$ of the domain is associated with at least one member of the codomain, and if every member $\vec{p}$ of the domain is associated with at most one member of the codomain. The first condition holds because the basis $B$ spans the domain—every $\vec{p}$ can be written as at least one linear combination of $\vec{β}$ ’s. The second condition holds because the basis $B$ is linearly independent—every member $\vec{p}$ of the domain can be written as at most one linear combination of the $\vec{β}$ ’s.
2. For the one-to-one argument, if ${Rep}_{B} (\vec{p}) = {Rep}_{B} (\vec{q})$ , that is, if ${Rep}_{B} (p_{1} {\vec{β}}_{1} + p_{2} {\vec{β}}_{2} + p_{3} {\vec{β}}_{3}) = {Rep}_{B} (q_{1} {\vec{β}}_{1} + q_{2} {\vec{β}}_{2} + q_{3} {\vec{β}}_{3})$ then
  
  $(\begin{matrix} p_{1} \\ p_{2} \\ p_{3} \end{matrix}) = (\begin{matrix} q_{1} \\ q_{2} \\ q_{3} \end{matrix})$
  
  and so $p_{1} = q_{1}$ and $p_{2} = q_{2}$ and $p_{3} = q_{3}$ , which gives the conclusion that $\vec{p} = \vec{q}$ . Therefore this map is one-to-one.
  
  For onto, we can just note that
  
  $(\begin{matrix} a \\ b \\ c \end{matrix})$
  
  equals ${Rep}_{B} (a {\vec{β}}_{1} + b {\vec{β}}_{2} + c {\vec{β}}_{3})$ , and so any member of the codomain $ℝ^{3}$ is the image of some member of the domain $𝒫_{2}$ .
3. This map respects addition and scalar multiplication because it respects combinations of two members of the domain (that is, we are using item (2) of Lemma 1.11): where $\vec{p} = p_{1} {\vec{β}}_{1} + p_{2} {\vec{β}}_{2} + p_{3} {\vec{β}}_{3}$ and $\vec{q} = q_{1} {\vec{β}}_{1} + q_{2} {\vec{β}}_{2} + q_{3} {\vec{β}}_{3}$ , we have this.
  
  $\begin{aligned} {Rep}_{B} (c \cdot \vec{p} + d \cdot \vec{q}) & = {Rep}_{B} ((c p_{1} + d q_{1}) {\vec{β}}_{1} + (c p_{2} + d q_{2}) {\vec{β}}_{2} + (c p_{3} + d q_{3}) {\vec{β}}_{3}) \\ = (\begin{array}{c} c p_{1} + d q_{1} \\ c p_{2} + d q_{2} \\ c p_{3} + d q_{3} \end{array}) \\ = c \cdot (\begin{array}{c} p_{1} \\ p_{2} \\ p_{3} \end{array}) + d \cdot (\begin{array}{c} q_{1} \\ q_{2} \\ q_{3} \end{array}) \\ = {Rep}_{B} (\vec{p}) + {Rep}_{B} (\vec{q}) \end{aligned}$
4. Use any basis $B$ for $𝒫_{2}$ whose first two members are $x + x^{2}$ and $1 - x$ , say $B = ⟨ x + x^{2}, 1 - x, 1 ⟩$ .
Exercise 1.38 Worked answer

Prove that a space is $n$ -dimensional if and only if it is isomorphic to $ℝ^{n}$ . Hint. Fix a basis $B$ for the space and consider the map sending a vector over to its representation with respect to $B$ .

Back to Exercise 1.38

Answer. See the next subsection.
Exercise 1.39 Worked answer

(Requires the subsection on Combining Subspaces, which is optional.) Let $U$ and $W$ be vector spaces. Define a new vector space, consisting of the set $U \times W = {(\vec{u}, \vec{w}) ∣ \vec{u} \in U and \vec{w} \in W}$ along with these operations.

$({\vec{u}}_{1}, {\vec{w}}_{1}) + ({\vec{u}}_{2}, {\vec{w}}_{2}) = ({\vec{u}}_{1} + {\vec{u}}_{2}, {\vec{w}}_{1} + {\vec{w}}_{2}) and r \cdot (\vec{u}, \vec{w}) = (r \vec{u}, r \vec{w})$

This is a vector space, the external direct sum of $U$ and $W$ .
1. Check that it is a vector space.
2. Find a basis for, and the dimension of, the external direct sum $𝒫_{2} \times ℝ^{2}$ .
3. What is the relationship among $\dim (U)$ , $\dim (W)$ , and $\dim (U \times W)$ ?
4. Suppose that $U$ and $W$ are subspaces of a vector space $V$ such that $V = U \oplus W$ (in this case we say that $V$ is the internal direct sum of $U$ and $W$ ). Show that the map $f : U \times W \to V$ given by
  
  $(\vec{u}, \vec{w}) \overset{f}{⟼} \vec{u} + \vec{w}$
  
  is an isomorphism. Thus if the internal direct sum is defined then the internal and external direct sums are isomorphic.
Back to Exercise 1.39

Answer.
1. Most of the conditions in the definition of a vector space are routine. We here sketch the verification of part (1) of that definition.
  
  For closure of $U \times W$ , note that because $U$ and $W$ are closed, we have that ${\vec{u}}_{1} + {\vec{u}}_{2} \in U$ and ${\vec{w}}_{1} + {\vec{w}}_{2} \in W$ and so $({\vec{u}}_{1} + {\vec{u}}_{2}, {\vec{w}}_{1} + {\vec{w}}_{2}) \in U \times W$ . Commutativity of addition in $U \times W$ follows from commutativity of addition in $U$ and $W$ .
  
  $({\vec{u}}_{1}, {\vec{w}}_{1}) + ({\vec{u}}_{2}, {\vec{w}}_{2}) = ({\vec{u}}_{1} + {\vec{u}}_{2}, {\vec{w}}_{1} + {\vec{w}}_{2}) = ({\vec{u}}_{2} + {\vec{u}}_{1}, {\vec{w}}_{2} + {\vec{w}}_{1}) = ({\vec{u}}_{2}, {\vec{w}}_{2}) + ({\vec{u}}_{1}, {\vec{w}}_{1})$
  
  The check for associativity of addition is similar. The zero element is $({\vec{0}}_{U}, {\vec{0}}_{W}) \in U \times W$ and the additive inverse of $(\vec{u}, \vec{w})$ is $(- \vec{u}, - \vec{w})$ .
  
  The checks for the second part of the definition of a vector space are also straightforward.
2. This is a basis
  
  $⟨ (1, (\begin{matrix} 0 \\ 0 \end{matrix})), (x, (\begin{matrix} 0 \\ 0 \end{matrix})), (x^{2}, (\begin{matrix} 0 \\ 0 \end{matrix})), (1, (\begin{matrix} 1 \\ 0 \end{matrix})), (1, (\begin{matrix} 0 \\ 1 \end{matrix})) ⟩$
  
  because there is one and only one way to represent any member of $𝒫_{2} \times ℝ^{2}$ with respect to this set; here is an example.
  
  $(3 + 2 x + x^{2}, (\begin{matrix} 5 \\ 4 \end{matrix})) = 3 \cdot (1, (\begin{matrix} 0 \\ 0 \end{matrix})) + 2 \cdot (x, (\begin{matrix} 0 \\ 0 \end{matrix})) + (x^{2}, (\begin{matrix} 0 \\ 0 \end{matrix})) + 5 \cdot (1, (\begin{matrix} 1 \\ 0 \end{matrix})) + 4 \cdot (1, (\begin{matrix} 0 \\ 1 \end{matrix}))$
  
  The dimension of this space is five.
3. We have $\dim (U \times W) = \dim (U) + \dim (W)$ as this is a basis.
  
  $⟨ ({\vec{μ}}_{1}, {\vec{0}}_{W}), \dots, ({\vec{μ}}_{\dim (U)}, {\vec{0}}_{W}), ({\vec{0}}_{U}, {\vec{ω}}_{1}), \dots, ({\vec{0}}_{U}, {\vec{ω}}_{\dim (W)}) ⟩$
4. We know that if $V = U \oplus W$ then each $\vec{v} \in V$ can be written as $\vec{v} = \vec{u} + \vec{w}$ in one and only one way. This is just what we need to prove that the given function an isomorphism.
  
  First, to show that $f$ is one-to-one we can show that if $f (({\vec{u}}_{1}, {\vec{w}}_{1})) = (({\vec{u}}_{2}, {\vec{w}}_{2}))$ , that is, if ${\vec{u}}_{1} + {\vec{w}}_{1} = {\vec{u}}_{2} + {\vec{w}}_{2}$ then ${\vec{u}}_{1} = {\vec{u}}_{2}$ and ${\vec{w}}_{1} = {\vec{w}}_{2}$ . But the statement ‘each $\vec{v}$ is such a sum in only one way’ is exactly what is needed to make this conclusion. Similarly, the argument that $f$ is onto is completed by the statement that ‘each $\vec{v}$ is such a sum in at least one way’.
  
  This map also preserves linear combinations
  
  $\begin{aligned} f (c_{1} \cdot ({\vec{u}}_{1}, {\vec{w}}_{1}) + c_{2} \cdot ({\vec{u}}_{2}, {\vec{w}}_{2})) & = f ((c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2}, c_{1} {\vec{w}}_{1} + c_{2} {\vec{w}}_{2})) \\ = c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2} + c_{1} {\vec{w}}_{1} + c_{2} {\vec{w}}_{2} \\ = c_{1} {\vec{u}}_{1} + c_{1} {\vec{w}}_{1} + c_{2} {\vec{u}}_{2} + c_{2} {\vec{w}}_{2} \\ = c_{1} \cdot f (({\vec{u}}_{1}, {\vec{w}}_{1})) + c_{2} \cdot f (({\vec{u}}_{2}, {\vec{w}}_{2})) \end{aligned}$
  
  and so it is an isomorphism.

Dimension Characterizes Isomorphism

In the prior subsection, after stating the definition of isomorphism, we gave some results supporting our sense that such a map describes spaces as “the same.” Here we will develop this intuition. When two (unequal) spaces are isomorphic we think of them as almost equal, as equivalent. We shall make that precise by proving that the relationship ‘is isomorphic to’ is an equivalence relation.

Lemma 2.1 The inverse of an isomorphism is also an isomorphism.

Proof Suppose that $V$ is isomorphic to $W$ via $f : V \to W$ . An isomorphism is a correspondence between the sets so $f$ has an inverse function $f^{- 1} : W \to V$ that is also a correspondence.²

We will show that because $f$ preserves linear combinations, so also does $f^{- 1}$ . Suppose that ${\vec{w}}_{1}, {\vec{w}}_{2} \in W$ . Because it is an isomorphism, $f$ is onto and there are ${\vec{v}}_{1}, {\vec{v}}_{2} \in V$ such that ${\vec{w}}_{1} = f ({\vec{v}}_{1})$ and ${\vec{w}}_{2} = f ({\vec{v}}_{2})$ . Then

$\begin{matrix} f^{- 1} (c_{1} \cdot {\vec{w}}_{1} + c_{2} \cdot {\vec{w}}_{2}) = f^{- 1} (c_{1} \cdot f ({\vec{v}}_{1}) + c_{2} \cdot f ({\vec{v}}_{2})) \\ = f^{- 1} (f (c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2})) = c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} = c_{1} \cdot f^{- 1} ({\vec{w}}_{1}) + c_{2} \cdot f^{- 1} ({\vec{w}}_{2}) \end{matrix}$

since $f^{- 1} ({\vec{w}}_{1}) = {\vec{v}}_{1}$ and $f^{- 1} ({\vec{w}}_{2}) = {\vec{v}}_{2}$ . With that, by Lemma 1.11’s second statement, this map preserves structure.

QED

Theorem 2.2 Isomorphism is an equivalence relation between vector spaces.

Proof We must prove that the relation is symmetric, reflexive, and transitive.

To check reflexivity, that any space is isomorphic to itself, consider the identity map. It is clearly one-to-one and onto. This shows that it preserves linear combinations.

$id (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2}) = c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} = c_{1} \cdot id ({\vec{v}}_{1}) + c_{2} \cdot id ({\vec{v}}_{2})$

Symmetry, that if $V$ is isomorphic to $W$ then also $W$ is isomorphic to $V$ , holds by Lemma 2.1 since each isomorphism map from $V$ to $W$ is paired with an isomorphism from $W$ to $V$ .

To finish we must check transitivity, that if $V$ is isomorphic to $W$ and $W$ is isomorphic to $U$ then $V$ is isomorphic to $U$ . Let $f : V \to W$ and $g : W \to U$ be isomorphisms. Consider their composition $g \circ f : V \to U$ . Because the composition of correspondences is a correspondence, we need only check that the composition preserves linear combinations.

$\begin{aligned} g \circ f (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2}) & = g (f (c_{1} \cdot {\vec{v}}_{1} + c_{2} \cdot {\vec{v}}_{2})) \\ = g (c_{1} \cdot f ({\vec{v}}_{1}) + c_{2} \cdot f ({\vec{v}}_{2})) \\ = c_{1} \cdot g (f ({\vec{v}}_{1})) + c_{2} \cdot g (f ({\vec{v}}_{2})) \\ = c_{1} \cdot (g \circ f) ({\vec{v}}_{1}) + c_{2} \cdot (g \circ f) ({\vec{v}}_{2}) \end{aligned}$

Thus the composition is an isomorphism.

QED

Since it is an equivalence, isomorphism partitions the universe of vector spaces into classes: each space is in one and only one isomorphism class.

All finite dimensional

vector spaces:

The collection of finite-dimensional vector spaces is partitioned into isomorphism classes. Points V and W lie in one region; dots indicate further classes.

V ≅ W

The next result characterizes these classes by dimension. That is, we can describe each class simply by giving the number that is the dimension of all of the spaces in that class.

Theorem 2.3 Vector spaces are isomorphic if and only if they have the same dimension.

In this double implication statement the proof of each half involves a significant idea so we will do the two separately.

Lemma 2.4 If spaces are isomorphic then they have the same dimension.

Proof We shall show that an isomorphism of two spaces gives a correspondence between their bases. That is, we shall show that if $f : V \to W$ is an isomorphism and a basis for the domain $V$ is $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ then its image $D = ⟨ f ({\vec{β}}_{1}), \dots, f ({\vec{β}}_{n}) ⟩$ is a basis for the codomain $W$ . (The other half of the correspondence, that for any basis of $W$ the inverse image is a basis for $V$ , follows from the fact that $f^{- 1}$ is also an isomorphism and so we can apply the prior sentence to $f^{- 1}$ .)

To see that $D$ spans $W$ , fix any $\vec{w} \in W$ . Because $f$ is an isomorphism it is onto and so there is a $\vec{v} \in V$ with $\vec{w} = f (\vec{v})$ . Expand $\vec{v}$ as a combination of basis vectors.

$\vec{w} = f (\vec{v}) = f (v_{1} {\vec{β}}_{1} + \dots + v_{n} {\vec{β}}_{n}) = v_{1} \cdot f ({\vec{β}}_{1}) + \dots + v_{n} \cdot f ({\vec{β}}_{n})$

For linear independence of $D$ , if

${\vec{0}}_{W} = c_{1} f ({\vec{β}}_{1}) + \dots + c_{n} f ({\vec{β}}_{n}) = f (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n})$

then, since $f$ is one-to-one and so the only vector sent to ${\vec{0}}_{W}$ is ${\vec{0}}_{V}$ , we have that ${\vec{0}}_{V} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ , which implies that all of the $c$ ’s are zero.

QED

Lemma 2.5 If spaces have the same dimension then they are isomorphic.

Proof We will prove that any space of dimension $n$ is isomorphic to $ℝ^{n}$ . Then we will have that all such spaces are isomorphic to each other by transitivity, which was shown in Theorem 2.2.

Let $V$ be $n$ -dimensional. Fix a basis $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ for the domain $V$ . Consider the operation of representing the members of $V$ with respect to $B$ as a function from $V$ to $ℝ^{n}$ .

$\vec{v} = v_{1} {\vec{β}}_{1} + \dots + v_{n} {\vec{β}}_{n} \overset{{Rep}_{B}}{⟼} (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix})$

It is well-defined³ since every $\vec{v}$ has one and only one such representation (see Remark 2.7 following this proof).

This function is one-to-one because if

${Rep}_{B} (u_{1} {\vec{β}}_{1} + \dots + u_{n} {\vec{β}}_{n}) = {Rep}_{B} (v_{1} {\vec{β}}_{1} + \dots + v_{n} {\vec{β}}_{n})$

then

$(\begin{matrix} u_{1} \\ ⋮ \\ u_{n} \end{matrix}) = (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix})$

and so $u_{1} = v_{1}$ , …, $u_{n} = v_{n}$ , implying that the original arguments $u_{1} {\vec{β}}_{1} + \dots + u_{n} {\vec{β}}_{n}$ and $v_{1} {\vec{β}}_{1} + \dots + v_{n} {\vec{β}}_{n}$ are equal.

This function is onto; any member of $ℝ^{n}$

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

is the image of some $\vec{v} \in V$ , namely $\vec{w} = {Rep}_{B} (w_{1} {\vec{β}}_{1} + \dots + w_{n} {\vec{β}}_{n})$ .

Finally, this function preserves structure.

$\begin{aligned} {Rep}_{B} (r \cdot \vec{u} + s \cdot \vec{v}) & = {Rep}_{B} ((r u_{1} + s v_{1}) {\vec{β}}_{1} + \dots + (r u_{n} + s v_{n}) {\vec{β}}_{n}) \\ = (\begin{array}{c} r u_{1} + s v_{1} \\ ⋮ \\ r u_{n} + s v_{n} \end{array}) \\ = r \cdot (\begin{array}{c} u_{1} \\ ⋮ \\ u_{n} \end{array}) + s \cdot (\begin{array}{c} v_{1} \\ ⋮ \\ v_{n} \end{array}) \\ = r \cdot {Rep}_{B} (\vec{u}) + s \cdot {Rep}_{B} (\vec{v}) \end{aligned}$

Therefore ${Rep}_{B}$ is an isomorphism. Consequently any $n$ -dimensional space is isomorphic to $ℝ^{n}$ .

QED

Remark 2.6 When we introduced the ${Rep}_{B}$ notation for vectors in Remark 1.17, we noted that it is not standard and said that one advantage it has is that it is harder to overlook. Here we see its other advantage: this notation makes explicit that ${Rep}_{B}$ is a function from $V$ to $ℝ^{n}$ .

Remark 2.7 The proof has a sentence about ‘well-defined.’ Its point is that to be an isomorphism ${Rep}_{B}$ must be a function. The definition of function requires that for all inputs the associated output must exists and must be determined by the input. So we must check that every $\vec{v}$ is associated with at least one ${Rep}_{B} (\vec{v})$ , and with no more than one.

In the proof we express elements $\vec{v}$ of the domain space as combinations of members of the basis $B$ and then associate $\vec{v}$ with the column vector of coefficients. That there is at least one expansion of each $\vec{v}$ holds because $B$ is a basis and so spans the space.

The worry that there is no more than one associated member of the codomain is subtler. A contrasting example, where an association fails this unique output requirement, illuminates the issue. Let the domain be $𝒫_{2}$ and consider a set that is not a basis (it is not linearly independent, although it does span the space).

$A = {1 + 0 x + 0 x^{2}, 0 + 1 x + 0 x^{2}, 0 + 0 x + 1 x^{2}, 1 + 1 x + 2 x^{2}}$

Call those polynomials ${\vec{α}}_{1}$ , …, ${\vec{α}}_{4}$ . In contrast to the situation when the set is a basis, here there can be more than one expression of a domain vector in terms of members of the set. For instance, consider $\vec{v} = 1 + x + x^{2}$ . Here are two different expansions.

$\vec{v} = 1 {\vec{α}}_{1} + 1 {\vec{α}}_{2} + 1 {\vec{α}}_{3} + 0 {\vec{α}}_{4} \vec{v} = 0 {\vec{α}}_{1} + 0 {\vec{α}}_{2} - 1 {\vec{α}}_{3} + 1 {\vec{α}}_{4}$

So this input vector $\vec{v}$ is associated with more than one column.

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

Thus, with $A$ the association is not well-defined. (The issue is that $A$ is not linearly independent; to show uniqueness Theorem Two.III.1.12’s proof uses only linear independence.)

In general, any time that we define a function we must check that output values are well-defined. Most of the time that condition is perfectly obvious but in the above proof it needs verification. See Exercise 2.22.

Corollary 2.8 Each finite-dimensional vector space is isomorphic to one and only one of the $ℝ^{n}$ .

This gives us a collection of representatives of the isomorphism classes.

All finite dimensional

vector spaces:

The same partition, with one representative in each class: stars mark R^0, R^1, R^2 and R^3, and dots indicate further dimensions.

One representative

per class

The proofs above pack many ideas into a small space. Through the rest of this chapter we’ll consider these ideas again, and fill them out. As a taste of this we will expand here on the proof of Lemma 2.5.

Example 2.9 The space $ℳ_{2 \times 2}$ of $2 \times 2$ matrices is isomorphic to $ℝ^{4}$ . With this basis for the domain

$B = ⟨ (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) ⟩$

the isomorphism given in the lemma, the representation map $f_{1} = {Rep}_{B}$ , carries the entries over.

$(\begin{matrix} a & b \\ c & d \end{matrix}) \overset{f_{1}}{⟼} (\begin{matrix} a \\ b \\ c \\ d \end{matrix})$

One way to think of the map $f_{1}$ is: fix the basis $B$ for the domain, use the standard basis $ℰ_{4}$ for the codomain, and associate ${\vec{β}}_{1}$ with ${\vec{e}}_{1}$ , ${\vec{β}}_{2}$ with ${\vec{e}}_{2}$ , etc. Then extend this association to all of the members of two spaces.

$(\begin{matrix} a & b \\ c & d \end{matrix}) = a {\vec{β}}_{1} + b {\vec{β}}_{2} + c {\vec{β}}_{3} + d {\vec{β}}_{4} \overset{f_{1}}{⟼} a {\vec{e}}_{1} + b {\vec{e}}_{2} + c {\vec{e}}_{3} + d {\vec{e}}_{4} = (\begin{matrix} a \\ b \\ c \\ d \end{matrix})$

We can do the same thing with different bases, for instance, taking this basis for the domain.

$A = ⟨ (\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 2 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 2 \end{matrix}) ⟩$

Associating corresponding members of $A$ and $ℰ_{4}$ gives this.

$\begin{matrix} (\begin{matrix} a & b \\ c & d \end{matrix}) = (a / 2) {\vec{α}}_{1} + (b / 2) {\vec{α}}_{2} + (c / 2) {\vec{α}}_{3} + (d / 2) {\vec{α}}_{4} \\ \overset{f_{2}}{⟼} (a / 2) {\vec{e}}_{1} + (b / 2) {\vec{e}}_{2} + (c / 2) {\vec{e}}_{3} + (d / 2) {\vec{e}}_{4} = (\begin{matrix} a / 2 \\ b / 2 \\ c / 2 \\ d / 2 \end{matrix}) \end{matrix}$

gives rise to an isomorphism that is different than $f_{1}$ .

The prior map arose by changing the basis for the domain. We can also change the basis for the codomain. Go back to the basis $B$ above and use this basis for the codomain.

$D = ⟨ (\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}) ⟩$

Associate ${\vec{β}}_{1}$ with ${\vec{δ}}_{1}$ , etc. Extending that gives another isomorphism.

$(\begin{matrix} a & b \\ c & d \end{matrix}) = a {\vec{β}}_{1} + b {\vec{β}}_{2} + c {\vec{β}}_{3} + d {\vec{β}}_{4} \overset{f_{3}}{⟼} a {\vec{δ}}_{1} + b {\vec{δ}}_{2} + c {\vec{δ}}_{3} + d {\vec{δ}}_{4} = (\begin{matrix} a \\ b \\ d \\ c \end{matrix})$

We close with a recap. Recall that the first chapter defines two matrices to be row equivalent if they can be derived from each other by row operations. There we showed that relation is an equivalence and so the collection of matrices is partitioned into classes, where all the matrices that are row equivalent together fall into a single class. Then for insight into which matrices are in each class we gave representatives for the classes, the reduced echelon form matrices.

In this section we have followed that pattern except that the notion here of “the same” is vector space isomorphism. We defined it and established some properties, including that it is an equivalence. Then, as before, we developed a list of class representatives to help us understand the partition— it classifies vector spaces by dimension.

In Chapter Two, with the definition of vector spaces, we seemed to have opened up our studies to many examples of new structures besides the familiar $ℝ^{n}$ ’s. We now know that isn’t the case. Any finite-dimensional vector space is actually “the same” as a real space.

Exercises

Exercise 2.10 Worked answer

Recommended. Decide if the spaces are isomorphic.
1. $ℝ^{2}$ , $ℝ^{4}$
2. $𝒫_{5}$ , $ℝ^{5}$
3. $ℳ_{2 \times 3}$ , $ℝ^{6}$
4. $𝒫_{5}$ , $ℳ_{2 \times 3}$
5. $ℳ_{2 \times k}$ , $ℳ_{k \times 2}$
Back to Exercise 2.10

Answer. Each pair of spaces is isomorphic if and only if the two have the same dimension. We can, when there is an isomorphism, state a map, but it isn’t strictly necessary.
1. No, they have different dimensions.
2. No, they have different dimensions.
3. Yes, they have the same dimension. One isomorphism is this.
  
  $(\begin{matrix} a & b & c \\ d & e & f \end{matrix}) \mapsto (\begin{matrix} a \\ ⋮ \\ f \end{matrix})$
4. Yes, they have the same dimension. This is an isomorphism.
  
  $a + b x + \dots + f x^{5} \mapsto (\begin{matrix} a & b & c \\ d & e & f \end{matrix})$
5. Yes, both have dimension $2 k$ .
Exercise 2.11 Worked answer

Which of these spaces are isomorphic to each other?
1. $ℝ^{3}$
2. $ℳ_{2 \times 2}$
3. $𝒫_{3}$
4. $ℝ^{4}$
5. $𝒫_{2}$
Back to Exercise 2.11

Answer. Just find the dimension of each space, for instance by finding a basis, and then spaces with the same dimension are isomorphic. This lists the dimension of each space.
1. $3$
2. $4$
3. $4$
4. $4$
5. $3$
Exercise 2.12 Worked answer

Recommended. Consider the isomorphism ${Rep}_{B} (\cdot) : 𝒫_{1} \to ℝ^{2}$ where $B = ⟨ 1, 1 + x ⟩$ . Find the image of each of these elements of the domain.
1. $3 - 2 x$ ;
2. $2 + 2 x$ ;
3. $x$
Back to Exercise 2.12

Answer.
1. ${Rep}_{B} (3 - 2 x) = (\begin{matrix} 5 \\ - 2 \end{matrix})$
2. $(\begin{matrix} 0 \\ 2 \end{matrix})$
3. $(\begin{matrix} - 1 \\ 1 \end{matrix})$
Exercise 2.13 Worked answer

For which $n$ is the space isomorphic to $ℝ^{n}$ ?
1. $𝒫_{4}$
2. $𝒫_{1}$
3. $ℳ_{2 \times 3}$
4. the plane $2 x - y + z = 0$ subset of $ℝ^{3}$
5. the vector space of linear combinations of three letters ${a x + b y + c z ∣ a, b, c \in ℝ}$
Back to Exercise 2.13

Answer. For each, the simplest thing is to find the dimension of the space by finding a basis. For each basis given below, We will omit the verification that it is a basis.
1. It is isomorphic to $ℝ^{5}$ . One basis for $𝒫_{4}$ is ${x^{4}, x^{3}, x^{2}, x, 1}$ so the space has dimension $5$ .
2. It is isomorphic to $ℝ^{2}$ since one basis for the space $𝒫_{1} = {a + b x ∣ a, b \in ℝ}$ is ${1, x}$ .
3. It is isomorphic to $ℝ^{6}$ . One basis has these six matrices.
  
  $(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}), \dots (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix})$
4. It is a plane so it is isomorphic to $ℝ^{2}$ . For a more extensive answer, parametrizing the plane gives this vector description
  
  ${(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 / 2 \\ 1 \\ 0 \end{matrix}) y + (\begin{matrix} - 1 / 2 \\ 0 \\ 1 \end{matrix}) z ∣ y, z \in ℝ}$
  
  and so it has a basis consisting of those two vectors.
5. It is isomorphic to $ℝ^{3}$ . One basis is the set of linear combinations ${x, y, z}$ , that is, ${x + 0 y + 0 z, 0 x + y + 0 z, 0 x + 0 y + z}$ .
Exercise 2.14 Worked answer

Recommended. Show that if $m \neq n$ then $ℝ^{m} ≇ ℝ^{n}$ .

Back to Exercise 2.14

Answer. They have different dimensions.
Exercise 2.15 Worked answer

Recommended. Is $ℳ_{m \times n} ≅ ℳ_{n \times m}$ ?

Back to Exercise 2.15

Answer. Yes, both are $m n$ -dimensional.
Exercise 2.16 Worked answer

Recommended. Are any two planes through the origin in $ℝ^{3}$ isomorphic?

Back to Exercise 2.16

Answer. Yes, any two (nondegenerate) planes are both two-dimensional vector spaces.
Exercise 2.17 Worked answer

Find a set of equivalence class representatives other than the set of $ℝ^{n}$ ’s.

Back to Exercise 2.17

Answer. There are many answers, one is the set of $𝒫_{k}$ (taking $𝒫_{- 1}$ to be the trivial vector space).
Exercise 2.18 Worked answer

True or false: between any $n$ -dimensional space and $ℝ^{n}$ there is exactly one isomorphism.

Back to Exercise 2.18

Answer. False (except when $n = 0$ ). For instance, if $f : V \to ℝ^{n}$ is an isomorphism then multiplying by any nonzero scalar, gives another, different, isomorphism. (Between trivial spaces the isomorphisms are unique; the only map possible is ${\vec{0}}_{V} \mapsto 0_{W}$ .)
Exercise 2.19 Worked answer

Can a vector space be isomorphic to one of its proper subspaces?

Back to Exercise 2.19

Answer. No. A proper subspace has a strictly lower dimension than it’s superspace; if $U$ is a proper subspace of $V$ then any linearly independent subset of $U$ must have fewer than $\dim (V)$ members or else that set would be a basis for $V$ , and $U$ wouldn’t be proper.
Exercise 2.20 Worked answer

Recommended. This subsection shows that for any isomorphism, the inverse map is also an isomorphism. This subsection also shows that for a fixed basis $B$ of an $n$ -dimensional vector space $V$ , the map ${Rep}_{B} : V \to ℝ^{n}$ is an isomorphism. Find the inverse of this map.

Back to Exercise 2.20

Answer. Where $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ , the inverse is this.

$(\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix}) \mapsto c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$
Exercise 2.21 Worked answer

Recommended. Prove these facts about matrices.
1. The row space of a matrix is isomorphic to the column space of its transpose.
2. The row space of a matrix is isomorphic to its column space.
Back to Exercise 2.21

Answer. All three spaces have dimension equal to the rank of the matrix.
Exercise 2.22 Worked answer

Show that the function from Theorem 2.3 is well-defined.

Back to Exercise 2.22

Answer. We must show that if $\vec{a} = \vec{b}$ then $f (\vec{a}) = f (\vec{b})$ . So suppose that $a_{1} {\vec{β}}_{1} + \dots + a_{n} {\vec{β}}_{n} = b_{1} {\vec{β}}_{1} + \dots + b_{n} {\vec{β}}_{n}$ . Each vector in a vector space (here, the domain space) has a unique representation as a linear combination of basis vectors, so we can conclude that $a_{1} = b_{1}$ , …, $a_{n} = b_{n}$ . Thus,

$f (\vec{a}) = (\begin{matrix} a_{1} \\ ⋮ \\ a_{n} \end{matrix}) = (\begin{matrix} b_{1} \\ ⋮ \\ b_{n} \end{matrix}) = f (\vec{b})$

and so the function is well-defined.
Exercise 2.23 Worked answer

Is the proof of Theorem 2.3 valid when $n = 0$ ?

Back to Exercise 2.23

Answer. Yes, because a zero-dimensional space is a trivial space.
Exercise 2.24 Worked answer

For each, decide if it is a set of isomorphism class representatives.
1. ${ℂ^{k} ∣ k \in ℕ}$
2. ${𝒫_{k} ∣ k \in {- 1, 0, 1, \dots}}$
3. ${ℳ_{m \times n} ∣ m, n \in ℕ}$
Back to Exercise 2.24

Answer.
1. No, this collection has no spaces of odd dimension.
2. Yes, because $𝒫_{k} ≅ ℝ^{k + 1}$ .
3. No, for instance, $ℳ_{2 \times 3} ≅ ℳ_{3 \times 2}$ .
Exercise 2.25 Worked answer

Let $f$ be a correspondence between vector spaces $V$ and $W$ (that is, a map that is one-to-one and onto). Show that the spaces $V$ and $W$ are isomorphic via $f$ if and only if there are bases $B \subset V$ and $D \subset W$ such that corresponding vectors have the same coordinates: ${Rep}_{B} (\vec{v}) = {Rep}_{D} (f (\vec{v}))$ .

Back to Exercise 2.25

Answer. One direction is easy: if the two are isomorphic via $f$ then for any basis $B \subseteq V$ , the set $D = f (B)$ is also a basis (this is shown in Lemma 2.4). The check that corresponding vectors have the same coordinates: $f (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) = c_{1} f ({\vec{β}}_{1}) + \dots + c_{n} f ({\vec{β}}_{n}) = c_{1} {\vec{δ}}_{1} + \dots + c_{n} {\vec{δ}}_{n}$ is routine.

For the other half, assume that there are bases such that corresponding vectors have the same coordinates with respect to those bases. Because $f$ is a correspondence, to show that it is an isomorphism, we need only show that it preserves structure. Because ${Rep}_{B} (\vec{v}) = {Rep}_{D} (f (\vec{v}))$ , the map $f$ preserves structure if and only if representations preserve addition: ${Rep}_{B} ({\vec{v}}_{1} + {\vec{v}}_{2}) = {Rep}_{B} ({\vec{v}}_{1}) + {Rep}_{B} ({\vec{v}}_{2})$ and scalar multiplication: ${Rep}_{B} (r \cdot \vec{v}) = r \cdot {Rep}_{B} (\vec{v})$ The addition calculation is this: $(c_{1} + d_{1}) {\vec{β}}_{1} + \dots + (c_{n} + d_{n}) {\vec{β}}_{n} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n} + d_{1} {\vec{β}}_{1} + \dots + d_{n} {\vec{β}}_{n}$ , and the scalar multiplication calculation is similar.
Exercise 2.26 Worked answer

Consider the isomorphism ${Rep}_{B} : 𝒫_{3} \to ℝ^{4}$ .
1. Vectors in a real space are orthogonal if and only if their dot product is zero. Give a definition of orthogonality for polynomials.
2. The derivative of a member of $𝒫_{3}$ is in $𝒫_{3}$ . Give a definition of the derivative of a vector in $ℝ^{4}$ .
Back to Exercise 2.26

Answer.
1. Pulling the definition back from $ℝ^{4}$ to $𝒫_{3}$ gives that $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}$ is orthogonal to $b_{0} + b_{1} x + b_{2} x^{2} + b_{3} x^{3}$ if and only if $a_{0} b_{0} + a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = 0$ .
2. A natural definition is this.
  
  $D ((\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix})) = (\begin{matrix} a_{1} \\ 2 a_{2} \\ 3 a_{3} \\ 0 \end{matrix})$
Exercise 2.27 Worked answer

Recommended. Does every correspondence between bases, when extended to the spaces, give an isomorphism? That is, suppose that $V$ is a vector space with basis $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ and that $f : B \to W$ is a correspondence such that $D = ⟨ f ({\vec{β}}_{1}), \dots, f ({\vec{β}}_{n}) ⟩$ is basis for $W$ . Must $\hat{f} : V \to W$ sending $\vec{v} = c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ to $\hat{f} (\vec{v}) = c_{1} f ({\vec{β}}_{1}) + \dots + c_{n} f ({\vec{β}}_{n})$ be an isomorphism?

Back to Exercise 2.27

Answer. Yes.

First, $\hat{f}$ is well-defined because every member of $V$ has one and only one representation as a linear combination of elements of $B$ .

Second we must show that $\hat{f}$ is one-to-one and onto. It is one-to-one because every member of $W$ has only one representation as a linear combination of elements of $D$ , since $D$ is a basis. And $\hat{f}$ is onto because every member of $W$ has at least one representation as a linear combination of members of $D$ .

Finally, preservation of structure is routine to check. For instance, here is the preservation of addition calculation.

$\begin{matrix} \hat{f} ((c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) + (d_{1} {\vec{β}}_{1} + \dots + d_{n} {\vec{β}}_{n})) \\ \begin{aligned} = \hat{f} ((c_{1} + d_{1}) {\vec{β}}_{1} + \dots + (c_{n} + d_{n}) {\vec{β}}_{n}) \\ = (c_{1} + d_{1}) f ({\vec{β}}_{1}) + \dots + (c_{n} + d_{n}) f ({\vec{β}}_{n}) \\ = c_{1} f ({\vec{β}}_{1}) + \dots + c_{n} f ({\vec{β}}_{n}) + d_{1} f ({\vec{β}}_{1}) + \dots + d_{n} f ({\vec{β}}_{n}) \\ = \hat{f} (c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}) + + \hat{f} (d_{1} {\vec{β}}_{1} + \dots + d_{n} {\vec{β}}_{n}) \end{aligned} \end{matrix}$

(The second equality is the definition of $\hat{f}$ .) Preservation of scalar multiplication is similar.
Exercise 2.28 Worked answer

(Requires the subsection on Combining Subspaces, which is optional.) Suppose that $V = V_{1} \oplus V_{2}$ and that $V$ is isomorphic to the space $U$ under the map $f$ . Show that $U = f (V_{1}) \oplus f (V_{2})$ .

Back to Exercise 2.28

Answer. Because $V_{1} \cap V_{2} = {{\vec{0}}_{V}}$ and $f$ is one-to-one we have that $f (V_{1}) \cap f (V_{2}) = {{\vec{0}}_{U}}$ . To finish, count the dimensions: $\dim (U) = \dim (V) = \dim (V_{1}) + \dim (V_{2}) = \dim (f (V_{1})) + \dim (f (V_{2}))$ , as required.
Exercise 2.29 Worked answer

Show that this is not a well-defined function from the rational numbers to the integers: with each fraction, associate the value of its numerator.

Back to Exercise 2.29

Answer. Rational numbers have many representations, e.g., $1 / 2 = 3 / 6$ , and the numerators can vary among representations.

References cited in this section

More information on correspondences is in the appendix.↩︎
More information on inverse functions is in the appendix.↩︎
More information on well-defined is in the appendix.↩︎