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2.4 Invertibility and Isomorphisms

The concept of invertibility is introduced quite early in the study of functions. Fortunately, many of the intrinsic properties of functions are shared by their inverses. For example, in calculus we learn that the properties of being continuous or differentiable are generally retained by the inverse functions. We see in this section (Theorem 2.17) that the inverse of a linear transformation is also linear. This result greatly aids us in the study of inverses of matrices. As one might expect from Section 2.3, the inverse of the left-multiplication transformation LA $L_{A}$ (when it exists) can be used to determine properties of the inverse of the matrix A.

In the remainder of this section, we apply many of the results about in- vertibility to the concept of isomorphism. We will see that finite-dimensional vector spaces (over F) of equal dimension may be identified. These ideas will be made precise shortly.

The facts about inverse functions presented in Appendix B are, of course, true for linear transformations. Nevertheless, we repeat some of the definitions for use in this section.

Definition.

Let V and W be vector spaces, and let T:V→W $T : V \to W$ be linear. A function U:W→V $U : W \to V$ is said to be an inverse of T if TU=IW $TU = I_{W}$ and UT=IV $UT = I_{V}$ . If T has an inverse, then T is said to be invertible. As noted in Appendix B, if T is invertible, then the inverse of T is unique and is denoted by T−1 $T^{- 1}$ .

The following facts hold for invertible functions T and U.

(TU)−1=U−1T−1 ${(TU)}^{- 1} = U^{- 1} T^{- 1}$ .
(T−1)−1=T ${(T^{- 1})}^{- 1} = T$ in particular, T−1 $T^{- 1}$ is invertible.

We often use the fact that a function is invertible if and only if it is both one-to-one and onto. We can therefore restate Theorem 2.5 as follows.
Let T:V→W $T : V \to W$ be a linear transformation, where V and W are finite-dimensional spaces of equal dimension. Then T is invertible if and only if rank(T)=dim(V) $rank (T) = dim (V)$ .

Example 1

Let T:P1(R)→R2 $T : P_{1} (R) \to R^{2}$ be the linear transformation defined by T(a+bx)=(a, a+b) $T (a + b x) = (a, a + b)$ . The reader can verify directly that T−1:R2→P1(R) $T^{- 1} : R^{2} \to P_{1} (R)$ is defined by T−1(c, d)=c+(d−c)x $T^{- 1} (c, d) = c + (d - c) x$ . Observe that T−1 $T^{- 1}$ is also linear. As Theorem 2.17 demonstrates, this is true in general.

Theorem 2.17.

Let V and W be vector spaces, and let T:V→W $T : V \to W$ be linear and invertible. Then T−1:W→V $T^{- 1} : W \to V$ is linear.

Proof.

Let y1, y2∈W $y_{1}, y_{2} \in W$ and c∈F $c \in F$ . Since T is onto and one-to-one, there exist unique vectors x1 $x_{1}$ and x2 $x_{2}$ such that T(x1)=y1 $T (x_{1}) = y_{1}$ and T(x2)=y2 $T (x_{2}) = y_{2}$ . Thus x1=T−1(y1) $x_{1} = T^{- 1} (y_{1})$ and x2=T−1(y2) $x_{2} = T^{- 1} (y_{2})$ so

T - 1 (c y 1 + y 2) = T - 1 [c T (x 1) + T (x 2)] = T - 1 [T (c x 1 + x 2)] = c x 1 + x 2 = c T - 1 (y 1) + T - 1 (y 2) .

$\begin{array}{l} T^{- 1} (c y_{1} + y_{2}) & = T^{- 1} [c T (x_{1}) + T (x_{2})] = T^{- 1} [T (c x_{1} + x_{2})] \\ {= c x}_{1} + x_{2} = c T^{- 1} (y_{1}) + T^{- 1} (y_{2}) . \end{array}$

Corollary.

Let T be an invertible linear transformation from V to W. Then V is finite-dimensional if and only if W is finite-dimensional. In this case, dim(V)=dim(W) $dim (V) = dim (W)$ .

Proof.

Suppose that V is finite-dimensional. Let β={x1, x2, …, xn} $β = {x_{1}, x_{2}, \dots, x_{n}}$ be a basis for V. By Theorem 2.2 (p. 68), T(β) $T (β)$ spans R(T)=W $R (T) = W$ hence W is finite-dimensional by Theorem 1.9 (p. 45). Conversely, if W is finite-dimensional, then so is V by a similar argument, using T−1 $T^{- 1}$ .

Now suppose that V and W are finite-dimensional. Because T is one-to-one and onto, we have

nullity (T) = 0 and rank (T) = dim (R (T)) = dim (W) .

$nullity (T) = 0 and rank (T) = dim (R (T)) = dim (W) .$

So by the dimension theorem (p. 70), it follows that dim(V)=dim(W) $dim (V) = dim (W)$ .

It now follows immediately from Theorem 2.5 (p. 71) that if T is a linear transformation between vector spaces of equal (finite) dimension, then the conditions of being invertible, one-to-one, and onto are all equivalent.

We are now ready to define the inverse of a matrix. The reader should note the analogy with the inverse of a linear transformation.

Definition.

Let A be an n×n $n \times n$ matrix. Then A is invertible if there exists an n×n $n \times n$ matrix B such that AB=BA=I $A B = B A = I$ .

If A is invertible, then the matrix B such that AB=BA=I $A B = B A = I$ is unique. (If C were another such matrix, then C=CI=C(AB)=(CA)B=IB=B $C = C I = C (A B) = (C A) B = I B = B$ .) The matrix B is called the inverse of A and is denoted by A−1 $A^{- 1}$ .

Example 2

The reader should verify that the inverse of

(5273) is (3 - 2 - 7 5) .

$(\begin{matrix} 5 & 7 \\ 2 & 3 \end{matrix}) is (\begin{array}{r} 3 & - 7 \\ - 2 & 5 \end{array}) .$

In Section 3.2, we will learn a technique for computing the inverse of a matrix. At this point, we develop a number of results that relate the inverses of matrices to the inverses of linear transformations.

Theorem 2.18.

Let V and W be finite-dimensional vector spaces with ordered bases β $β$ and γ $γ$ , respectively. Let T:V→W $T : V \to W$ be linear. Then T is invertible if and only if [T]γβ ${[T]}_{β}^{γ}$ is invertible. Furthermore, [T−1]βγ=([T]γβ)−1 ${[T^{- 1}]}_{γ}^{β} = {({[T]}_{β}^{γ})}^{- 1}$ .

Proof.

Suppose that T is invertible. By the Corollary to Theorem 2.17, we have dim(V)=dim(W) $dim (V) = dim (W)$ . Let n=dim(V) $n = dim (V)$ . So [T]γβ ${[T]}_{β}^{γ}$ is an n×n $n \times n$ matrix. Now T−1:W→V $T^{- 1} : W \to V$ satisfies TT−1=IW ${TT}^{- 1} = I_{W}$ and T−1T=IV $T^{- 1} T = I_{V}$ . Thus

I n = [I V] β = [T - 1 T] β = [T - 1] β γ [T] γ β .

$I_{n} = {[I_{V}]}_{β} = {[T^{- 1} T]}_{β} = {[T^{- 1}]}_{γ}^{β} {[T]}_{β}^{γ} .$

Similarly, [T]γβ[T−1]βγ=In ${[T]}_{β}^{γ} {[T^{- 1}]}_{γ}^{β} = I_{n}$ . So [T]γβ ${[T]}_{β}^{γ}$ is invertible and ([T]γβ)−1=[T−1]βγ ${({[T]}_{β}^{γ})}^{- 1} = {[T^{- 1}]}_{γ}^{β}$ .

Now suppose that A=[T]γβ $A = {[T]}_{β}^{γ}$ is invertible. Then there exists an n×n $n \times n$ matrix B such that AB=BA=In $A B = B A = I_{n}$ . By Theorem 2.6 (p. 73), there exists U∈L(W, V) $U \in L (W, V)$ such that

U (w j) = \sum i = 1 n B i j v i for j = 1, 2, \dots, n,

$U (w_{j}) = \sum_{i = 1}^{n} B_{i j} v_{i} for j = 1, 2, \dots, n,$

where γ={w1, w2, …, wn} $γ = {w_{1}, w_{2}, \dots, w_{n}}$ and β={v1, v2, …, vn} $β = {v_{1}, v_{2}, \dots, v_{n}}$ . It follows that [U]βγ=B ${[U]}_{γ}^{β} = B$ . To show that U=T−1 $U = T^{- 1}$ observe that

[UT] β = [U] β γ [T] γ β = B A = I n = [I V] β

${[UT]}_{β} = {[U]}_{γ}^{β} {[T]}_{β}^{γ} = B A = I_{n} = {[I_{V}]}_{β}$

by Theorem 2.11 (p. 89). So UT=IV $UT = I_{V}$ , and similarly, TU=IW $TU = I_{W}$

Example 3

Let β $β$ and γ $γ$ be the standard ordered bases of P1(R) $P_{1} (R)$ and R2 $R^{2}$ , respectively. For T as in Example 1, we have

[T] γ β = (1101) and [T - 1] β γ = (1 - 1 01) .

${[T]}_{β}^{γ} = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) and {[T^{- 1}]}_{γ}^{β} = (\begin{array}{r} 1 & 0 \\ - 1 & 1 \end{array}) .$

It can be verified by matrix multiplication that each matrix is the inverse of the other.

Corollary 1.

Let V be a finite-dimensional vector space with an ordered basis β $β$ , and let T:V→V $T : V \to V$ be linear. Then T is invertible if and only if [T]β ${[T]}_{β}$ is invertible. Furthermore, [T−1]β=([T]β)−1 ${[T^{- 1}]}_{β} = {({[T]}_{β})}^{- 1}$ .

Proof.

Exercise.

Corollary 2.

Let A be an n×n $n \times n$ matrix. Then A is invertible if and only if LA $L_{A}$ is invertible. Furthermore, (LA)−1=LA−1 ${(L_{A})}^{- 1} = L_{A^{- 1}}$ .

Proof.

Exercise.

The notion of invertibility may be used to formalize what may already have been observed by the reader, that is, that certain vector spaces strongly resemble one another except for the form of their vectors. For example, in the case of M2×2(F) $M_{2 \times 2} (F)$ and F4 $F^{4}$ , if we associate to each matrix

(a c b d)

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

the 4-tuple (a, b, c, d), we see that sums and scalar products associate in a similar manner; that is, in terms of the vector space structure, these two vector spaces may be considered identical or isomorphic.

Definitions.

Let V and W be vector spaces. We say that V is isomorphic to W if there exists a linear transformation T:V→W $T : V \to W$ that is invertible. Such a linear transformation is called an isomorphism from V onto W.

We leave as an exercise (see Exercise 13) the proof that “is isomorphic to” is an equivalence relation. (See Appendix A.) So we need only say that V and W are isomorphic.

Example 4

Define T:F2→P1(F) $T : F^{2} \to P_{1} (F)$ by T(a1, a2)=a1+a2x $T (a_{1}, a_{2}) = a_{1} + a_{2} x$ . It is easily checked that T is an isomorphism; so F2 $F^{2}$ is isomorphic to P1(F) $P_{1} (F)$ .

Example 5

Define

T : P 3 (R) \to M 2 \times 2 (R) by T (f) = (f (1) f (3) f (2) f (4)) .

$T : P_{3} (R) \to M_{2 \times 2} (R) by T (f) = (\begin{matrix} f (1) & f (2) \\ f (3) & f (4) \end{matrix}) .$

It is easily verified that T is linear. By use of the Lagrange interpolation formula in Section 1.6, it can be shown (compare with Exercise 22) that T(f)=O $T (f) = O$ only when f is the zero polynomial. Thus T is one-to-one (see Exercise 11). Moreover, because dim(P3(R))=dim(M2×2(R)) $dim (P_{3} (R)) = dim (M_{2 \times 2} (R))$ , it follows that T is invertible by Theorem 2.5 (p. 71). We conclude that P3(R) $P_{3} (R)$ is isomorphic to M2×2(R) $M_{2 \times 2} (R)$ .

In each of Examples 4 and 5, the reader may have observed that isomor-phic vector spaces have equal dimensions. As the next theorem shows, this is no coincidence.

Theorem 2.19.

Let V and W be finite-dimensional vector spaces (over the same field). Then V is isomorphic to W if and only if dim(V)=dim(W) $dim (V) = dim (W)$ .

Proof.

Suppose that V is isomorphic to W and that T:V→W $T : V \to W$ is an isomorphism from V to W. By the lemma preceding Theorem 2.18, we have that dim(V)=dim(W) $dim (V) = dim (W)$ .

Now suppose that dim(V)=dim(W) $dim (V) = dim (W)$ , and let β={v1, v2, …, vn} $β = {v_{1}, v_{2}, \dots, v_{n}}$ and γ={w1, w2, …, wn} $γ = {w_{1}, w_{2}, \dots, w_{n}}$ be bases for V and W, respectively. By Theorem 2.6 (p. 73), there exists T:V→W $T : V \to W$ such that T is linear and T(vi)=wi $T (v_{i}) = w_{i}$ for i=1, 2, …, n $i = 1, 2, \dots, n$ Using Theorem 2.2 (p. 68), we have

R (T) = span (T (β)) = span (γ) = W .

$R (T) = span (T (β)) = span (γ) = W .$

So T is onto. From Theorem 2.5 (p. 71), we have that T is also one-to-one. Hence T is an isomorphism.

By the lemma to Theorem 2.18, if V and W are isomorphic, then either both of V and W are finite-dimensional or both are infinite-dimensional.

Corollary.

Let V be a vector space over F. Then V is isomorphic to Fn $F^{n}$ if and only if dim(V)=n $dim (V) = n$ .

Up to this point, we have associated linear transformations with their matrix representations. We are now in a position to prove that, as a vector space, the collection of all linear transformations between two given vector spaces may be identified with the appropriate vector space of m×n $m \times n$ matrices.

Theorem 2.20.

Let V and W be finite-dimensional vector spaces over F of dimensions n and m, respectively, and let β $β$ and γ $γ$ be ordered bases for V and W, respectively. Then the function Φγβ:L(V, W)→Mm×n(F) $Φ_{β}^{γ} : L (V, W) \to M_{m \times n} (F)$ , defined by Φγβ(T)=[T]γβ $Φ_{β}^{γ} (T) = {[T]}_{β}^{γ}$ for T∈L(V, W) $T \in L (V, W)$ , is an isomorphism.

Proof.

By Theorem 2.8 (p. 83), Φγβ $Φ_{β}^{γ}$ is linear. Hence we must show that Φγβ $Φ_{β}^{γ}$ is one-to-one and onto. This is accomplished if we show that for every m×n $m \times n$ matrix A, there exists a unique linear transformation T:V→W $T : V \to W$ such that Φγβ(T)=A $Φ_{β}^{γ} (T) = A$ . Let β={v1, v2, …, vn}, γ={w1, w2, …, wm} $β = {v_{1}, v_{2}, \dots, v_{n}}, γ = {w_{1}, w_{2}, \dots, w_{m}}$ , and let A be a given m×n $m \times n$ matrix. By Theorem 2.6 (p. 73), there exists a unique linear transformation T:V→W $T : V \to W$ such that

T (v j) = \sum i = 1 m A i j w i for 1 \leq j \leq n .

$T (v_{j}) = \sum_{i = 1}^{m} A_{i j} w_{i} for 1 \leq j \leq n .$

But this means that [T]γβ=A, or Φγβ(T)=A ${[T]}_{β}^{γ} = A, or Φ_{β}^{γ} (T) = A$ . So Φγβ $Φ_{β}^{γ}$ is an isomorphism.

Corollary.

Let V and W be finite-dimensional vector spaces of dimensions n and m, respectively. Then L(V, W) $L (V, W)$ is finite-dimensional of dimension mn.

Proof.

The proof follows from Theorems 2.20 and 2.19 and the fact that dim(Mm×n(F))=mn $dim (M_{m \times n} (F)) = m n$ .

We conclude this section with a result that allows us to see more clearly the relationship between linear transformations defined on abstract finite- dimensional vector spaces and linear transformations from Fn $F^{n}$ to Fm $F^{m}$ .

We begin by naming the transformation x→[x]β $x \to {[x]}_{β}$ introduced in Section 2.2.

Definition.

Let β $β$ be an ordered basis for an n-dimensional vector space V over the field F. The standard representation of V with respect to β $β$ is the function ϕβ:V→Fn $ϕ_{β} : V \to F^{n}$ defined by ϕβ(x)=[x]β $ϕ_{β} (x) = {[x]}_{β}$ for each x∈V $x \in V$ .

Example 6

Let β={(1, 0), (0, 1)} $β = {(1, 0), (0, 1)}$ and γ={(1, 2), (3, 4)} $γ = {(1, 2), (3, 4)}$ . It is easily observed that β $β$ and γ $γ$ are ordered bases for R2 $R^{2}$ . For x=(1, −2) $x = (1, - 2)$ , we have

ϕ β (x) = [x] β = (1 - 2) and ϕ γ (x) = [x] γ = (- 5 2) .

$ϕ_{β} (x) = {[x]}_{β} = (\begin{array}{r} 1 \\ - 2 \end{array}) and ϕ_{γ} (x) = {[x]}_{γ} = (\begin{array}{r} - 5 \\ 2 \end{array}) .$

We observed earlier that ϕβ $ϕ_{β}$ is a linear transformation. The next theorem tells us much more.

Theorem 2.21.

For any finite-dimensional vector space V with ordered basis β, ϕβ $β, ϕ_{β}$ is an isomorphism.

Proof.

Exercise.

This theorem provides us with an alternate proof that an n-dimensional vector space is isomorphic to Fn $F^{n}$ (see the corollary to Theorem 2.19).

Let V and W be vector spaces of dimension n and m, respectively, and let T:V→W $T : V \to W$ be a linear transformation. Define A=[T]γβ $A = {[T]}_{β}^{γ}$ , where β $β$ and γ $γ$ are arbitrary ordered bases of V and W, respectively. We are now able to use ϕβ $ϕ_{β}$ and ϕγ $ϕ_{γ}$ to study the relationship between the linear transformations T and LA:Fn→Fm $L_{A} : F^{n} \to F^{m}$ .

Let us first consider Figure 2.2. Notice that there are two composites of linear transformations that map V into Fm $F^{m}$ :

Map V into Fn $F^{n}$ with ϕβ $ϕ_{β}$ and follow this transformation with LA $L_{A}$ this yields the composite LAϕβ $L_{A} ϕ_{β}$ .
Map V into W with T and follow it by ϕγ $ϕ_{γ}$ to obtain the composite ϕγT $ϕ_{γ} T$ .

A rectangular diagram of two composites of linear transformations that map V into F. — Figure 2.2

Figure 2.2 Full Alternative Text

These two composites are depicted by the dashed arrows in the diagram. By a simple reformulation of Theorem 2.14 (p. 92), we may conclude that

L A ϕ β = ϕ γ T;

$L_{A} ϕ_{β} = ϕ_{γ} T;$

that is, the diagram “commutes.” Heuristically, this relationship indicates that after V and W are identified with Fn $F^{n}$ and Fm $F^{m}$ via ϕβ $ϕ_{β}$ and ϕγ $ϕ_{γ}$ , respectively, we may “identify” T with LA $L_{A}$ . This diagram allows us to transfer operations on abstract vector spaces to ones on Fn $F^{n}$ and Fm $F^{m}$ .

Example 7

Recall the linear transformation T:P3(R)→P2(R) $T : P_{3} (R) \to P_{2} (R)$ defined in Example 4 of Section 2.2 (T(f(x))=f′(x)) $(T (f (x)) = f^{'} (x))$ . Let β $β$ and γ $γ$ be the standard ordered bases for P3(R) $P_{3} (R)$ and P2(R) $P_{2} (R)$ , respectively, and let ϕβ:P3(R)→R4 $ϕ_{β} : P_{3} (R) \to R^{4}$ and ϕγ:P2(R)→R3 $ϕ_{γ} {:P}_{2} (R) \to R^{3}$ be the corresponding standard representations of P3(R) $P_{3} (R)$ and P2(R) $P_{2} (R)$ . If A=[T]γβ $A = {[T]}_{β}^{γ}$ , then

A = ⎛ ⎝ ⎜ 000100020003 ⎞ ⎠ ⎟ .

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

Consider the polynomial p(x)=2+x−3x2+5x3 $p (x) = 2 + x - 3 x^{2} + 5 x^{3}$ . We show that LAϕβ(p(x))=ϕγT(p(x)) $L_{A} ϕ_{β} (p (x)) = ϕ_{γ} T (p (x))$ . Now

L A ϕ β (p (x)) = ⎛ ⎝ ⎜ 000100020003 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎜ ⎜ 21 - 3 5 ⎞ ⎠ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ 1 - 6 15 ⎞ ⎠ ⎟ .

$L_{A} ϕ_{β} (p (x)) = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) (\begin{array}{r} 2 \\ 1 \\ - 3 \\ 5 \end{array}) = (\begin{array}{r} 1 \\ - 6 \\ 15 \end{array}) .$

But since T(p(x))=p′(x)=1−6x+15x2 $T (p (x)) = p^{'} (x) = 1 - 6 x + 15 x^{2}$ , we have

ϕ γ T (p (x)) = ⎛ ⎝ ⎜ 1 - 6 15 ⎞ ⎠ ⎟ .

$ϕ_{γ} T (p (x)) = (\begin{array}{r} 1 \\ - 6 \\ 15 \end{array}) .$

So LAϕβ(p(x))=ϕγT(p(x)) $L_{A} ϕ_{β} (p (x)) = ϕ_{γ} T (p (x))$ .

Try repeating Example 7 with different polynomials p(x).

Exercises

Label the following statements as true or false. In each part, V and W are vector spaces with ordered (finite) bases α $α$ and β $β$ , respectively, T:V→W $T : V \to W$ is linear, and A and B are matrices.
1. (a) ([T]βα)−1=[T−1]βα ${({[T]}_{α}^{β})}^{- 1} = {[T^{- 1}]}_{α}^{β}$
2. (b) T is invertible if and only if T is one-to-one and onto.
3. (c) T=LA, $T = L_{A},$ where A=[T]βα $A = {[T]}_{α}^{β}$
4. (d) M2×3(F) $M_{2 \times 3} (F)$ is isomorphic to F5 $F^{5}$ .
5. (e) Pn(F) $P_{n} (F)$ is isomorphic to Pm(F) $P_{m} (F)$ if and only if n=m $n = m$ .
6. (f) AB=I $A B = I$ implies that A and B are invertible.
7. (g) If A is invertible, then (A−1)−1=A ${(A^{- 1})}^{- 1} = A$ .
8. (h) A is invertible if and only if LA $L_{A}$ is invertible.
9. (i) A must be square in order to possess an inverse.
For each of the following linear transformations T, determine whether T is invertible and justify your answer.
1. (a) T:R2→R3 $T : R^{2} \to R^{3}$ defined by T(a1, a2)=(a1−2a2, a2, 3a1+4a2) $T (a_{1}, a_{2}) = (a_{1} - 2 a_{2}, a_{2}, 3 a_{1} + 4 a_{2})$ .
2. (b) T:R2→R3 $T : R^{2} \to R^{3}$ defined by T(a1, a2)=(3a1−a2, a2, 4a1) $T (a_{1}, a_{2}) = (3 a_{1} - a_{2}, a_{2}, 4 a_{1})$ .
3. (c) T:R3→R3 $T : R^{3} \to R^{3}$ defined by T(a1, a2, a3)=(3a1−2a3, a2, 3a1+4a2) $T (a_{1}, a_{2}, a_{3}) = (3 a_{1} - 2 a_{3}, a_{2}, 3 a_{1} + 4 a_{2})$ .
4. (d) T:P3(R)→P2(R) $T : P_{3} (R) \to P_{2} (R)$ defined by T(p(x))=p′(x) $T (p (x)) = p^{'} (x)$ .
5. (e) T:M2×2(R)→P2(R) $T : M_{2 \times 2} (R) \to P_{2} (R)$ defined by T(acbd)=a+2bx+(c+d)x2 $T (\begin{matrix} a & b \\ c & d \end{matrix}) = a + 2 b x + (c + d) x^{2}$ .
6. (f) T:M2×2(R)→M2×2(R) $T : M_{2 \times 2} (R) \to M_{2 \times 2} (R)$ defined by T(acbd)=(a+bcac+d) $T (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} a + b & a \\ c & c + d \end{matrix})$ .
Which of the following pairs of vector spaces are isomorphic? Justify your answers.
1. (a) F3 $F^{3}$ and P3(F) $P_{3} (F)$ .
2. (b) F4 $F^{4}$ and P3(F) $P_{3} (F)$ .
3. (c) M2×2(R) $M_{2 \times 2} (R)$ and P3(R) $P_{3} (R)$ .
4. (d) V={A∈M2×2(R):tr(A)=0} $V = {A \in M_{2 \times 2} (R) : tr (A) = 0}$ and R4 $R^{4}$ .
Let A and B be n×n $n \times n$ invertible matrices. Prove that AB is invertible and (AB)−1=B−1A−1 ${(A B)}^{- 1} = B^{- 1} A^{- 1}$ .
^† Let A be invertible. Prove that At $A^{t}$ is invertible and (At)−1=(A−1)t ${(A^{t})}^{- 1} = {(A^{- 1})}^{t}$ . Visit goo.gl/suFm6V for a solution.
Prove that if A is invertible and AB=O $A B = O$ , then B=O $B = O$ .
Let A be an n×n $n \times n$ matrix.
1. (a) Suppose that A2=O $A^{2} = O$ . Prove that A is not invertible.
2. (b) Suppose that AB=O $A B = O$ for some nonzero n×n $n \times n$ matrix B. Could A be invertible? Explain.
Prove Corollaries 1 and 2 of Theorem 2.18.
^† Let A and B be n×n $n \times n$ matrices such that AB is invertible.
1. (a) Prove that A and B are invertible. Hint: See Exercise 12 of Section 2.3.
2. (b) Give an example to show that a product of nonsquare matrices can be invertible even though the factors, by definition, are not.
^† Let A and B be n×n $n \times n$ matrices such that AB=In $A B = I_{n}$ .
1. (a) Use Exercise 9 to conclude that A and B are invertible.
2. (b) Prove A=B−1 $A = B^{- 1}$ (and hence B=A−1 $B = A^{- 1}$ ). (We are, in effect, saying that for square matrices, a “one-sided” inverse is a “two-sided” inverse.)
3. (c) State and prove analogous results for linear transformations defined on finite-dimensional vector spaces.
Verify that the transformation in Example 5 is one-to-one.
Prove Theorem 2.21.
Let ∼ $\sim$ mean “is isomorphic to.” Prove that ∼ $\sim$ is an equivalence relation on the class of vector spaces over F.
Let

$V = {(a 0 a + b c) : a, b, c \in F} .$ $V = {(\begin{matrix} a & a + b \\ 0 & c \end{matrix}) : a, b, c \in F} .$

Construct an isomorphism from V to F3 $F^{3}$ .
Let V and W be n-dimensional vector spaces, and let T:V→W $T : V \to W$ be a linear transformation. Suppose that β $β$ is a basis for V. Prove that T is an isomorphism if and only if T(β) $T (β)$ is a basis for W.
Let B be an n×n $n \times n$ invertible matrix. Define Φ:Mn×n(F)→Mn×n(F) $Φ : M_{n \times n} (F) \to M_{n \times n} (F)$ by Φ(A)=B−1AB $Φ (A) = B^{- 1} A B$ . Prove that Φ $Φ$ is an isomorphism.
^† Let V and W be finite-dimensional vector spaces and T:V→W $T : V \to W$ be an isomorphism. Let V0 $V_{0}$ be a subspace of V.
1. (a) Prove that T(V0) $T (V_{0})$ is a subspace of W.
2. (b) Prove that dim(V0)=dim(T(V0)) $dim (V_{0}) = dim (T (V_{0}))$ .
Repeat Example 7 with the polynomial p(x)=1+x+2x2+x3 $p (x) = 1 + x + 2 x^{2} + x^{3}$ .
In Example 5 of Section 2.1, the mapping T:M2×2(R)→M2×2(R) $T : M_{2 \times 2} (R) \to M_{2 \times 2} (R)$ defined by T(M)=Mt $T (M) = M^{t}$ for each M∈M2×2(R) $M \in M_{2 \times 2} (R)$ is a linear transformation. Let β={E11, E12, E21, E22} $β = {E^{11}, E^{12}, E^{21}, E^{22}}$ , which is a basis for M2×2(R) $M_{2 \times 2} (R)$ , as noted in Example 3 of Section 1.6.
1. (a) Compute [T]β ${[T]}_{β}$ .
2. (b) Verify that LAϕβ(M)=ϕβT(M) $L_{A} ϕ_{β} (M) = ϕ_{β} T (M)$ for A=[T]β $A = {[T]}_{β}$ and
  
  $M = (1324) .$ $M = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) .$
^† Let T:V→W $T : V \to W$ be a linear transformation from an n-dimensional vector space V to an m-dimensional vector space W. Let β $β$ and γ $γ$ be ordered bases for V and W, respectively. Prove that rank(T)=rank(LA) $rank (T) = rank (L_{A})$ and that nullity(T)=nullity(LA) $nullity (T) = nullity (L_{A})$ , where A=[T]γβ $A = {[T]}_{β}^{γ}$ . Hint: Apply Exercise 17 to Figure 2.2.
Let V and W be finite-dimensional vector spaces with ordered bases β={v1, v2, …, vn} $β = {v_{1}, v_{2}, \dots, v_{n}}$ and γ={w1, w2, …, wm} $γ = {w_{1}, w_{2}, \dots, w_{m}}$ , respectively. By Theorem 2.6 (p. 73), there exist linear transformations Tij:V→W $T_{i j} : V \to W$ such that

$T i j (v k) = {w i if k = j 0 if k \neq j .$ $T_{i j} (v_{k}) = {\begin{cases} w_{i} if k = j \\ 0 if k \neq j . \end{cases}$

First prove that {Tij:1≤i≤m, 1≤j≤n} ${T_{i j} : 1 \leq i \leq m, 1 \leq j \leq n}$ is a basis for L(V, W) $L (V, W)$ . Then let Mij $M^{i j}$ be the m×n $m \times n$ matrix with 1 in the ith row and jth column and 0 elsewhere, and prove that [Tij]γβ=Mij ${[T_{i j}]}_{β}^{γ} = M^{i j}$ . Again by Theorem 2.6, there exists a linear transformation Φγβ:L(V, W)→Mm×n(F) $Φ_{β}^{γ} : L (V, W) \to M_{m \times n} (F)$ such that Φγβ(Tij)=Mij $Φ_{β}^{γ} (T_{i j}) = M^{i j}$ . Prove that Φγβ $Φ_{β}^{γ}$ is an isomorphism.
Let c0, c1, …, cn $c_{0}, c_{1}, \dots, c_{n}$ be distinct scalars from an infinite field F. Define T:Pn(F)→Fn+1 $T : P_{n} (F) \to F^{n + 1}$ by T(f)=(f(c0), f(c1), …, f(cn)) $T (f) = (f (c_{0}), f (c_{1}), \dots, f (c_{n}))$ . Prove that T is an isomorphism. Hint: Use the Lagrange polynomials associated with c0, c1, …, cn $c_{0}, c_{1}, \dots, c_{n}$
Let W denote the vector space of all sequences in F that have only a finite number of nonzero terms (defined in Exercise 18 of Section 1.6), and let Z=P(F) $Z = P (F)$ . Define

$T : W \to Z by T (σ) = \sum i = 0 n σ (i) x i,$ $T : W \to Z by T (σ) = \sum_{i = 0}^{n} σ (i) x^{i},$

where n is the largest integer such that σ(n)≠0 $σ (n) \neq 0$ . Prove that T is an isomorphism.

The following exercise requires familiarity with the concept of quotient space defined in Exercise 31 of Section 1.3 and with Exercise 42 of Section 2.1.

Let V and Z be vector spaces and T:V→Z $T : V \to Z$ be a linear transformation that is onto. Define the mapping

$T ¯ ¯ ¯ : V / N (T) \to Z by T ¯ ¯ ¯ (v + N (T)) = T (v)$ $\bar{T} : V / N (T) \to Z by \bar{T} (v + N (T)) = T (v)$

for any coset v+N(T) $v + N (T)$ in V/N(T) $V / N (T)$ .
1. (a) Prove that T¯¯¯ $\bar{T}$ is well-defined; that is, prove that if v+N(T)=v′+N(T) $v + N (T) = v^{'} + N (T)$ , then T(v)=T(v′) $T (v) = T (v^{'})$ .
2. (b) Prove that T¯¯¯ $\bar{T}$ is linear.
3. (c) Prove that T¯¯¯ $\bar{T}$ is an isomorphism.
4. (d) Prove that the diagram shown in Figure 2.3 commutes; that is, prove that T=T¯¯¯η $T = \bar{T} η$ .
  
  Figure 2.3
  
  Figure 2.3 Full Alternative Text
Let V be a nonzero vector space over a field F, and suppose that S is a basis for V. (By the corollary to Theorem 1.13 (p. 61) in Section 1.7, every vector space has a basis.) Let C(S, F) denote the vector space of all functions f∈F(S, F) $f \in F (S, F)$ such that f(s)=0 $f (s) = 0$ for all but a finite number of vectors in S. (See Exercise 14 of Section 1.3.) Let Ψ:C(S, F)→V $Ψ : C (S, F) \to V$ be defined by Ψ(f)=0 $Ψ (f) = 0$ if f is the zero function, and

$Ψ (f) = \sum s \in S, f (s) \neq 0 f (s) s,$ $Ψ (f) = \sum_{s \in S, f (s) \neq 0} f (s) s,$

otherwise. Prove that Ψ $Ψ$ is an isomorphism. Thus every nonzero vector space can be viewed as a space of functions.

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Table of Contents for 2.4 Invertibility and Isomorphisms

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Table of Contents for
2.4 Invertibility and Isomorphisms