6.10.3E: Isomorphisms and Composition Exercises

Exercises for 1

Exercise \(\PageIndex{1}\)

Verify that each of the following is an isomorphism (Theorem [thm:022192] is useful).

1. \(T : \mathbb{R}^3 \to \mathbb{R}^3\); \(T(x, y, z) = (x + y, y + z, z + x)\)
2. \(T : \mathbb{R}^3 \to \mathbb{R}^3\); \(T(x, y, z) = (x, x + y, x + y + z)\)
3. \(T : \mathbb{C} \to \mathbb{C}\); \(T(z) = \overline{z}\)
4. \(T :\mathbf{M}_{mn} \to\mathbf{M}_{mn}\); \(T(X) = UXV\), \(U\) and \(V\) invertible
5. \(T :\mathbf{P}_{1} \to \mathbb{R}^2\); \(T\left[p(x)\right] = \left[p(0), p(1)\right]\)
6. \(T : V \to V\); \(T(\mathbf{v}) = k\mathbf{v}\), \(k \neq 0\) a fixed number, \(V\) any vector space
7. \(T :\mathbf{M}_{22} \to \mathbb{R}^4\); \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = (a + b, d, c, a - b)\)
8. \(T :\mathbf{M}_{mn} \to\mathbf{M}_{nm}\); \(T(A) = A^{T}\)

Answer

2. \(T\) is onto because \(T(1, -1, 0) = (1, 0, 0)\), \(T(0, 1, -1) = (0, 1, 0)\), and \(T(0, 0, 1) = (0, 0, 1)\). Use Theorem [thm:022192].
3. \(T\) is one-to-one because \(0 = T(X) = UXV\) implies that \(X = 0\) (\(U\) and \(V\) are invertible). Use Theorem [thm:022192].
4. \(T\) is one-to-one because \(\mathbf{0} = T(\mathbf{v}) = k\mathbf{v}\) implies that \(\mathbf{v} = \mathbf{0}\) (because \(k \neq 0\)).
5. \(T\) is onto because \(T\left(\frac{1}{k}\mathbf{v}\right) = \mathbf{v}\) for all \(\mathbf{v}\). [Here Theorem [thm:022192] does not apply if \(dim \;V\) is not finite.]
6. \(T\) is one-to-one because \(T(A) = 0\) implies \(A^{T} = 0\), whence \(A = 0\). Use Theorem [thm:022192].

Exercise \(\PageIndex{2}\)

Show that

\[\{a + bx + cx^{2}, a_{1} + b_{1}x + c_{1}x^{2}, a_{2} + b_{2}x + c_{2}x^{2}\} \nonumber \]

is a basis of \(\mathbf{P}_{2}\) if and only if
\(\{(a, b, c), (a_{1}, b_{1}, c_{1}), (a_{2}, b_{2}, c_{2})\}\) is a basis of \(\mathbb{R}^3\).

Exercise \(\PageIndex{3}\)

If \(V\) is any vector space, let \(V^{n}\) denote the space of all \(n\)-tuples \((\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n})\), where each \(\mathbf{v}_{i}\) lies in \(V\). (This is a vector space with component-wise operations; see Exercise [ex:ex6_1_17].) If \(C_{j}(A)\) denotes the \(j\)th column of the \(m \times n\) matrix \(A\), show that \(T :\mathbf{M}_{mn} \to (\mathbb{R}^m)^{n}\) is an isomorphism if
\(T(A) = \left[ \begin{array}{cccc} C_{1}(A) & C_{2}(A) & \cdots & C_{n}(A) \end{array} \right]\). (Here \(\mathbb{R}^m\) consists of columns.)

Exercise \(\PageIndex{4}\)

In each case, compute the action of \(ST\) and \(TS\), and show that \(ST \neq TS\).

1. \(S : \mathbb{R}^2 \to \mathbb{R}^2\) with \(S(x, y) = (y, x)\); \(T : \mathbb{R}^2 \to \mathbb{R}^2\) with \(T(x, y) = (x, 0)\)

2. \(S : \mathbb{R}^3 \to \mathbb{R}^3\) with \(S(x, y, z) = (x, 0, z)\);
   \(T : \mathbb{R}^3 \to \mathbb{R}^3\) with \(T(x, y, z) = (x + y, 0, y + z)\)

3. \(S :\mathbf{P}_{2} \to\mathbf{P}_{2}\) with \(S(p) = p(0) + p(1)x + p(2)x^{2}\); \(T :\mathbf{P}_{2} \to\mathbf{P}_{2}\) with \(T(a + bx + cx^{2}) = b + cx + ax^{2}\)

4. \(S :\mathbf{M}_{22} \to\mathbf{M}_{22}\) with \(S\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a & 0 \\ 0 & d \end{array} \right]\);
   \(T :\mathbf{M}_{22} \to\mathbf{M}_{22}\) with \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} c & a \\ d & b \end{array} \right]\)

Answer

2. \(ST(x, y, z) = (x + y, 0, y + z)\), \(TS(x, y, z) = (x, 0, z)\)
3. \(ST\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} c & 0 \\ 0 & d \end{array} \right]\), \(TS\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} 0 & a \\ d & 0 \end{array} \right]\)

Exercise \(\PageIndex{5}\)

In each case, show that the linear transformation \(T\) satisfies \(T^{2} = T\).

1. \(T : \mathbb{R}^4 \to \mathbb{R}^4\); \(T(x, y, z, w) = (x, 0, z, 0)\)
2. \(T : \mathbb{R}^2 \to \mathbb{R}^2\); \(T(x, y) = (x + y, 0)\)

3. \(T :\mathbf{P}_{2} \to\mathbf{P}_{2}\);
   \(T(a + bx + cx^{2}) = (a + b - c) + cx + cx^{2}\)

4. \(T :\mathbf{M}_{22} \to\mathbf{M}_{22}\);
   \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \frac{1}{2}\left[ \begin{array}{cc} a + c & b + d \\ a + c & b + d \end{array} \right]\)

Answer

2. \(T^{2}(x, y) = T(x + y, 0) = (x + y, 0) = T(x, y)\). Hence \(T^{2} = T\).
3. \(T^2 \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] = \frac{1}{2}T\left[ \begin{array}{cc} a + c & b + d \\ a + c & b + d \end{array} \right] = \frac{1}{2}\left[ \begin{array}{cc} a + c & b + d \\ a + c & b + d \end{array} \right]\)

Exercise \(\PageIndex{6}\)

Determine whether each of the following transformations \(T\) has an inverse and, if so, determine the action of \(T^{-1}\).

1. \(T : \mathbb{R}^3 \to \mathbb{R}^3\);
   \(T(x, y, z) = (x + y, y + z, z + x)\)

2. \(T : \mathbb{R}^4 \to \mathbb{R}^4\);
   \(T(x, y, z, t) = (x + y, y + z, z + t, t + x)\)

3. \(T :\mathbf{M}_{22} \to\mathbf{M}_{22}\);
   \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a - c & b - d \\ 2a - c & 2b - d \end{array} \right]\)

4. \(T :\mathbf{M}_{22} \to\mathbf{M}_{22}\);
   \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a + 2c & b + 2d \\ 3c - a & 3d - b \end{array} \right]\)

5. \(T :\mathbf{P}_{2} \to \mathbb{R}^3\); \(T(a + bx + cx^{2}) = (a - c, 2b, a + c)\)
6. \(T :\mathbf{P}_{2} \to \mathbb{R}^3\); \(T(p) = \left[p(0), p(1), p(-1)\right]\)

Answer

2. No inverse; \((1, -1, 1, -1)\) is in \(\text{ker }T\).
3. \(T^{-1}\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \frac{1}{5}\left[ \begin{array}{cc} 3a - 2c & 3b - 2d \\ a + c & b + d \end{array} \right]\)
4. \(T^{-1}(a, b, c) = \frac{1}{2}\left[2a + (b - c)x - (2a - b - c)x^{2}\right]\)

Exercise \(\PageIndex{7}\)

In each case, show that \(T\) is self-inverse, that is: \(T^{-1} = T\).

1. \(T : \mathbb{R}^4 \to \mathbb{R}^4\); \(T(x, y, z, w) = (x, -y, -z, w)\)
2. \(T : \mathbb{R}^2 \to \mathbb{R}^2\); \(T(x, y) = (ky - x, y)\), \(k\) any fixed number
3. \(T :\mathbf{P}_{n} \to\mathbf{P}_{n}\); \(T(p(x)) = p(3 - x)\)

4. \(T :\mathbf{M}_{22} \to\mathbf{M}_{22}\); \(T(X) = AX\) where
   \(A = \frac{1}{4}\left[ \begin{array}{rr} 5 & -3 \\ 3 & -5 \end{array} \right]\)

Answer

2. \(T^{2}(x, y) = T(ky - x, y) = (ky - (ky - x), y) = (x, y)\)
3. \(T^{2}(X) = A^{2}X = IX = X\)

Exercise \(\PageIndex{8}\)

In each case, show that \(T^{6} = 1_{R^4}\) and so determine \(T^{-1}\).

1. \(T : \mathbb{R}^4 \to \mathbb{R}^4\); \(T(x, y, z, w) = (-x, z, w, y)\)
2. \(T : \mathbb{R}^4 \to \mathbb{R}^4\); \(T(x, y, z, w) = (-y, x - y, z, -w)\)

Answer

b. \(T^{3}(x, y, z, w) = (x, y, z, -w)\) so \(T^{6}(x, y, z, w) = T^{3}\left[T^{3}(x, y, z, w)\right] = (x, y, z, w)\). Hence \(T^{-1} = T^{5}\). So \(T^{-1}(x, y, z, w) = (y - x, -x, z, -w)\).

Exercise \(\PageIndex{9}\)

In each case, show that \(T\) is an isomorphism by defining \(T^{-1}\) explicitly.

1. \(T :\mathbf{P}_{n} \to\mathbf{P}_{n}\) is given by \(T\left[p(x)\right] = p(x + 1)\).
2. \(T :\mathbf{M}_{nn} \to\mathbf{M}_{nn}\) is given by \(T(A) = UA\) where \(U\) is invertible in \(\mathbf{M}_{nn}\).

Answer

b. \(T^{-1}(A) = U^{-1} A\).

Exercise \(\PageIndex{10}\)

Given linear transformations
\(V \xrightarrow{T} W \xrightarrow{S} U\):

1. If \(S\) and \(T\) are both one-to-one, show that \(ST\) is one-to-one.
2. If \(S\) and \(T\) are both onto, show that \(ST\) is onto.

Answer

b. Given \(\mathbf{u}\) in \(U\), write \(\mathbf{u} = S(\mathbf{w})\), \(\mathbf{w}\) in \(W\) (because \(S\) is onto). Then write \(\mathbf{w} = T(\mathbf{v})\), \(\mathbf{v}\) in \(V\) (\(T\) is onto). Hence \(\mathbf{u} = ST(\mathbf{v})\), so \(ST\) is onto.

Exercise \(\PageIndex{11}\)

Let \(T : V \to W\) be a linear transformation.

1. If \(T\) is one-to-one and \(TR = TR_{1}\) for transformations \(R\) and \(R_{1} : U \to V\), show that \(R = R_{1}\).
2. If \(T\) is onto and \(ST = S_{1}T\) for transformations \(S\) and \(S_{1} : W \to U\), show that \(S = S_{1}\).

Exercise \(\PageIndex{12}\)

Consider the linear transformations \(V \xrightarrow{T} W \xrightarrow{R} U\).

1. Show that \(\text{ker }T \subseteq \text{ker }RT\).
2. Show that \(im \;RT \subseteq im \;R\).

Answer

b. For all \(\mathbf{v}\) in \(V\), \((RT)(\mathbf{v}) = R\left[T(\mathbf{v})\right]\) is in \(im \;(R)\).

Exercise \(\PageIndex{13}\)

Let \(V \xrightarrow{T} U \xrightarrow{S} W\) be linear transformations.

1. If \(ST\) is one-to-one, show that \(T\) is one-to-one and that \(dim \;V \leq dim \;U\).

2. If \(ST\) is onto, show that \(S\) is onto and that
   \(dim \;W \leq dim \;U\).

Answer

b. Given \(\mathbf{w}\) in \(W\), write \(\mathbf{w} = ST(\mathbf{v})\), \(\mathbf{v}\) in \(V\) (\(ST\) is onto). Then \(\mathbf{w} = S\left[T(\mathbf{v})\right]\), \(T(\mathbf{v})\) in \(U\), so \(S\) is onto. But then \(im \;S = W\), so \(dim \;U = dim \;(\text{ker }S) + dim \;(im \;S) \geq dim \;(im \;S) = dim \;W\).

Exercise \(\PageIndex{14}\)

Let \(T : V \to V\) be a linear transformation. Show that \(T^{2} = 1_{V}\) if and only if \(T\) is invertible and \(T = T^{-1}\)

Exercise \(\PageIndex{15}\)

Let \(N\) be a nilpotent \(n \times n\) matrix (that is, \(N^{k} = 0\) for some \(k\)). Show that \(T :\mathbf{M}_{nm} \to\mathbf{M}_{nm}\) is an isomorphism if \(T(X) = X - NX\). [Hint: If \(X\) is in \(\text{ker }T\), show that \(X = NX = N^{2}X = \cdots\). Then use Theorem \(\PageIndex{3}\).]

Exercise \(\PageIndex{16}\)

Let \(T : V \to W\) be a linear transformation, and let \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) be any basis of \(V\) such that \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) is a basis of \(\text{ker }T\). Show that \(im \;T \cong span \;\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}\}\). [Hint: See Theorem \(\PageIndex{3}\).]

Answer

\(\{T(\mathbf{e}_{1}), T(\mathbf{e}_{2}), \dots, T(\mathbf{e}_{r})\}\) is a basis of \(im \;T\) by Theorem [thm:021572]. So \(T : span \;\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}\} \to im \;T\) is an isomorphism by Theorem \(\PageIndex{3}\).

Exercise \(\PageIndex{17}\)

Is every isomorphism \(T :\mathbf{M}_{22} \to\mathbf{M}_{22}\) given by an invertible matrix \(U\) such that \(T(X) = UX\) for all \(X\) in \(\mathbf{M}_{22}\)? Prove your answer.

Exercise \(\PageIndex{18}\)

Let \(\mathbf{D}_{n}\) denote the space of all functions \(f\) from \(\{1, 2, \dots, n\}\) to \(\mathbb{R}\) (see Exercise [ex:ex6_3_35]). If \(T :\mathbf{D}_{n} \to \mathbb{R}^n\) is defined by

\[T(f) = (f(1), f(2), \dots, f(n)), \nonumber \]

show that \(T\) is an isomorphism

Exercise \(\PageIndex{19}\)

1. Let \(V\) be the vector space of Exercise6.1.3. Find an isomorphism \(T : V \to \mathbb{R}^1\).
2. Let \(V\) be the vector space of Exercise 6.1.4. Find an isomorphism \(T : V \to \mathbb{R}^2\).

Answer

b. \(T(x, y) = (x, y + 1)\)

Exercise \(\PageIndex{20}\)

Let \(V \xrightarrow{T} W \xrightarrow{S} V\) be linear transformations such that \(ST = 1_{V}\). If \(dim \;V = dim \;W = n\), show that \(S = T^{-1}\) and \(T = S^{-1}\). [Hint: Exercise \(\PageIndex{13}\) and Theorem \(\PageIndex{3}\), Theorem \(\PageIndex{4}\), and Theorem \(\PageIndex{5}\).]

Exercise \(\PageIndex{21}\)

Let \(V \xrightarrow{T} W \xrightarrow{S} V\) be functions such that \(TS = 1_{W}\) and \(ST = 1_{V}\). If \(T\) is linear, show that \(S\) is also linear

Exercise \(\PageIndex{22}\)

Let \(A\) and \(B\) be matrices of size \(p \times m\) and \(n \times q\). Assume that \(mn = pq\). Define \(R :\mathbf{M}_{mn} \to\mathbf{M}_{pq}\) by \(R(X) = AXB\).

1. Show that \(\mathbf{M}_{mn} \cong\mathbf{M}_{pq}\) by comparing dimensions.
2. Show that \(R\) is a linear transformation.
3. Show that if \(R\) is an isomorphism, then \(m = p\) and \(n = q\). [Hint: Show that \(T :\mathbf{M}_{mn} \to\mathbf{M}_{pn}\) given by \(T(X) = AX\) and \(S :\mathbf{M}_{mn} \to\mathbf{M}_{mq}\) given by \(S(X) = XB\) are both one-to-one, and use the dimension theorem.]

Exercise \(\PageIndex{23}\)

Let \(T : V \to V\) be a linear transformation such that \(T^{2} = 0\) is the zero transformation.

1. If \(V \neq \{\mathbf{0}\}\), show that \(T\) cannot be invertible.
2. If \(R : V \to V\) is defined by \(R(\mathbf{v}) = \mathbf{v} + T(\mathbf{v})\) for all \(\mathbf{v}\) in \(V\), show that \(R\) is linear and invertible.

Exercise \(\PageIndex{24}\)

Let \(V\) consist of all sequences \([x_{0}, x_{1}, x_{2}, \dots)\) of numbers, and define vector operations

\begin{aligned}
{\left[x_o, x_1, \ldots\right)+\left[y_0, y_1, \ldots\right) } & =\left[x_0+y_0, x_1+y_1, \ldots\right) \\
r\left[x_0, x_1, \ldots\right) & =\left[r x_0, r x_1, \ldots\right)
\end{aligned}

1. Show that \(V\) is a vector space of infinite dimension.

2. Define \(T : V \to V\) and \(S : V \to V\) by \(T[x_{0}, x_{1}, \dots) = [x_{1}, x_{2}, \dots)\) and
   \(S[x_{0}, x_{1}, \dots) = [0, x_{0}, x_{1}, \dots)\). Show that \(TS = 1_{V}\), so \(TS\) is one-to-one and onto, but that \(T\) is not one-to-one and \(S\) is not onto.

Answer

b. \(TS[x_{0}, x_{1}, \dots) = T[0, x_{0}, x_{1}, \dots) = [x_{0}, x_{1}, \dots)\), so \(TS = 1_{V}\). Hence \(TS\) is both onto and one-to-one, so \(T\) is onto and \(S\) is one-to-one by Exercise [ex:ex7_3_13]. But \([1, 0, 0, \dots)\) is in \(\text{ker }T\) while \([1, 0, 0, \dots)\) is not in \(im \;S\).

Exercise \(\PageIndex{25}\)

Prove (1) and (2) of Theorem \(\PageIndex{4}\)

Exercise \(\PageIndex{26}\)

Define \(T :\mathbf{P}_{n} \to\mathbf{P}_{n}\) by
\(T(p) = p(x) + xp^\prime(x)\) for all \(p\) in \(\mathbf{P}_{n}\).

1. Show that \(T\) is linear.
2. Show that \(\text{ker }T = \{\mathbf{0}\}\) and conclude that \(T\) is an isomorphism. [Hint: Write \(p(x) = a_{0} + a_{1}x + \cdots + a_{n}x^{n}\) and compare coefficients if \(p(x) = -xp^\prime(x)\).]

3. Conclude that each \(q(x)\) in \(\mathbf{P}_{n}\) has the form
   \(q(x) = p(x) + xp^\prime(x)\) for some unique polynomial \(p(x)\).

4. Does this remain valid if \(T\) is defined by
   \(T[p(x)] = p(x) - xp^\prime(x)\)? Explain.

Answer

If \(T(p) = 0\), then \(p(x) = -xp^\prime(x)\). We write \(p(x) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots + a_{n}x^{n}\), and this becomes \(a_{0} + a_{1}x + a_{2}x^{2} + \cdots + a_{n}x^{n} = -a_{1}x - 2a_{2}x^{2} - \cdots - na_{n}x^{n}\). Equating coefficients yields \(a_{0} = 0, 2a_{1} = 0, 3a_{2} = 0, \dots, (n + 1)a_{n} = 0\), whence \(p(x) = 0\). This means that \(\text{ker }T = 0\), so \(T\) is one-to-one. But then \(T\) is an isomorphism .

Exercise \(\PageIndex{27}\)

Let \(T : V \to W\) be a linear transformation, where \(V\) and \(W\) are finite dimensional.

1. Show that \(T\) is one-to-one if and only if there exists a linear transformation \(S : W \to V\) with \(ST = 1_{V}\). [Hint: If \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{n}\}\) is a basis of \(V\) and \(T\) is one-to-one, show that \(W\) has a basis \(\{T(\mathbf{e}_{1}), \dots, T(\mathbf{e}_{n}), \mathbf{f}_{n+1}, \dots, \mathbf{f}_{n+k}\}\) and use Theorem 7.1.2 and Theorem 7.1.3.]
2. Show that \(T\) is onto if and only if there exists a linear transformation \(S : W \to V\) with \(TS = 1_{W}\). [Hint: Let \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \dots, \mathbf{e}_{n}\}\) be a basis of \(V\) such that \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) is a basis of \(\text{ker }T\). Use Theorem 7.2.5, Theorem \7.1.2 and Theorem 7.1.3.]

Answer

If \(ST = 1_{V}\) for some \(S\), then \(T\) is onto by Exercise [ex:ex7_3_13]. If \(T\) is onto, let \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \dots, \mathbf{e}_{n}\}\) be a basis of \(V\) such that \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) is a basis of \(\text{ker }T\). Since \(T\) is onto, \(\{T(\mathbf{e}_{1}), \dots, T(\mathbf{e}_{r})\}\) is a basis of \(im \;T = W\) by Theorem [thm:021572]. Thus \(S : W \to V\) is an isomorphism where by \(S\{T(\mathbf{e}_{i})] = \mathbf{e}_{i}\) for \(i = 1, 2, \dots, r\). Hence \(TS[T(\mathbf{e}_{i})] = T(\mathbf{e}_{i})\) for each \(i\), that is \(TS[T(\mathbf{e}_{i})] = 1_{W}[T(\mathbf{e}_{i})]\). This means that \(TS = 1_{W}\) because they agree on the basis \(\{T(\mathbf{e}_{1}), \dots, T(\mathbf{e}_{r})\}\) of \(W\).

Exercise \(\PageIndex{28}\)

Let \(S\) and \(T\) be linear transformations \(V \to W\), where \(dim \;V = n\) and \(dim \;W = m\).

1. Show that \(\text{ker }S = \text{ker }T\) if and only if \(T = RS\) for some isomorphism \(R : W \to W\). [Hint: Let \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \dots, \mathbf{e}_{n}\}\) be a basis of \(V\) such that \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) is a basis of \(\text{ker }S = \text{ker }T\). Use Theorem [thm:021572] to extend \(\{S(\mathbf{e}_{1}), \dots, S(\mathbf{e}_{r})\}\) and \(\{T(\mathbf{e}_{1}), \dots, T(\mathbf{e}_{r})\}\) to bases of \(W\).]
2. Show that \(im \;S = im \;T\) if and only if \(T = SR\) for some isomorphism \(R : V \to V\). [Hint: Show that \(dim \;(\text{ker }S) = dim \;(\text{ker }T)\) and choose bases \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \dots, \mathbf{e}_{n}\}\) and \(\{\mathbf{f}_{1}, \dots, \mathbf{f}_{r}, \dots, \mathbf{f}_{n}\}\) of \(V\) where \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) and \(\{\mathbf{f}_{r+1}, \dots, \mathbf{f}_{n}\}\) are bases of \(\text{ker }S\) and \(\text{ker }T\), respectively. If \(1 \leq i \leq r\), show that \(S(\mathbf{e}_{i}) = T(\mathbf{g}_{i})\) for some \(\mathbf{g}_{i}\) in \(V\), and prove that \(\{\mathbf{g}_{1}, \dots, \mathbf{g}_{r}, \mathbf{f}_{r+1}, \dots, \mathbf{f}_{n}\}\) is a basis of \(V\).]

Answer

Conversely, assume that \(im \;S = im \;T\). Then \(dim \;(\text{ker }S) = dim \;(\text{ker }T)\) by the dimension theorem. Let \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) and \(\{\mathbf{f}_{1}, \dots, \mathbf{f}_{r}, \mathbf{f}_{r+1}, \dots, \mathbf{f}_{n}\}\) be bases of \(V\) such that \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) and \(\{\mathbf{f}_{r+1}, \dots, \mathbf{f}_{n}\}\) are bases of \(\text{ker }S\) and \(\text{ker }T\), respectively. By Theorem [thm:021572], \(\{S(\mathbf{e}_{1}), \dots, S(\mathbf{e}_{r})\}\) and \(\{T(\mathbf{f}_{1}), \dots, T(\mathbf{f}_{r})\}\) are both bases of \(im \;S = im \;T\). So let \(\mathbf{g}_{1}, \dots, \mathbf{g}_{r}\) in \(V\) be such that \(S(\mathbf{e}_{i}) = T(\mathbf{g}_{i})\) for each \(i = 1, 2, \dots, r\). Show that

\[B = \{\mathbf{g}_1, \dots, \mathbf{g}_r, \mathbf{f}_{r+1}, \dots, \mathbf{f}_n\} \mbox{ is a basis of } V. \nonumber \]

Then define \(R : V \to V\) by \(R(\mathbf{g}_{i}) = \mathbf{e}_{i}\) for \(i = 1, 2, \dots, r\), and \(R(\mathbf{f}_{j}) = \mathbf{e}_{j}\) for \(j = r + 1, \dots, n\). Then \(R\) is an isomorphism by Theorem \(\PageIndex{25}\). Finally \(SR = T\) since they have the same effect on the basis \(B\).

Exercise \(\PageIndex{29}\)

If \(T : V \to V\) is a linear transformation where \(dim \;V = n\), show that \(TST = T\) for some isomorphism \(S : V \to V\). [Hint: Let \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\)} be as in Theorem [thm:021572]. Extend \(\{T(\mathbf{e}_{1}), \dots, T(\mathbf{e}_{r})\}\) to a basis of \(V\), and use Theorem \(\PageIndex{25}\), Theorem [thm:020878] and Theorem \(\PageIndex{25}\).]

Answer

Let \(B = \{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) be a basis of \(V\) with \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) a basis of \(\text{ker }T\). If \(\{T(\mathbf{e}_{1}), \dots, T(\mathbf{e}_{r}), \mathbf{w}_{r+1}, \dots, \mathbf{w}_{n}\}\) is a basis of \(V\), define \(S\) by \(S[T(\mathbf{e}_{i})] = \mathbf{e}_{i}\) for \(1 \leq i \leq r\), and \(S(\mathbf{w}_{j}) = \mathbf{e}_{j}\) for \(r + 1 \leq j \leq n\). Then \(S\) is an isomorphism by Theorem \(\PageIndex{25}\), and \(TST(\mathbf{e}_{i}) = T(\mathbf{e}_{i})\) clearly holds for \(1 \leq i \leq r\). But if \(i \geq r + 1\), then \(T(\mathbf{e}_{i}) = \mathbf{0} = TST(\mathbf{e}_{i})\), so \(T = TST\) by Theorem \(\PageIndex{25}\).

Exercise \(\PageIndex{30}\)

Let \(A\) and \(B\) denote \(m \times n\) matrices. In each case show that (1) and (2) are equivalent.

1. (1) \(A\) and \(B\) have the same \(\func{null}\) space. (2) \(B = PA\) for some invertible \(m \times m\) matrix \(P\).
2. (1) \(A\) and \(B\) have the same range. (2) \(B = AQ\) for some invertible \(n \times n\) matrix \(Q\).

[Hint: Use Exercise \(\PageIndex{28}\).]