If you find any mistakes, please make a comment! Thank you.

## Solution to Linear Algebra Hoffman & Kunze Chapter 3.4

#### Exercise 3.4.1

Let $T$ be the linear operator on $\mathbb C^2$ defined by $T(x_1,x_2)=(x_1,0)$. Let $\mathcal B$ be the standard ordered basis for $\mathbb C^2$ and let $\mathcal B'=\{\alpha_1,\alpha_2\}$ be the ordered basis defined by $\alpha_1=(1,i)$, $\alpha_2=(-i,2)$.

(a) What is the matrix of $T$ relative to the pair $\mathcal B$, $\mathcal B'$?
(b) What is the matrix of $T$ relative to the pair $\mathcal B'$, $\mathcal B$?
(c) What is the matrix of $T$ in the ordered basis $\mathcal B'$?
(d) What is the matrix of $T$ in the ordered basis $\{\alpha_2,\alpha_1\}$?

Solution:

(a) According to the comments at the bottom of page 87, the $i$-th column of the matrix is given by $[T\epsilon_i]_{\mathcal B'}$, where $\epsilon_1=(1,0)$ and $\epsilon_2=(0,1)$, the standard basis vectors of $\mathbb C^2$. Now $T\epsilon_1=(1,0)$ and $T\epsilon_2=(0,0)$. To write these in terms of $\alpha_1$ and $\alpha_2$ we use the approach of row-reducing the augmented matrix
$$\left[\begin{array}{cc|cc}1&-i&1&0\\i&2&0&0\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&-i&1&0\\0&1&-i&0\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&0&2&0\\0&1&-i&0\end{array}\right].$$ Thus $T\epsilon_1=2\alpha_1-i\alpha_2$ and $T\epsilon_2=0\cdot\alpha_1+0\cdot\alpha_2$ and the matrix of $T$ relative to $\mathcal B$, $\mathcal B'$ is
$$\left[\begin{array}{cc}2&0\\-i&0\end{array}\right].$$(b) In this case we have to write $T\alpha_1$ and $T\alpha_2$ as linear combinations of $\epsilon_1,\epsilon_2$.
$$T\alpha_1=(1,0)=1\cdot\epsilon_1+0\cdot\epsilon_2$$ $$T\alpha_2=(-i,0)=-i\cdot\epsilon_1+0\cdot\epsilon_2.$$ Thus the matrix of $T$ relative to $\mathcal B'$, $\mathcal B$ is $$\left[\begin{array}{cc}1&-i\\0&0\end{array}\right].$$(c) In this case we need to write $T\alpha_1$ and $T\alpha_2$ as linear combinations of $\alpha_1$ and $\alpha_2$. $T\alpha_1=(1,0)$, $T\alpha_2=(-i,0)$. We row-reduce the augmented matrix:
$$\left[\begin{array}{cc|cc}1&-i&1&-i\\i&2&0&0\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&-i&1&-i\\0&1&-i&-1\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&0&2&-2i\\0&1&-i&-1\end{array}\right].$$
Thus the matrix of $T$ in the ordered basis $\mathcal B'$ is
$$\left[\begin{array}{cc}2&-2i\\-i&-1\end{array}\right].$$(d) In this case we need to write $T\alpha_2$ and $T\alpha_1$ as linear combinations of $\alpha_2$ and $\alpha_1$. In this case the matrix we need to row-reduce is just the same as in (c) but with columns switched:
$$\left[\begin{array}{cc|cc}-i&1&-i&1\\2&i&0&0\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&i&1&i\\2&i&0&0\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&i&1&i\\0&-i&-2&-2i\end{array}\right]$$ $$\rightarrow\left[\begin{array}{cc|cc}1&i&1&i\\0&1&-2i&2\end{array}\right] \rightarrow\left[\begin{array}{cc|cc}1&0&-1&-i\\0&1&-2i&2\end{array}\right]$$Thus the matrix of $T$ in the ordered basis $\{\alpha_2,\alpha_1\}$ is
$$\left[\begin{array}{cc}-1&-i\\-2i&2\end{array}\right].$$

#### Exercise 3.4.2

Let $T$ be the linear transformation from $\Bbb R^3$ to $\Bbb R^2$ defined by $$T(x_1,x_2,x_3)=(x_1+x_2,2x_3-x_1).$$(a) If $\mathcal B$ is the standard ordered basisfor $\Bbb R^3$ and $\mathcal B'$ is the standard ordered basis for $\Bbb R^2$, what is the matrix of $T$ relative to the pair $\mathcal B$, $\mathcal B'$?
(b) If $\mathcal B=\{\alpha_1,\alpha_2,\alpha_3\}$ and $\mathcal B'=\{\beta_1,\beta_2\}$, where $$\alpha_1=(1,0,-1),\quad\alpha_2=(1,1,1),\quad\alpha_3=(1,0,0),\quad\beta_1=(0,1),\quad\beta_2=(1,0)$$ what is the matrix of $T$ relative to the pair $\mathcal B$, $\mathcal B'$?

Solution:

(a) With respect to the standard bases, the matrix is simply
$$\left[\begin{array}{ccc}1&1&0\\-1&0&2\end{array}\right].$$(b) We must write $T\alpha_1,T\alpha_2,T\alpha_3$ in terms of $\beta_1,\beta_2$. $$T\alpha_1=(1,-3)$$ $$T\alpha_2=(2,1)$$ $$T\alpha_3=(1,0).$$ We row-reduce the augmented matrix
$$\left[\begin{array}{cc|ccc}0&1&1&2&1\\1&0&-3&1&0\end{array}\right] \rightarrow\left[\begin{array}{cc|ccc}1&0&-3&1&0\\0&1&1&2&1\end{array}\right].$$ Thus the matrix of $T$ with respect to $\mathcal B$, $\mathcal B'$ is
$$\left[\begin{array}{ccc}-3&1&0\\1&2&1\end{array}\right].$$

#### Exercise 3.4.3

Let $T$ be a linear operator on $F^n$, let $A$ be the matrix of $T$ in the standard ordered basis for $F^n$, and let $W$ be the subspace of $F^n$ spanned by the column vectors of $A$. What does $W$ have to do with $T$?

Solution: Since $\{\alpha_1,\dots,\alpha_n\}$ is a basis of $F^n$, we know $\{T\epsilon_1,\dots,T\epsilon_n\}$ generate the range of $T$. But $T\epsilon_i$ equals the $i$-th column vector of $A$. Thus the column vectors of $A$ generate the range of $T$ (where we identify $F^n$ with $F^{n\times1}$). We can also conclude that a subset of the columns of $A$ give a basis for the range of $T$.

#### Exercise 3.4.4

Let $V$ be a two-dimensional vector space over the field $F$, and let $\mathcal B$ be an ordered basis for $V$. If $T$ is a linear operator on $V$ and
$$[T]_{\mathcal B}=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$$ prove that $T^2-(a+d)T+(ad-bc)I=0.$

Solution: The coordinate matrix of $T^2-(a+d)T+(ad-bc)I$ with respect to $\mathcal B$ is $$[T^2-(a+d)T+(ad-bc)I]_{\mathcal B}=\ \ \left[\begin{array}{cc}a&b\\c&d\end{array}\right]^2-(a+d)\left[\begin{array}{cc}a&b\\c&d\end{array}\right]+(ad-bc)\left[\begin{array}{cc}1&0\\0&1\end{array}\right]\$$
Expanding gives
$$=\left[\begin{array}{cc}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{array}\right]-\left[\begin{array}{cc}a^2+ad&ab+bd\\ac+cd&ad+d^2\end{array}\right]+\left[\begin{array}{cc}ad-bc&0\\0&ad-bc\end{array}\right]$$
$$=\left[\begin{array}{cc}0&0\\0&0\end{array}\right].$$
Thus $T^2-(a+d)T+(ad-bc)I$ is represented by the zero matrix with respect to $\mathcal B$. Thus $T^2-(a+d)T+(ad-bc)I=0$.

#### Exercise 3.4.5

Let $T$ be the linear operator on $\mathbb R^3$, the matrix of which in the standard ordered basis is
$$\left[\begin{array}{ccc}1&2&1\\0&1&1\\-1&3&4\end{array}\right].$$ Find a basis for the range of $T$ and a basis for the null space of $T$.

Solution: The range is the column-space, which is the row-space of the following matrix (the transpose):
$$\left[\begin{array}{ccc}1&0&-1\\2&1&3\\1&1&4\end{array}\right]$$which we can easily determine a basis of by putting it in row-reduced echelon form.
$$\left[\begin{array}{ccc}1&0&-1\\2&1&3\\1&1&4\end{array}\right] \rightarrow\left[\begin{array}{ccc}1&0&-1\\0&1&5\\0&1&5\end{array}\right] \rightarrow\left[\begin{array}{ccc}1&0&-1\\0&1&5\\0&0&0\end{array}\right].$$So a basis of the range is $\{(1,0,-1), (0,1,5)\}$.

The null space can be found by row-reducing the matrix
$$\left[\begin{array}{ccc}1&2&1\\0&1&1\\-1&3&4\end{array}\right] \rightarrow\left[\begin{array}{ccc}1&2&1\\0&1&1\\0&5&5\end{array}\right] \rightarrow\left[\begin{array}{ccc}1&0&-1\\0&1&1\\0&0&0\end{array}\right]$$So
$$\left\{\begin{array}{c}x-z=0\\y+z=0\end{array}\right.$$which implies
$$\left\{\begin{array}{c}x=z\\y=-z\end{array}\right.$$The solutions are parameterized by the one variable $z$, thus the null space has dimension equal to one. A basis is obtained by setting $z=1$. Thus $\{(1,-1,1)\}$ is a basis for the null space.

#### Exercise 3.4.6

Let $T$ be the linear operator on $\mathbb R^2$ defined by
$$T(x_1,x_2)=(-x_2,x_1).$$(a) What is the matrix of $T$ in the standard ordered basis for $\mathbb R^2$?
(b) What is the matrix of $T$ in the ordered basis $\mathcal B=\{\alpha_1,\alpha_2\}$, where $\alpha_1=(1,2)$ and $\alpha_2=(1,-1)$?
(c) Prove that for every real number $c$ the operator $(T-cI)$ is invertible.
(d) Prove that if $\mathcal B$ is any ordered basis for $\mathbb R^2$ and $[T]_{\mathcal B}=A$, then $A_{12}A_{21}\not=0$.

Solution:

(a) We must write $T\epsilon_1=(0,1)$ and $T\epsilon_2=(-1,0)$ in terms of $\epsilon_1$ and $\epsilon_2$. Clearly $T\epsilon_1=\epsilon_2$ and $T\epsilon_2=-\epsilon_1$. Thus the matrix is
$$\left[\begin{array}{cc}0&-1\\1&0\end{array}\right].$$(b) We must write $T\alpha_1=(-2,1)$ and $T\alpha_2=(1,1)$ in terms of $\alpha_1,\alpha_2$. We can do this by row-reducing the augmented matrix
$$\left[\begin{array}{cc|cc}1&1&-2&1\\2&-1&1&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{cc|cc}1&1&-2&1\\0&-3&5&-1\end{array}\right]$$$$\rightarrow\left[\begin{array}{cc|cc}1&1&-2&1\\0&1&-5/3&1/3\end{array}\right]$$$$\rightarrow\left[\begin{array}{cc|cc}1&0&-1/3&2/3\\0&1&-5/3&1/3\end{array}\right]$$Thus the matrix of $T$ in the ordered basis $\mathcal B$ is
$$[T]_{\mathcal B}=\left[\begin{array}{cc}-1/3&2/3\\-5/3&1/3\end{array}\right].$$(c) The matrix of $T-cI$ with respect to the standard basis is
$$\left[\begin{array}{cc}0&-1\\1&0\end{array}\right]-c\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$$$\left[\begin{array}{cc}0&-1\\1&0\end{array}\right]-\left[\begin{array}{cc}c&0\\0&c\end{array}\right]$$$$\left[\begin{array}{cc}-c&-1\\1&-c\end{array}\right].$$Row-reducing the matrix
$$\left[\begin{array}{cc}-c&-1\\1&-c\end{array}\right] \rightarrow\left[\begin{array}{cc}1&-c\\-c&-1\end{array}\right] \rightarrow\left[\begin{array}{cc}1&-c\\0&-1-c^2\end{array}\right].$$Now $-1-c^2\not=0$ (since $c^2\geq0$). Thus we can continue row-reducing by dividing the second row by $-1-c^2$ to get
$$\rightarrow\left[\begin{array}{cc}1&-c\\0&1\end{array}\right] \rightarrow\left[\begin{array}{cc}1&0\\0&1\end{array}\right].$$Thus the matrix has rank two, thus $T$ is invertible.

(d) Let $\{\alpha_1,\alpha_2\}$ be any basis. Write $\alpha_1=(a,b)$, $\alpha_2=(c,d)$. Then $T\alpha_1=(-b,a)$, $T\alpha_2=(-d,c)$. We need to write $T\alpha_1$ and $T\alpha_2$ in terms of $\alpha_1$ and $\alpha_2$. We can do this by row reducing the augmented matrix
$$\left[\begin{array}{cc|cc}a&c&-b&-d\\b&d&a&c\end{array}\right].$$Since $\{\alpha_1,\alpha_2\}$ is a basis, the matrix $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$ is invertible. Thus (recalling Exercise 1.6.8, page 27), $ad-bc\not=0$. Thus the matrix row-reduces to
$$\left[\begin{array}{cc|cc}1&0&\frac{ac+bd}{ad-bc}&\frac{c^2+d^2}{ad-bc}\\\rule{0mm}{5mm}0&1&\frac{a^2+b^2}{ad-bc}&\frac{ac+bd}{ad-bc}\end{array}\right].$$Assuming $a\not=0$ this can be shown as follows:
$$\rightarrow\left[\begin{array}{cc|cc}1&c/a&-b/a&-d/a\\b&d&a&c\end{array}\right].$$$$\rightarrow\left[\begin{array}{cc|cc}1&c/a&-b/a&-d/a\\0&\frac{ad-bc}{a}&\frac{a^2+b^2}{a}&\frac{ac+bd}{a}\end{array}\right].$$$$\rightarrow\left[\begin{array}{cc|cc}1&c/a&-b/a&-d/a\\0&1&\frac{a^2+b^2}{ad-bc}&\frac{ac+bd}{ad-bc}\end{array}\right].$$$$\rightarrow\left[\begin{array}{cc|cc}1&0&\frac{ac+bd}{ad-bc}&\frac{c^2+d^2}{ad-bc}\\\rule{0mm}{5mm}0&1&\frac{a^2+b^2}{ad-bc}&\frac{ac+bd}{ad-bc}\end{array}\right].$$If $b\not=0$ then a similar computation results in the same thing. Thus
$$[T]_{\mathcal B} = \left[\begin{array}{cc}\frac{ac+bd}{ad-bc}&\frac{c^2+d^2}{ad-bc}\\\rule{0mm}{5mm}\frac{a^2+b^2}{ad-bc}&\frac{ac+bd}{ad-bc}\end{array}\right].$$Now $ad-bc\not=0$ implies that at least one of $a$ or $b$ is non-zero and at least one of $c$ or $d$ is non-zero, it follows that $a^2+b^2>0$ and $c^2+d^2>0$. Thus $(a^2+b^2)(c^2+d^2)\not=0$. Thus$$\frac{a^2+b^2}{ad-bc}\cdot\frac{c^2+d^2}{ad-bc}\not=0.$$

#### Exercise 3.4.7

Let $T$ be the linear operator on $\mathbb R^3$ defined by
$$T(x_1,x_2,x_3)=(3x_1+x_3,\ \ -2x_1+x_2,\ \ -x_1+2x_2+4x_3).$$(a) What is the matrix of $T$ in the standard ordered basis for $\mathbb R^3$.
(b) What is the matrix of $T$ in the ordered basis
$$(\alpha_1,\alpha_2,\alpha_3)$$where $\alpha_1=(1,0,1)$, $\alpha_2=(-1,2,1)$, and $\alpha_3=(2,1,1)$?
(c) Prove that $T$ is invertible and give a rule for $T^{-1}$ like the one which defines $T$.

Solution:

(a) As usual we can read the matrix in the standard basis right off the definition of $T$:
$$[T]_{\{\epsilon_1,\epsilon_2,\epsilon_3\}}=\left[\begin{array}{ccc}3&0&1\\-2&1&0\\-1&2&4\end{array}\right].$$(b) $T\alpha_1=(4,-2,3)$, $T\alpha_2=(-2,4,9)$ and $T\alpha_3=(7,-3,4)$. We must write these in terms of $\alpha_1,\alpha_2,\alpha_3$. We do this by row-reducing the augmented matrix
$$\left[\begin{array}{ccc|ccc}1&-1&2&4&-2&7\\0&2&1&-2&4&-3\\1&1&1&3&9&4\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&2&4&-2&7\\0&2&1&-2&4&-3\\0&2&-1&-1&11&-3\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&2&4&-2&7\\0&2&1&-2&4&-3\\0&0&-2&1&7&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&2&4&-2&7\\0&1&1/2&-1&2&-3/2\\0&0&1&-1/2&-7/2&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&5/2&3&0&11/2\\0&1&1/2&-1&2&-3/2\\0&0&1&-1/2&-7/2&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&0&17/4&35/4&11/2\\0&1&0&-3/4&15/4&-3/2\\0&0&1&-1/2&-7/2&0\end{array}\right]$$Thus the matrix of $T$ in the basis $\{\alpha_1,\alpha_2,\alpha_3\}$ is
$$[T]_{\{\alpha_1,\alpha_2,\alpha_3\}}=\left[\begin{array}{ccc}17/4&35/4&11/2\\-3/4&15/4&-3/2\\-1/2&-7/2&0\end{array}\right].$$(c) We row reduce the augmented matrix (of $T$ in the standard basis). If we achieve the identity matrix on the left of the dividing line then $T$ is invertible and the matrix on the right will represent $T^{-1}$ in the standard basis, from which we will be able read the rule for $T^{-1}$ by inspection.
$$\left[\begin{array}{ccc|ccc}3&0&1 & 1&0&0\\-2&1&0 & 0&1&0\\-1&2&4 & 0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}-1&2&4 & 0&0&1\\3&0&1 & 1&0&0\\-2&1&0 & 0&1&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-2&-4 & 0&0&-1\\3&0&1 & 1&0&0\\-2&1&0 & 0&1&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-2&-4 & 0&0&-1\\0&6&13 & 1&0&3\\0&-3&-8 & 0&1&-2\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-2&-4 & 0&0&-1\\0&0&-3 & 1&2&-1\\0&-3&-8 & 0&1&-2\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-2&-4 & 0&0&-1\\0&-3&-8 & 0&1&-2\\0&0&-3 & 1&2&-1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-2&-4 & 0&0&-1\\0&1&8/3 & 0&-1/3&2/3\\0&0&-3 & 1&2&-1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&4/3& 0&-2/3&1/3\\0&1&8/3 & 0&-1/3&2/3\\0&0&-3 & 1&2&-1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&4/3& 0&-2/3&1/3\\0&1&8/3 & 0&-1/3&2/3\\0&0&1 & -1/3&-2/3&1/3\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&0& 4/9&2/9&-1/9\\0&1&0 & 8/9&13/9&-2/9\\0&0&1 & -1/3&-2/3&1/3\end{array}\right]$$Thus $T$ is invertible and the matrix for $T^{-1}$ in the standard basis is
$$\left[\begin{array}{ccc}4/9&2/9&-1/9\\8/9&13/9&-2/9\\-1/3&-2/3&1/3\end{array}\right].$$Thus $T^{-1}(x_1,x_2,x_3)=\left(\frac49 x_1+\frac29x_2-\frac19x_3,\frac89x_1+\frac{13}9x_2-\frac29x_3,-\frac13x_1-\frac23x_2+\frac13x_3\right)$.

#### Exercise 3.4.8

Let $\theta$ be a real number. Prove that the following two matrices are similar over the field of complex numbers:
$$\left[\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right],\quad\left[\begin{array}{cc}e^{i\theta}&0\\0&e^{-i\theta}\end{array}\right]$$(Hint: Let $T$ be the linear operator on $\Bbb C^2$ which is represented by the first matrix in the standard ordered basis. Then find vectors $\alpha_1$ and $\alpha_2$ such that $T\alpha_1=e^{i\theta}\alpha_1$, $T\alpha_2=e^{-i\theta}\alpha_2$, and $\{\alpha_1,\alpha_2\}$ is a basis.)

Solution: Let $\mathcal B$ be the standard basis. Following the hint, let $T$ be the linear operator on $\mathbb C^2$ which is represented by the first matrix in the standard ordered basis $\mathcal B$. Thus $[T]_{\mathcal B}$ is the first matrix above. Let $\alpha_1=(i,1)$, $\alpha_2=(i,-1)$. Then $\alpha_1,\alpha_2$ are clealry linearly independent so $\mathcal B'=\{\alpha_1,\alpha_2\}$ is a basis for $\mathbb C^2$ (as a vector space over $\mathbb C$). Since $e^{i\theta}=\cos\theta+i\sin\theta$, it follows that $$T\alpha_1=(i\cos\theta-\sin\theta,\ \ i\sin\theta+\cos\theta)=(\cos\theta+i\sin\theta)(i,1)=e^{i\theta}\alpha_1$$ and similarly since and $e^{-i\theta}=\cos\theta-i\sin\theta$, it follows that $T\alpha_2=e^{-i\theta}\alpha_2$. Thus the matrix of $T$ with respect to $\mathcal B'$ is
$$[T]_{\mathcal B'}=\left[\begin{array}{cc}e^{i\theta}&0\\0&e^{-i\theta}\end{array}\right].$$By Theorem 14, page 92, $[T]_{\mathcal B}$ and $[T]_{\mathcal B'}$ are similar.

#### Exercise 3.4.9

Let $V$ be a finite-dimensional vector space over the field $F$ and let $S$ and $T$ be linear operators on $V$. We ask: When do there exist ordered bases $\mathcal B$ and $\mathcal B'$ for $V$ such that $[S]_{\mathcal B}=[T]_{\mathcal B'}$? Prove that such bases exist if and only if there is an invertible linear operator $U$ on $V$ such that $T=USU^{-1}$. (Outline of proof: If $[S]_{\mathcal B}=[T]_{\mathcal B'}$, let $U$ be the operator which carries $\mathcal B$ onto $\mathcal B'$ and show that $S=UTU^{-1}$. Conversely, if $T=USU^{-1}$ for some invertible $U$, let $\mathcal B$ be any ordered basis for $V$ and let $\mathcal B'$ be its image under $U$. Then show that $[S]_{\mathcal B}=[T]_{\mathcal B'}$.)

Solution: We follow the hint. Suppose there exist bases $\mathcal B=\{\alpha_1,\dots,\alpha_n\}$ and $\mathcal B=\{\beta_1,\dots,\beta_n\}$ such that $[S]_{\mathcal B}=[T]_{\mathcal B'}$. Let $U$ be the operator which carries $\mathcal B$ onto $\mathcal B'$. Then by Theorem 14, page 92, $[USU^{-1}]_{\mathcal B'}=[U]_{\mathcal B}^{-1}[USU^{-1}]_{\mathcal B}[U]_{\mathcal B}$ and by the comments at the very bottom of page 90, this equals $[U]_{\mathcal B}^{-1}[U]_{\mathcal B}[S]_{\mathcal B}[U]_{\mathcal B}^{-1}[U]_{\mathcal B}$ which equals $[S]_{\mathcal B}$, which we've assumed equals $[T]_{\mathcal B'}$. Thus $[USU^{-1}]_{\mathcal B'}=[T]_{\mathcal B'}$. Thus $USU^{-1}=T$.

Conversely, assume $T=USU^{-1}$ for some invertible $U$. Let $\mathcal B$ be any ordered basis for $V$ and let $\mathcal B'$ be its image under $U$. Then $[T]_{\mathcal B'}=[USU^{-1}]_{\mathcal B'}=[U]_{\mathcal B'}[S]_{\mathcal B'}[U]_{\mathcal B'}^{-1}$, which by Theorem 14, page 92, equals $[S]_{\mathcal B}$ (because $U^{-1}$ carries $\mathcal B'$ into $\mathcal B$). Thus $[T]_{\mathcal B'}=[S]_{\mathcal B}$.

#### Exercise 3.4.10

We have seen that the linear operator $T$ on $\mathbb R^2$ defined by $T(x_1,x_2)=(x_1,0)$ is represented in the standard ordered basis by the matrix
$$A=\left[\begin{array}{cc}1&0\\0&0\end{array}\right].$$This operator satisfies $T^2=T$. Prove that if $S$ is a linear operator on $\mathbb R^2$ such that $S^2=S$, then $S=0$, or $S=I$, or there is an ordered basis $\mathcal B$ for $\mathbb R^2$ such that $[S]_{\mathcal B}=A$ (above).

Solution: Suppose $S^2=S$. Let $\epsilon_1,\epsilon_2$ be the standard basis vectors for $\Bbb R^2$. Consider $\{S\epsilon_1,S\epsilon_2\}$.

If both $S\epsilon_1=S\epsilon_2=0$ then $S=0$. Thus suppose WLOG that $S\epsilon_1\not=0$.

First note that if $x\in S(\Bbb R^2)$ then $x=S(y)$ for some $y\in\Bbb R^2$ and therefore $S(x)=S(S(y))=S^2(y)=S(y)=x$. In other words $S(x)=x$ for all $x\in S(\Bbb R^2)$.

Case 1: Suppose $\exists$ $c\in\mathbb R$ such that $S\epsilon_2=cS\epsilon_1$. Then $S(\epsilon_2-c\epsilon_1)=0$. In this case $S$ is singular because it maps a non-zero vector to zero. Thus since $S\epsilon_1\not=0$ we can conclude that $\dim(S(\mathbb R^2))=1$. Let $\alpha_1$ be a basis for $S(\mathbb R^2)$. Let $\alpha_2\in\mathbb R^2$ be such that $\{\alpha_1,\alpha_2\}$ is a basis for $\mathbb R^2$. Then $S\alpha_2=k\alpha_1$ for some $k\in\mathbb R$. Let $\alpha'_2=\alpha_2-k\alpha_1$. Then $\{\alpha_1,\alpha'_2\}$ span $\mathbb R^2$ because if $x=a\alpha_1+b\alpha_2$ then $x=(a+bk)\alpha_1+b\alpha'_2$. Thus $\{\alpha_1,\alpha'_2\}$ is a basis for $\mathbb R^2$. We now determine the matrix of $S$ with respect to this basis. Since $\alpha_1\in S(\mathbb R^2)$ and $S(x)=x$ $\forall$ $x\in S(\mathbb R^2)$, it follows that $S\alpha_1=\alpha_1$. And consequently $S(\alpha_1)=1\cdot\alpha_1 + 0\cdot\alpha'_2$. Thus the first column of the matrix of $S$ with respect to $\alpha_1,\alpha'_2$ is $[1,0]^{\text{T}}$. Also \begin{align*}S\alpha'_2&=S(\alpha_2-k\alpha_1)=S\alpha_2-kS\alpha_1\\&=S\alpha_2-k\alpha_1=k\alpha_1-k\alpha_1=0=0\cdot\alpha_1+0\cdot\alpha'_2.\end{align*} So the second column of the matrix is $[0,0]^{\text{T}}$. Thus the matrix of $S$ with respect to the basis $\{\alpha_1,\alpha'_2\}$ is exactly $A$.

Case 2: There does not exist $c\in\mathbb R$ such that $S\epsilon_2=cS\epsilon_1$. In this case $S\epsilon_1$ and $S\epsilon_2$ are linearly independent from each other. Thus if we let $\alpha_i=S\epsilon_i$ then $\{\alpha_1,\alpha_2\}$ is a basis for $\mathbb R^2$. Now by assumption $S(x)=x$ for all $x\in S(\mathbb R^2)$, thus $S\alpha_1=\alpha_1$ and $S\alpha_2=\alpha_2$. Thus the matrix of $S$ with respect to the basis $\{\alpha_1,\alpha_2\}$ is exactly the identity matrix $I$.

Exercise 3.4.11

Let $W$ be the space of all $n\times1$ column matrices over a field $F$. If $A$ is an $n\times n$ matrix over $F$, then $A$ defines a linear operator $L_A$ on $W$ through left multiplication: $L_A(X)=AX$. Prove that every linear operator on $W$ is left multiplication by some $n\times n$ matrix, i.e., is $L_A$ for some $A$.

Now suppose $V$ is an $n$-dimensional vector space over the field $F$, and let $\mathcal B$ be an ordered basis for $V$. For each $\alpha$ in $V$, define $U\alpha=[\alpha]_{\mathcal B}$. Prove that $U$ is an isomorphism of $V$ onto $W$. If $T$ is a linear operator on $V$, then $UTU^{-1}$ is a linear operator on $W$. Accordingly, $UTU^{-1}$ is left multiplication by some $n\times n$ matrix $A$. What is $A$?

Solution:

Part 1: I'm confused by the first half of this question because isn't this exactly Theorem 11, page 87 in the special case $V=W$ where $\mathcal B=\mathcal B'$ is the standard basis of $F^{n\times1}$. This special case is discussed on page 88 after Theorem 12, and in particular in Example 13. I don't know what we're supposed to add to that.

Part 2: Since $U(c\alpha_1+\alpha_2)=[c\alpha_1+\alpha_2]_{\mathcal B}=c[\alpha_1]_{\mathcal B}+[\alpha_2]_{\mathcal B}=cU(\alpha_1)+U(\alpha_2)$, $U$ is linear, we just must show it is invertible. Suppose $\mathcal B=\{\alpha_1,\dots,\alpha_n\}$. Let $T$ be the function from $W$ to $V$ defined as follows:
$$\left[\begin{array}{c}a_1\\a_2\\ \vdots\\ a_n\end{array}\right]\mapsto a_1\alpha_1+\cdots a_n\alpha_n.$$ Then $T$ is well defined and linear and it is also clear by inspection that $TU$ is the identity transformation on $V$ and $UT$ is the identity transformation on $W$. Thus $U$ is an isomorphism from $V$ to $W$.

It remains to deterine the matrix of $UTU^{-1}$. Now $U\alpha_i$ is the standard $n\times1$ matrix with all zeros except in the $i$-th place which equals one. Let $\mathcal B'$ be the standard basis for $W$. Then the matrix of $U$ with respect to $\mathcal B$ and $\mathcal B'$ is the identity matrix. Likewise the matrix of $U^{-1}$ with respect to $\mathcal B'$ and $\mathcal B$ is the identity matrix. Thus $[UTU^{-1}]_{\mathcal B'}=I[T]_{\mathcal B}I^{-1}=[T]_{\mathcal B}$. Therefore the matrix $A$ is simply $[T]_{\mathcal B}$, the matrix of $T$ with respect to $\mathcal B$.

#### Exercise 3.4.12

Let $V$ be an $n$-dimensional vector space over the field $F$, and let $\mathcal B=\{\alpha_1,\dots,\alpha_n\}$ be an ordered basis for $V$.

(a) According to Theorem 1, there is a unique linear operator $T$ on $V$ such that
$$T\alpha_j=\alpha_{j+1},\quad j=1,\dots,n-1,\quad T\alpha_n=0.$$What is the matrix $A$ of $T$ in the ordered basis $\mathcal B$.?
(b) Prove that $T^n=0$ but $T^{n-1}\not=0$.
(c) Let $S$ be any linear operator on $V$ such that $S^n=0$ but $S^{n-1}\not=0$. Prove that there is an ordered basis $\mathcal B'$ for $V$ such that the matrix of $S$ in the ordered basis $\mathcal B'$ is the matrix $A$ of part (a).
(d) Prove that if $M$ and $N$ are $n\times n$ matrices over $F$ such that $M^n=N^n=0$ but $M^{n-1}\not=0\not=N^{n-1}$, then $M$ and $N$ are similar.

Solution:

(a) The $i$-th column of $A$ is given by the coefficients obtained by writing $\alpha_i$ in terms of $\{\alpha_1,\dots,\alpha_n\}$. Since $T\alpha_i=\alpha_{i+1}$, $i<n$ and $T\alpha_n=0$, the matrix is therefore
$$A=\left[\begin{array}{ccccccc} 0&0&0&0&\cdots&0&0\\ 1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots&1&0\end{array}\right].$$(b) $A$ has all zeros except $1$'s along the diagonal one below the main diagonal. Thus $A^2$ has all zeros except $1$'s along the diagonal that is two diagonals below the main diagonal, as follows:
$$A^2=\left[\begin{array}{ccccccc} 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ 1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots&0&0\end{array}\right].$$Similarly $A^3$ has all zeros except the diagonal three below the main diagonal. Continuing we see that $A^{n-1}$ is the matrix that is all zeros except for the bottom left entry which is a $1$:
$$A^{n-1}=\left[\begin{array}{ccccccc} 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 1&0&0&0&\cdots&0&0\end{array}\right].$$Multiplying by $A$ one more time then yields the zero matrix, $A^n=0$. Since $A$ represents $T$ with respect to the basis $\mathcal B$, and $A^i$ represents $T^i$, we see that $T^{n-1}\not=0$ and $T^n=0$.

(c) We will first show that $\dim(S^{k}(V))=n-k$. Suppose $\dim(S(V))=n$. Then $\dim(S^k(V))=n$ $\forall$ $k=1,2,\dots$, which contradicts the fact that $S^n=0$. Thus it must be that $\dim(S(V))\leq n-1$. Now $\dim(S^2(V))$ cannot be greater than $\dim(S(V))$ because a linear transformation cannot map a space onto one with higher dimension. Thus $\dim(S^2(V))\leq n-1$. Suppose that $\dim(S^2(V))=n-1$. Thus $$n-1=\dim(S^2(V))\leq \dim(S(V))\leq n-1.$$ Thus it must be that $\dim(S(V))=n-1$. Thus $S$ is an isomorphism on $S(V)$ because $S(V)$ and $S(S(V))$ have the same dimension. It follows that $S^k$ is also an isomorphism on $S(V)$ $\forall$ $k\geq2$. Thus it follows that $\dim(S^k(V))=n-1$ for all $k=2,3,4,\dots$, another contradiction. Thus $\dim(S^2(V))\leq n-2$. Suppose that $\dim(S^3(V))=n-2$, then it must be that $\dim(S^2(V))=n-2$ and therefore $S$ is an isomorphism on $S^2(V)$, from which it follows that $\dim(S^k(V))=n-2$ for all $k=3,4,\dots$, a contradiction. Thus $\dim(S^3(V))\leq n-3$. Continuing in this way we see that $\dim(S^k(V))\leq n-k$. Thus $\dim(S^{n-1}(V))\leq 1$. Since we are assuming $S^{n-1}\not=0$ it follows that $\dim(S^{n-1}(V))=1$. We have seen that $\dim(S^k(V))$ cannot equal $\dim(S^{k+1}(V))$ for $k=1,2,\dots,n-1$, thus it follows that the dimension must go down by one for each application of $S$. In other words $\dim(S^{n-2}(V))$ must equal $2$, and then in turn $\dim(S^{n-3}(V))$ must equal $3$, and generally $\dim(S^{k}(V))=n-k$.

Now let $\alpha_1$ be any basis vector for $S^{n-1}(V)$ which we have shown has dimension one. Now $S^{n-2}(V)$ has dimension two and $S$ takes this space onto a space $S^{n-1}(V)$ of dimension one. Thus there must be $\alpha_2\in S^{n-2}(V)\setminus S^{n-1}(V)$ such that $S(\alpha_2)=\alpha_1$. Since $\alpha_2$ is not in the space generated by $\alpha_1$ and $\{\alpha_1,\alpha_2\}$ are in the space $S^{n-2}(V)$ of dimension two, it follows that $\{\alpha_1,\alpha_2\}$ is a basis for $S^{n-2}(V)$. Now $S^{n-3}(V)$ has dimension three and $S$ takes this space onto a space $S^{n-2}(V)$ of dimension two. Thus there must be $\alpha_3\in S^{n-3}(V)\setminus S^{n-2}(V)$ such that $S(\alpha_3)=\alpha_2$. Since $\alpha_3$ is not in the space generated by $\alpha_1$ and $\alpha_2$ and $\{\alpha_1,\alpha_2, \alpha_3\}$ are in the space $S^{n-3}(V)$ of dimension three, it follows that $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for $S^{n-3}(V)$. Continuing in this way we produce a sequence of elements $\{\alpha_1,\alpha_2,\dots,\alpha_k\}$ that is a basis for $S^{n-k}(V)$ and such that $S(\alpha_i)=\alpha_{i-1}$ for all $i=2,3,\dots,k$. In particular we have a basis $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ for $V$ and such that $S(\alpha_i)=\alpha_{i-1}$ for all $i=2,3,\dots,n$. Reverse the ordering of this bases to give $\mathcal B=\{\alpha_n,\alpha_{n-1},\dots,\alpha_1\}$. Then $\mathcal B$ therefore is the required basis for which the matrix of $S$ with respect to this basis will be the matrix given in part (a).

(d) Suppose $S$ is the transformation of $F^{n\times1}$ given by $v\mapsto Mv$ and similarly let $T$ be the transformation $v\mapsto Nv$. Then $S^n=T^n=0$ and $S^{n-1}\not=0\not=T^{n-1}$. Then we know from the previous parts of this problem that there is a basis $\mathcal B$ for which $S$ is represented by the matrix from part (a). By Theorem 14, page 92, it follows that $M$ is similar to the matrix in part (a). Likewise there's a basis $\mathcal B'$ for which $T$ is represented by the matrix from part (a) and thus the matrix $N$ is also similar to the matrix in part (a). Since similarity is an equivalence relation (see last paragraph page 94), it follows that since $M$ and $N$ are similar to the same matrix that they must be similar to each other.

Exercise 3.4.13

Let $V$ and $W$ be finite-dimensional vector spaces over the field $F$ and let $T$ be a linear transformation from $V$ into $W$. If
$$\mathcal B=\{\alpha_1,\dots,\alpha_n\}\ \ \ \text{and}\ \ \ \mathcal B'=\{\beta_1,\dots,\beta_n\}$$are ordered bases for $V$ and $W$, respectively, define the linear transformations $E^{p,q}$ as in the proof of Theorem 5: $E^{p,q}(\alpha_i)=\delta_{i,q}\beta_p$. Then the $E^{p,q}$, $1\leq p\leq m$, $1\leq q\leq n$, form a basis for $L(V,W)$, and so
$$T=\sum_{p=1}^m\sum_{q=1}^nA_{pq}E^{p,q}$$for certain scalars $A_{pq}$ (the coordinates of $T$ in this basis for $L(V,W)$). Show that the matrix $A$ with entries $A(p,q)=A_{pq}$ is precisely the matrix of $T$ relative to the pair $\mathcal B$, $\mathcal B'$.

Solution: Let $E^{p,q}_{\text M}$ be the matrix of the linear transformation $E^{p,q}$ with respect to the bases $\mathcal B$ and $\mathcal B'$. Then by the formula for a matrix associated to a linear transformation as given in the proof of Theorem 11, page 87, $E^{p,q}_{\text M}$ is the matrix all of whose entries are zero except for the $p,q$-the entry which is one. Thus $A=\sum_{p,q}A_{p,q}E^{p,q}_{\text M}$. Since the association between linear transformations and matrices is an isomorphism, $T\mapsto A$ implies $\sum_{p,q}A_{pq}E^{p,q}\mapsto\sum_{p,q}A_{pq}E^{p,q}_{\text M}$. And thus $A$ is exactly the matrix whose entries are the $A_{pq}$'s.