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## Solution to Linear Algebra Hoffman & Kunze Chapter 6.2

#### Exercise 6.2.1

In each of the following cases, let $T$ be the linear operator on $\mathbb R^2$ which is represented by the matrix $A$ in the standard ordered basis for $\mathbb R^2$, and let $U$ be the linear operator on $\mathbb C^2$ represented by $A$ in the standard ordered basis. Find the characyteristic polynomial for $T$ and that for $U$, find the characteristic values of each operator, and for each such charactersistic value $c$ find a basis for the corresponding space of characteristic vectors.
$$A=\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\quad\left[\begin{array}{cc}2&3\\-1&1\end{array}\right],\quad\left[\begin{array}{cc}1&1\\1&1\end{array}\right].$$Solution: We have

$$A=\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\quad xI-A=\left[\begin{array}{cc}x-1&0\\0&x\end{array}\right]$$The characteristic polynomial equals $|xI-A|=x(x-1)$. So $c_1=0$, $c_1=1$. A basis for $W_1$ is $\{(0,1)\}$, $\alpha_1=(0,1)$. A basis for $W_2$ is $\{(1,0)\}$, $\alpha_2=(1,0)$. This is the same whether the base field is $\mathbb R$ or $\mathbb C$ since the characteristic polynomial factors completely.

We have$$A=\left[\begin{array}{cc}2&3\\-1&1\end{array}\right],\quad |xI-A|=\left|\begin{array}{cc}x-2&-3\\1&x-1\end{array}\right|$$
$=(x-2)(x-1)+3=x^2-3x+5$. This is a parabola opening up with vertex $(3/2,11/4)$. Thus there are no real roots. Using the quadratic formula $c_1=\frac{3+\sqrt{11}i}{2}$ and $c_2=\frac{3-\sqrt{11}i}{2}$. To find the a characteristic vector for $c_1$ we solve
$$\left[\begin{array}{cc}\frac{-1+\sqrt{11}i}{2} & 3\\ -1 & \frac{1+\sqrt{11}i}{2}\end{array}\right]\left[\begin{array}{c} x\\y\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right].$$This gives the characteristic vector $\alpha_1=(\frac{1+\sqrt{11}i}{2},1)$.To find the a characteristic vector for $c_2$ we solve
$$\left[\begin{array}{cc}\frac{-1-\sqrt{11}i}{2} & 3\\ -1 & \frac{1-\sqrt{11}i}{2}\end{array}\right]\left[\begin{array}{c} x\\y\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right].$$This gives the characteristic vector $\alpha_2=(\frac{1-\sqrt{11}i}{2},1)$.

We have $$A=\left[\begin{array}{cc}1&1\\1&1\end{array}\right],$$$$|xI-A|=\left|\begin{array}{cc}x-1 & 1\\ 1 & x-1\end{array}\right|=(x-1)^2-1=x(x-2).$$ So $c_1=0$ for which $\alpha_1=(1,-1)$. And $c_2=2$ for which $\alpha_2=(1,1)$. This is the same in both $\mathbb R$ and $\mathbb C$ since the characteristic polynomial factors completely.

#### Exercise 6.2.2

Let $F$ be an $n$-dimensional vector space over $F$. What is the characteristic polynomial of the identity operator on $V$? What is the characteristic polynomial for the zero operator?

Solution: The identity operator can be represented by the $n\times n$ identity matrix $I$. The characteristic polynomial of the identity operator is therefore $(x-1)^n$. The zero operator is represented by the zero matrix in any basis. Thus the characteristic polynomial of the zero operator is $x^n$.

#### Exercise 6.2.3

Let $A$ be an $n\times n$ triangular matrix over the field $F$. Prove that the characteristic values of $A$ are the diagonal entries of $A$, i.e., the scalars $A_{ii}$.

Solution: The determinant of a triangular matrix is the product of the diagonal entries. Thus $|xI-A|=\prod(x-a_{ii})$.

#### Exercise 6.2.4

Let $T$ be the linear operator of $\mathbb R^3$ which is represented in the standard ordered basis by the matrix
$$\left[\begin{array}{ccc}-9 & 4 & 4\\-8 & 3 & 4\\-16 & 8 & 7\end{array}\right].$$Prove that $T$ is diagonalizable by exhibiting a basis for $\mathbb R^3$, each vector fo which is a characteristic vector of $T$.

Solution: We have$$|xI-A|=\left|\begin{array}{ccc}x+9 & -4 & -4\\8 & x-3 & -4\\16 & -8 & x-7\end{array}\right|$$$$=\left|\begin{array}{ccc}x+9 & 0 & -4\\8 & x+1 & -4\\16 & -x-1 & x-7\end{array}\right|$$$$=(x+1)\left|\begin{array}{ccc}x+9 & 0 & -4\\8 & 1 & -4\\16 & -1 & x-7\end{array}\right|$$$$=(x+1)\left|\begin{array}{ccc}x+9 & 0 & -4\\8 & 1 & -4\\24 & 0 & x-11\end{array}\right|$$$$=(x+1)\left|\begin{array}{cc}x+9 & -4\\24 & x-11\end{array}\right|$$$$=(x+1)[(x+9)(x-11)+96]=(x+1)(x^2-2x-3)$$$$=(x+1)(x-3)(x+1)=(x+1)^2(x-3).$$ Thus $c_1=-1$, $c_2=3$. For $c_1$, $xI-A$ equals
$$=\left[\begin{array}{ccc}8 & -4 & -4\\8 & -4 & -4\\16 & -8 & -8\end{array}\right]$$
This matrix evidently has rank one. Thus the null space has rank two. The two characteristic vectors $(1,2,0)$ and $(1,0,2)$ are independent, so they form a basis for $W_1$. For $c_2$, $xI-A$ equals
$$=\left[\begin{array}{ccc}-12 & -4 & -4\\8 & 0 & -4\\16 & -8 & -4\end{array}\right]$$This is row equivalent to
$$=\left[\begin{array}{ccc}1 & -0& -1/2\\0& 1 & -1/2\\0 & 0 & 0\end{array}\right]$$Thus the null space one dimensional and is given by $(z/2,z/2,z)$. So $(1,1,2)$ is a characteristic vector and a basis for $W_2$. By Theorem 2 (ii) $T$ is diagonalizable.

#### Exercise 6.2.5

Let
$$A=\left[\begin{array}{ccc}6& -3 & -2\\4 & -1 & -2\\10 & -5 & -3\end{array}\right].$$Is $A$ similar over the field $\mathbb R$ to a diagonal matrix? Is $A$ similar over the field $\mathbb C$ to a diagonal matrix?

Solution: $$|xI-A|=\left|\begin{array}{ccc}x-6& 3 & 2\\-4 & x+1 & 2\\-10 & 5 & x+3\end{array}\right|$$$$=\left|\begin{array}{ccc}x-6& 3 & 2\\-x+2 & x-2 & 0\\-10 & 5 & x+3\end{array}\right|$$$$=(x-2)\left|\begin{array}{ccc}x-6& 3 & 2\\-1 & 1 & 0\\-10 & 5 & x+3\end{array}\right|$$$$=(x-2)\left|\begin{array}{ccc}x-3& 3 & 2\\0 & 1 & 0\\-5 & 5 & x+3\end{array}\right|$$$$=(x-2)((x-3)(x+3)+10)=(x-2)(x^2+1).$$ Since this is not a product of linear factors over $\mathbb R$, by Theorem 2, page 187, $A$ is not diagonalizable over $\mathbb R$. Over $\mathbb C$ this factors to $(x-2)(x-i)(x+i)$. Thus over $\mathbb C$ the matrix $A$ has three distinct characteristic values. The space of characteristic vectors for a given characteristic value has dimension at least one. Thus the sum of the dimensions of the $W_i$’s must be at least $n$. It cannot be greater than $n$ so it must equal $n$ exactly. Thus $A$ is diagonalizable over $\mathbb C$.

#### Exercise 6.2.6

Let $T$ be the linear operator on $\mathbb R^4$ which is represented in the standard ordered basis by the matrix
$$\left[\begin{array}{cccc}0 & 0 & 0 & 0 \\ a & 0 & 0 & 0\\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \end{array}\right].$$
Under what conditions on $a$, $b$, and $c$ is $T$ diagonalizable?

Solution: We have$$|xI-A|=\left|\begin{array}{cccc}x & 0 & 0 & 0 \\ -a & x & 0 & 0\\ 0 & -b & x & 0 \\ 0 & 0 & -c & x \end{array}\right|=x^4.$$Therefore there is only one characteristic value $c_1=0$. Thus $c_1I-A=A$ and $W_1$ is the null space of $A$. So $A$ is diagonalizable $\Leftrightarrow$ $\dim(W)=4$ $\Leftrightarrow$ $A$ is the zero matrix $\Leftrightarrow$ $a=b=c=0$.

#### Exercise 6.2.7

Let $T$ be the linear operator on the $n$-dimensional vector space $V$, and suppose that $T$ has $n$ distinct characteristic values. Prove that $T$ is diagonalizable.

Solution: The space of characteristic vectors for a given characteristic value has dimension at least one. Thus the sum of the dimensions of the $W_i$’s must be at least $n$. It cannot be greater than $n$ so it must equal $n$ exactly. Thus by Theorem 2, $T$ is diagonalizable.

#### Exercise 6.2.8

Let $A$ and $B$ be $n\times n$ matrices over the field $F$. Prove that if $(I-AB)$ is invertible, then $I-BA$ is invertible and
$$(I-BA)^{-1}=I+B(I-AB)^{-1}A.$$Solution: We have$$(I-BA)(I+B(I-AB)^{-1}A)$$$$=I-BA+B(I-AB)^{-1}A-BAB(I-AB)^{-1}A$$$$=I-B(A-(I-AB)^{-1}A+AB(I-AB)^{-1}A)$$$$=I-B(I-(I-AB)^{-1}+AB(I-AB)^{-1}A$$$$=I-B(I-(I-AB)(I-AB)^{-1})A$$$$=I-B(I-I)A=I.$$

#### Exercise 6.2.9

Use the result of Exercise 8 to prove that, if $A$ and $B$ are $n\times n$ matrices over the field $F$, then $AB$ and $BA$ have precisely the same characteristic values in $F$.

Solution:  By Theorem 3, page 154, $\det(AB)=\det(A)\det(B)$. Thus $AB$ is singular $\Leftrightarrow$ $BA$ is singular. Therefore $0$ is a characteristic values of $AB$ $\Leftrightarrow$ $0$ is a characteristic value of $BA$. Now suppose the characteristic value $c$ of $AB$ is not equal to zero. Then $|cI-AB|=0$ $\Leftrightarrow$ $c^n|I-{\frac1c}AB| =0$ $\overset{\text{by \#8}}{\Leftrightarrow}$ $c^n|I-{\scriptstyle\frac1c}BA|=0$ $\Leftrightarrow$ $|cI-BA|=0$.

#### Exercise 6.2.10

Suppose that $A$ is a $2\times2$ matrix with real entries which is symmetrix $(A^t=A)$. Prove that $A$ is similar over $\mathbb R$ to a diagonal matrix.

Solution: $A=\left[\begin{array}{cc}a & b\\c & d\end{array}\right]$. So $|xI-A|=\left|\begin{array}{cc}x-a & -b\\ -b & x-a\end{array}\right|=(x-a)^2-b^2=(x-a-b)(x-a+b)$. So $c_1=a+b$, $c_2=a=b$. If $b=0$ then $A$ is already diagonal. If $b\not=0$ then $c_1\not=c_2$ so by Exercise 7 $A$ is diagonalizable.

#### Exercise 6.2.11

Let $N$ be a $2\times2$ complex matrix such that $N^2=0$. Prove that either $N=0$ or $N$ is similar over $\mathbb C$ to
$$\left[\begin{array}{cc}0 & 0\\1 & 0\end{array}\right].$$Solution:  Suppose $N=\left[\begin{array}{cc}a & b\\c & d\end{array}\right]$. Now $N^2=0$ $\Rightarrow$ $\left[\begin{array}{c}a\\c\end{array}\right]$, $\left[\begin{array}{c}b\\d\end{array}\right]$ are characteristic vectors for the characteristic value $0$. If $\left[\begin{array}{c}a\\c\end{array}\right]$, $\left[\begin{array}{c}b\\d\end{array}\right]$ are linearly independent then $W_1$ has rank two and $N$ is diagonalizable to $\left[\begin{array}{cc}0 & 0\\0 & 0\end{array}\right]$. If $PNP^{-1}=0$ then $N=P^{-1}0P=0$ so in this case $N$ itself is the zero matrix. This contradicts the assumption that $\left[\begin{array}{c}a\\c\end{array}\right]$, $\left[\begin{array}{c}b\\d\end{array}\right]$ are linearly independent.

So we can assume that $\left[\begin{array}{c}a\\c\end{array}\right]$, $\left[\begin{array}{c}b\\d\end{array}\right]$ are linearly dependent. If both equal the zero vector then $N=0$. So we can assume at least one vector is non-zero. If $\left[\begin{array}{c}b\\d\end{array}\right]$ is the zero vector then $N=\left[\begin{array}{cc}a & 0 \\ c & 0\end{array}\right]$. So $N^2=0$ $\Rightarrow$ $a^2=0$ $\rightarrow$ $a=0$. Thus $N=\left[\begin{array}{cc}a & 0 \\ c & 0\end{array}\right]$. In this case $N$ is similar to $N =\left[\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right]$ via the matrix $P=\left[\begin{array}{cc}c & 0 \\ 0 & 1\end{array}\right]$. Similary if $\left[\begin{array}{c}a\\c\end{array}\right]$ is the zero vector, then $N^2=0$ implies $d^2=0$ implies $d=0$ so $N=\left[\begin{array}{cc}0 & b \\ 0 & 0\end{array}\right]$. In this case $N$ is similar to $N=\left[\begin{array}{cc}0 & 0 \\ b & 0\end{array}\right]$ via the matrix $P=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$, which is simiilar to $\left[\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right]$ as above.

By the above we can assume neither $\left[\begin{array}{c}a\\c\end{array}\right]$ or $\left[\begin{array}{c}b\\d\end{array}\right]$ is the zero vector. Since they are linearly dependent we can assume $\left[\begin{array}{c}b\\d\end{array}\right]=x\left[\begin{array}{c}a\\c\end{array}\right]$ so $N=\left[\begin{array}{cc}a & ax \\ c & cx\end{array}\right]$. So $N^2=0$ implies
$$a(a+cx)=0,\quad c(a+cx)=0,$$$$ax(a+cx)=0,\quad cx(a+cx)=0.$$We know that at least one of $a$ or $c$ is not zero. If $a=0$ then since $c\not=0$ it must be that $x=0$. So in this case $N=\left[\begin{array}{cc}0 & 0 \\ c & 0\end{array}\right]$ which is similar to $\left[\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right]$ as before. If $a\not=0$ then $x\not=0$ else $a(a+cx)=0$ implies $a=0$. Thus $a+cx=0$ so $N=\left[\begin{array}{cc} a & ax\\ -a/x & -a\end{array}\right]$. This is similar to $\left[\begin{array}{cc} a & a\\ -a & -a\end{array}\right]$ via $P=\left[\begin{array}{cc}\sqrt{x} & 0\\0 & 1/\sqrt{x}\end{array}\right]$. And $\left[\begin{array}{cc} a & a\\ -a & -a\end{array}\right]$ is similar to $\left[\begin{array}{cc} 0 & 0\\ -a & 0\end{array}\right]$ via $P=\left[\begin{array}{cc}-1 & -1\\1 & 0\end{array}\right]$. And this finally is similar to $\left[\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right]$ as before.

#### Exercise 6.2.12

Use the result of Exercise 11 to prove the following: If $A$ is a $2\times2$ matrix with complex entries, then $A$ is similar over $\mathbb C$ to a matrix of one of the two types
$$\left[\begin{array}{cc}a & 0\\ 0 & b\end{array}\right]\quad\quad\left[\begin{array}{cc}a & 0\\ 1 & a\end{array}\right].$$Solution: Suppose $A=\left[\begin{array}{cc}a & b\\c & d\end{array}\right]$. Since the base field is $\Bbb C$ the characteristic polynomial $p(x)=(x-c_1)(x-c_2)$. If $c_1\not=c_2$ then $A$ is diagonalizable by Exercise 7. If $c_1=c_2$ then $p(x)=(x-c_1)^2$. If $W$ has dimension two then $A$ is diagonalizable by Theorem 2. Thus we will be done if we show that if $p(x)=(x-c_1)^2$ and $\dim(W_1)=1$ then $A$ is similar to $\left[\begin{array}{cc}a & 0\\1 & a\end{array}\right]$.

We will need the following three identities:
\begin{equation}
\left[\begin{array}{cc}a & 0\\c & d\end{array}\right]\sim\left[\begin{array}{cc}a & 0\\1 & d\end{array}\right]\quad\text{via $p=\left[\begin{array}{cc}c & 0\\0 & 1\end{array}\right]$}\label{eq2}
\end{equation}\begin{equation}
\left[\begin{array}{cc}a & b\\c & d\end{array}\right]\sim\left[\begin{array}{cc}a-b & c-d\\b & d\end{array}\right]\quad\text{via $p=\left[\begin{array}{cc}1 & 0\\-1 & 1\end{array}\right]$}
\label{eq3}
\end{equation}\begin{equation}
\left[\begin{array}{cc}a & b\\c & d\end{array}\right]\sim\left[\begin{array}{cc}a & xb\\c/x & d\end{array}\right]\quad\text{via $p=\left[\begin{array}{cc}\sqrt{x} & 0\\0 & 1/\sqrt{x}\end{array}\right]$ for $x\not=0$.}
\label{eq4}
\end{equation}Now we know in this case that $A$ is not diagonalizable. If $d\not=0$ then $\left[\begin{array}{cc}a & b\\c & d\end{array}\right]\sim\left[\begin{array}{cc}a & bc/d\\d & d\end{array}\right]$ by (\ref{eq4}) with $x=c/d$ and this in turn is similar to $\left[\begin{array}{cc}a-bc/d & 0\\-a+2bc/d & d\end{array}\right]$ by (\ref{eq3}).

Now we know the diagonal entries are the characteristic values, which are equal. Thus $a-\frac{bc}{d}=d$. So this equals $\left[\begin{array}{cc}d & 0\\x & d\end{array}\right]$ where $x=-a+\frac{2bc}{d}$ and we know $x\not=0$ since $A$ is not diagonalizable. Thus $A\sim\left[\begin{array}{cc}d & 0\\1 & d\end{array}\right]$ by (\ref{eq2}). Now suppose $d=0$. Then $A=\left[\begin{array}{cc}a & b\\c &0\end{array}\right]\sim\left[\begin{array}{cc}0 & c\\b & a\end{array}\right]$ via $p=\left[\begin{array}{cc}0 & 1\\1 & 0\end{array}\right]$. If $b=0$ then $A=\left[\begin{array}{cc}a & 0\\c & 0\end{array}\right]$ and again since $A$ has equal characteristic values it must be that $a=0$. So $A=\left[\begin{array}{cc}0& 0\\c & 0\end{array}\right]$ which is similar to $\left[\begin{array}{cc}0 & 0\\1 & 0\end{array}\right]$ via $P=\left[\begin{array}{cc}c & 0\\0 & 1\end{array}\right]$. So assume $b\not=0$. Then $A\sim\left[\begin{array}{cc}0 & c\\b & a\end{array}\right]$ and we can argue exact as above were $d\not=0$.

#### Exercise 6.2.13

Let $V$ be the vector space of all functions from $\mathbb R$ into $\mathbb R$ which are continuous, i.e., the space of continuous real-valued functions on the real line. Let $T$ be the linear operator on $V$ defined by
$$(Tf)(x)=\int_0^xf(t)dt.$$Prove that $T$ has no characteristic values.

Solution: Suppose $c$ is a characteristic value of $T$, then there exists a non-zero continuous function $f(x)$ on $\mathbb R$ such that $T(f)=cf(x),\quad\text{for all }x\in \mathbb R.$That is$cf(x)=\int_{0}^xf(x)dx$for all $x\in \mathbb R$. Define a function$F(x)=\int_0^xf(x)dx.$Then $F(x)$ is differentiable on $\mathbb R$ and $F'(x)=f(x)$ by fundamental theorem of calculus. Moreover, $F(0)=0$. Therefore, we have$cF'(x)=F(x).$If $c=0$, then $F(x)\equiv 0$ and hence $f(x)=\equiv 0$ which is impossible by assumption. Thus $c\ne 0$, we must have$\left(e^{-x/c}F(x)\right)’=-e^{-x/c}(F(x)-cF'(x))/c=0.$Hence $e^{-x/c}F(x)$ is a constant, namely $F(x)=ke^{x/c}$. However, because $F(0)=0$, we have $k=0$. Therefore $F(x)\equiv 0$ and we obtain a contradiction again. Therefore $T$ has no characteristic values.

#### Exercise 6.2.14

Let $A$ be an $n\times n$  diagonal matrix with characteristic polynomial
$$(x-c_1)^{d_1}\cdots(x-c_k)^{d_k},$$ where $c_1,\dots,c_k$ are distinct. Let $V$ be the space of $n\times n$ matrices $B$ such that $AB=BA$. Prove that the dimension of $V$ is $d_1^2+\cdots+ d_k^2$.

Solution: Write
$$A=\left[\begin{array}{cccc} c_1I & & & \\ & c_2I & \Large 0 &\\ & \Large 0 & \ddots & \\ & & & c_kI\end{array}\right].$$Write
$$B=\left[\begin{array}{cccc} B_{11} & B_{12} & \cdots & B_{1k}\\ B_{21} & B_{22} & \cdots & B_{2k}\\ \vdots & \vdots & \ddots & \vdots\\ B_{k1} & B_{k2} & \cdots & B_{kk}\\ \end{array}\right]$$where $B_{ij}$ has dimenson $d_i\times d_j$. Then $AB=BA$ implies
$$\left[\begin{array}{cccc} c_1B_{11} & c_1B_{12} & \cdots & c_1B_{1k}\\ c_2B_{21} & c_2B_{22} & \cdots & c_2B_{2k}\\ \vdots & \vdots & \ddots & \vdots\\ c_kB_{kk} & c_kB_{k2} & \cdots & c_kB_{kk}\\ \end{array}\right] = \left[\begin{array}{cccc} c_1B_{11} & c_2B_{12} & \cdots & c_kB_{1k}\\ c_1B_{21} & c_2B_{22} & \cdots & c_kB_{2k}\\ \vdots & \vdots & \ddots & \vdots\\ c_1B_{k1} & c_2B_{k2} & \cdots & c_kB_{kk}\\ \end{array}\right]$$Thus $c_i\not=c_j$ for $i\not=j$ implies $B_{ij}=0$ for $i\not=j$, while $B_{11},B_{22},\dots,B_{kk}$ can be arbitrary. The dimension of $B_{ii}$ is therefore $d_i^2$ thus the dimension of the space of all such $B_{ii}$’s is $d_1^2+d_2^2+\cdots+d_k^2$.

#### Exercise 6.2.15

Let $V$ be the space of $n\times n$ matrices over $F$. Let $A$ be a fixed $n\times n$ matrix over $F$. Let $T$ be the linear operator `left multiplication by $A$’ on $V$. Is it true that $A$ and $T$ have the same characteristic values?

Solution: Yes. Represent an element of $V$ as a column vector by stacking the columns of $V$ on top of each other, with the first column on top. Then the matrix for $T$ is given by $$\left[\begin{array}{cccc}A & & & \\& A & \Large 0 & \\ & \Large 0 & \ddots & \\ & & & A\end{array}\right].$$ By the argument on page 157 the determinant of this matrix is $\det(A)^n$. Thus if $p$ is the characteristic polynomial of $A$ then $p^n$ is the characteristic polynomial of $T$. Thus they have exactly the same roots and thus they have exactly the same characteristic values.