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## Solution to Linear Algebra Hoffman & Kunze Chapter 6.3

Exercise 6.3.1

Solution: The minimal polynomial for the identity operator is $x-1$. It annihilates the identity operator and the monic zero degree polynomial $p(x)=1$ does not, so it must be the minimal polynomial. The minimal polynomial for the zero operator is $x$. It is a monic polynomial that annihilates the zero operator and again the monic zero degree polynomial $p(x)=1$ does not, so it must be the minimal polynomial.

Exercise 6.3.2

Solution: The characteristic polynomial is
$$\left|\begin{array}{ccc} x & 0 & -c\\-1 & x & -b\\0 & -1 & x-a\end{array}\right|=\left|\begin{array}{ccc} x & 0 & -c\\-1 & 0 & x^2-ax-b\\0 & -1 & x-a\end{array}\right|$$$$=1\cdot\left|\begin{array}{cc}x & -c\\ -1 & x^2-ax-b\end{array}\right|=x^3-ax^2-bx-c.$$Now for any $r,s\in F$
$$A^2+rA+s=\left[\begin{array}{ccc}0 & c & ac\\0 & b & c+ba\\1 & a & b+a^2\end{array}\right] + \left[\begin{array}{ccc}0 & 0 & rc\\r& 0 &rb\\0 & r& ra\end{array}\right] + \left[\begin{array}{ccc}s & 0& 0\\0 & s & 0\\0 & 0 & s\end{array}\right]$$$$=\left[\begin{array}{ccc}s & c & ac+rc\\r & b +s& c+ba+br\\1 & a+r & b+a^2+ra+s\end{array}\right]\not=0.$$Thus $f(A)\not=0$ for all $f\in F[x]$ such that $\deg(F)=2$. Thus the minimum polynomial cannot have degree two, it must therefore have degree three. Since it divides $x^3-ax^2-bx-c$ it must equal $x^3-ax^2-bx-c$.

Exercise 6.3.3

Solution: The characteristic polynomial equals
$$\left|\begin{array}{cccc}x-1 & -1 & 0 & 0\\ 1 &x+ 1 & 0 & 0\\ 2 & 2 & x-2 & -1\\ -1 & -1 & 1 & x\end{array}\right|=\left|\begin{array}{cc}x-1&-1\\1&x+1\end{array}\right|\cdot\left|\begin{array}{cc}x-2&-1\\1&x\end{array}\right|,$$$$=x^2(x^2-2x+1)=x^2(x-1)^2.$$Here we used (5-20) page 158. The minimum polynomial is clearly not linear, thus the minimal polynomial is one of $x^2(x-1)^2$, $x^2(x-1)$, $x(x-1)^2$ or $x(x-1)$. We will plug $A$ in to the first three and show it is not zero. It will follow that the minimum polynomial must be $x^2(x-1)^2$.
$$A^2= \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ -3 & -3 & 3 & 2\\ 2 & 2 & -2 & -1 \end{array}\right]$$$$A-I= \left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ -1 & -2 & 0 & 0\\ -2 & -2 & 1 & 1\\ 1 & 1 & -1 & -1 \end{array}\right]$$and
$$(A-I)^2= \left[\begin{array}{cccc} -1 & -2 & 0 & 0\\ 2 & 3 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]$$Thus
$$A^2(A-I)= \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ -1 & -1 & 1 & 1\\ 1 & 1 & -1 & -1 \end{array}\right]\not=0$$$$A(A-I)^2= \left[\begin{array}{cccc} 1 & 1 & 0 & 0\\ -1 & -1& 0 & 0\\ 0 & 0 & 0 & 0\\ 0& 0& 0 & 0 \end{array}\right]\not=0$$and
$$A(A-I)= \left[\begin{array}{cccc} -1 & -1 & 0 & 0\\ 1 & 1& 0 & 0\\ -1 & -1 & 1 & 1\\ 1& 1& -2 & -2 \end{array}\right]\not=0.$$Thus the minimal polynomial must be $x^2(x-1)^2$.

Exercise 6.3.4

Solution: Not diagonalizable, because for characteristic value $c=0$ the matrix $A-cI=A$ and $A$ is row equivalent to
$$\left[\begin{array}{cccc}1&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{array}\right]$$which has rank three. So the null space has dimension one. So if $W$ is the null space for $A-cI$ then $W$ has dimension one, which is less than the power of $x$ in the characteristic polynomial. So by Theorem 2, page 187, $A$ is not diagonalizable.

Exercise 6.3.5

Solution: $T^k=0$ $\Rightarrow$ the only characteristic value is zero. We know the minimal polynomial divides this so the minimal polynomial is of the form $t^r$ for some $1\leq r\leq n$. Thus by Theorem 3, page 193, the characteristic polynomial's only root is zero, and the characteristic polynomial has degree $n$. So the characteristic polynomial equals $t^n$. By Theorem 4 (Caley-Hamilton) $T^n=0$.

Exercise 6.3.6

Solution: If $A^2=0$ and $A\not=0$ then the minimal polynomial is $x$ or $x^2$. So any $A\not=0$ such that $A^2=0$ has minimal polynomial $x^2$. E.g.
$$A=\left[\begin{array}{ccc}0&0&0\\1&0&0\\0&0&0\end{array}\right].$$

Exercise 6.3.7

Solution: $1,x,x^2,\dots,x^n$ is a basis.
$$1\mapsto0$$$$x\mapsto1$$$$x^2\mapsto2x$$$$\vdots$$$$\rule{4mm}{0mm}x^n\mapsto nx^{n-1}$$The matrix for $D$ is therefore
$$\left[\begin{array}{cccccc} 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 2 & 0 & \cdots & 0\\ 0 & 0 & 0 & 3 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & 0 & \cdots &n\\0 & 0 & 0 & 0 & \cdots &0\end{array}\right]$$Suppose $A$ is a matrix such that $a_{ij}=0$ except when $j=i+1$. Then $A^2$ has $a_{ij}=0$ except when $j=i+2$. $A^3$ has $a_{ij}=0$ except when $j=i+3$. Etc., where finally $A^{n+1}=0$. Thus if $a_{ij}\not=0$ $\forall$ $j=i+1$ then $A^k\not=0$ for $k<{n+1}$ and $A^{n+1}=0$. Thus the minimum polynomial divides $x^{n+1}$ and cannot be $x^k$ for $k<{n+1}$. Thus the minimum polynomial is $x^{n+1}$.

Exercise 6.3.8

Solution: $P$ can be given in the standard basis by left multiplication by $A=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$. Since $P$ is given by left multiplication by a matrix, $P$ is clearly linear. Since $A$ is diagonal, the characteristic values are the diagonal values. Thus the characteristic values of $A$ are $0$ and $1$. The characteristic polynomial is a degree two monic polynomial for which both $0$ and $1$ are roots. Therefore the characteristic polynomial is $x(x-1)$. If the characteristic polynomial is a product of distinct linear terms then it must equal the minimal polynomial. Thus the minimal polynomial is also $x(x-1)$.

Exercise 6.3.9

Solution: Suppose $A$ is $n\times n$. Claim: $|xI-A|=x^n+\text{trace}(A)x^{n-1}+\cdots$. Proof by induction: case $n=2$. $A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$. $|xI-A|=x^2-(a+d)x+(ad-bc)$. The trace of $A$ is $a+d$ so we have established the claim for the case $n=2$. Suppose true for up to $n-1$. Let $r=a_{22}+a_{33}+\cdots+a_{nn}$. Then
$$\left|\begin{array}{cccc} x-a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & x-a_{22} & \cdots & a_{2n}\\ \ddots & \ddots & \cdots & \ddots\\ a_{n1} & a_{n2} & \cdots & x-a_{nn} \end{array}\right|$$Now expanding by minors using the first column, and using induction, we get that this equals
$$(x-a_{11})(x^{n-1}-rx^{n-2}+\cdots)$$$$-a_{21}(\text{polynomial of degree n-2})$$$$+a_{31}(\text{polynomial of degree n-2})$$$$+\cdots$$$$=x^n+(r+a_{11})x^{n-1}+\text{polynomial of degree at most n-2}$$$$=x^n-\text{tr(A)}x^{n-1} +\cdots$$Now if $f(x)=(x-c_1)^{d_1}\cdots(x-c_k)^{d_k}$ then the coefficient of $x^{n-1}$ is $c_1d_1+\cdots c_kd_k$ so it must be that $c_1d_1+\cdots c_kd_k=\text{tr}(A)$.

Exercise 6.3.10

Solution: If we represent a $n\times n$ matrix as a column vector by stacking the columns of the matrix on top of each other, with the first column on the top, then the transformation $T$ is represented in the standard basis by the matrix
$$M=\left[\begin{array}{cccc}A & & & \\& A & \Large 0 & \\ & \Large 0 & \ddots & \\ & & & A\end{array}\right].$$And since
$$f(M)=\left[\begin{array}{cccc}f(A) & & & \\& f(A) & \Large 0 & \\ & \Large 0 & \ddots & \\ & & & f(A)\end{array}\right]$$it is evident that $f(M)=0$ $\Leftrightarrow$ $f(A)=0$.

Exercise 6.3.11

Solution: We have
\begin{align*}\begin{vmatrix}I_n & A\\ B & xI_n\end{vmatrix}=&\ \begin{vmatrix}I_n & A\\ B & xI_n\end{vmatrix}\begin{vmatrix}I_n & -A\\ 0 & I_n\end{vmatrix}\\=&\ \begin{vmatrix}I_n & 0\\ B & xI_n-BA\end{vmatrix}\\=&\ |xI_n-BA|\end{align*}Similarly,\begin{align*}\begin{vmatrix}I_n & A\\ B & xI_n\end{vmatrix}=&\ \begin{vmatrix}I_n & A\\ B & xI_n\end{vmatrix}\begin{vmatrix}I_n & 0\\ -Bx^{-1} & I_n\end{vmatrix}\\=&\ \begin{vmatrix}I_n-x^{-1}AB & A\\ 0 & xI_n\end{vmatrix}\\=&\ x^n|I_n-x^{-1}AB|=|xI_n-AB|.\end{align*}So $AB$ and $BA$ have the same characteristic polynomials.

But the minimum polynomials need not be equal. To see this let $A=\left[\begin{array}{cc}0&0\\1&0\end{array}\right]$ and $B=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$. Then $AB=\left[\begin{array}{cc}0&0\\1&0\end{array}\right]$ and $BA=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$ so the minimal polynomial of $BA$ is $x$ and the minimal polynomial of $AB$ is clearly not $x$ (it is in fact $x^2$).