Exercise 6.5.1
Find an invertible real matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal, where $A$ and $B$ are the real matrices
$$\text{(a)}\quad A=\left[\begin{array}{cc} 1&2\\0&2\end{array}\right],\quad B=\left[\begin{array}{cc} 3&-8\\0&-1\end{array}\right]$$$$\text{(b)} \quad A=\left[\begin{array}{cc} 1&1\\1&1\end{array}\right],\quad B=\left[\begin{array}{cc} 1&a\\a&1\end{array}\right].$$Solution: The proof of Theorem 8 shows that if a $2\times2$ matrix has two characteristic values then the $P$ that diagonalizes $A$ will necessarily also diagonalize any $B$ that commutes with $A$.
(a) Characteristic polynomial equals $(x-1)(x-2)$. So $c_1=1$, $c_2=2$.
$$c_1:\quad\left[\begin{array}{cc}0&-2\\0&-2\end{array}\right]\left[\begin{array}{c}1\\0\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$$$$c_2:\quad\left[\begin{array}{cc}1&-2\\0&0\end{array}\right]\left[\begin{array}{c}2\\1\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$$So $P=\left[\begin{array}{cc}1&2\\0&1\end{array}\right]$ and $P^{-1}=\left[\begin{array}{cc}1&-2\\0&1\end{array}\right]$.
$$P^{-1}AP=\left[\begin{array}{cc}1&0\\0&2\end{array}\right],\quad P^{-1}BP=\left[\begin{array}{cc}3&0\\0&-1\end{array}\right]$$(b) Characteristic polynomial equals $x(x-2)$. So $c_1=0$, $c_2=2$. $$c_1:\quad\left[\begin{array}{cc}-1&-1\\-1&-1\end{array}\right]\left[\begin{array}{c}-1\\1\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$$$$c_2:\quad\left[\begin{array}{cc}1&-1\\-1&1\end{array}\right]\left[\begin{array}{c}1\\1\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$$So $P=\left[\begin{array}{cc}-1&1\\1&1\end{array}\right]$ and $P^{-1}=\left[\begin{array}{cc}-1/2&1/2\\1/2&1/2\end{array}\right]$.$$P^{-1}AP=\left[\begin{array}{cc}0&0\\0&2\end{array}\right],\quad P^{-1}BP=\left[\begin{array}{cc}1-a&0\\0&1+a\end{array}\right].$$
Exercise 6.5.2
Let $\mathcal F$ be a commuting family of $3\times3$ complex matrices. How many linearly independent matrices can $\mathcal F$ contain? What about the $n\times n$ case?
Solution: This turns out to be quite a hard question, so I’m not sure what Hoffman & Kunze had in mind. But there’s a general theorem from 1905 by I. Schur which says the answer is $\left \lfloor{\frac{n^2}{4}}\right \rfloor+1$. A simpler proof was published in 1998 by M. Mirzakhani in the American Mathematical Monthly.
Exercise 6.5.3
Let $T$ be a linear operator on an $n$-dimensional space, and suppose that $T$ has $n$ distinct characteristic values. Prove that any linear operator which commutes with $T$ is a polynomial in $T$.
Solution: Since $T$ has $n$ distinct characteristic values, $T$ is diagonalizable (exercise 6.2.7, page 190). Choose a basis $\mathcal B$ for which $T$ is represented by a diagonal matrix $A$. Suppose the linear transformation $S$ commutes with $T$. Let $B$ be the matrix of $S$ in the basis $\mathcal B$. Then the $ij$-th entry of $AB$ is $a_{ii}b_{ij}$ and the $ij$-th entry of $BA$ is $a_{jj}b_{ij}$. Therefore if $a_{ii}b_{ij}=a_{jj}b_{ij}$ and $a_{ii}\not=a_{jj}$, then it must be that $b_{ij}=0$. So we have shown that $B$ must also be diagonal. So we have to show there exists a polynomial such that $f(a_{ii})=b_{ii}$ for all $i=1,\dots,n$. By Section 4.3 there exists a polynomial with this property.
Exercise 6.5.4
Let $A$, $B$, $C$, and $D$ be $n\times n$ complex matrices which commute. Let $E$ be the $2n\times2n$ matrix
$$E=\left[\begin{array}{cc} A & B\\ C & D\end{array}\right].$$Prove that $\det E=\det(AD-BC)$.
Solution: If $A$ is invertible, then\[\left[\begin{array}{cc} A & B\\ C & D\end{array}\right]\cdot \left[\begin{array}{cc} I_n & -A^{-1}B\\ 0 & I_n\end{array}\right]=\left[\begin{array}{cc} A & 0\\ C & D-CA^{-1}B\end{array}\right].\]Hence $$\det E=\det A\det (D-CA^{-1}B)=\det(AD-ACA^{-1}B).$$Since $A$ and $C$ commute, we have $AD-ACA^{-1}B=AD-CB$. Thus $$\det E=\det(AD-ACA^{-1}B)=\det(AD-BC).$$If $A$ is not invertible, then we set $A_t=A+tI_n$, where $t$ is a complex number. Let$$E_t=\left[\begin{array}{cc} A_t & B\\ C & D\end{array}\right].$$Clearly, for enough small $t$, $A_t$ is invertible ($\det A_t$ is a polynomial of $t$ which can only have finitely many solutions). Moreover, $A_t$ and $C$ commute. Hence by the case we proved, we have $\det E_t=\det(A_tD-BC)$ for sufficiently small $t$. But both sides are polynomials in $t$, hence we can take $t\to 0$ and the limit should be the same. That is $\det E=\det(AD-BC)$.
Exercise 6.5.5
Let $F$ be a field, $n$ a positive integer, and let $V$ be the space of $n\times n$ matrices over $F$. If $A$ is a fixed $n\times n$ matrix over $F$, let $T_A$ be the linear operator on $V$ defined by $T_A(B)=AB-BA$. Consider the family of linear operators $T_A$ obtained by letting $A$ vary over all diagonal matrices. Prove that the operators in that family are simultaneously diagonalizable.
Solution: If we stack the cloumns of an $n\times n$ matrix on top of each other with column one at the top, the matrix of $T_A$ in the standard basis is then given by
$$\left[\begin{array}{cccc}
A & & & \\
& A & &\\
& & \ddots & \\
& & & A
\end{array}\right].$$Thus if $A$ is diagonal then $T_A$ is diagonalizable.
Now $T_AT_B(C)=ABC-ACB-BCA+CBA$ and $T_BT_A(C)=BAC-BCA-ACB+CAB$. Therefore we must show that $BAC+CAB=ABC+CBA$. The $i,j$-th entry of $BAC+CAB$ is $c_{ij}(a_{ii}b_{ii}+a_{jj}b_{jj})$. And this is exactly the same as the $i,j$-th entry of $ABC+CBA$. Thus $T_A$ and $T_B$ commute. Thus by Theorem 8 the family can be simultaneously diagonalized.
