1.5 Matrix Multiplication

矩阵可以理解为线性变换在一组基下的表示，故矩阵乘法就是两个线性变换结合后的表示。Hoffman把矩阵更一般地作为函数，故其给矩阵乘法的引入是：矩阵乘法是对矩阵的行进行线性组合时产生的东西。在定义和后面的例子（Example 10）中，都体现了矩阵乘法的这一个思想：A某一行的每一个数与B的行向量进行线性组合，得到AB的这一行的结果。特别提到了矩阵乘法是not commutative的。
列矩阵（column matrix）引致一个有用的notation（可以从定义证得）：如果有矩阵 $B_{n\times p}$，其中列向量为 $B_1,\dots,B_p$，则 $B=\begin{bmatrix}B_1,\dots,B_p\end{bmatrix}$并且 $AB=\begin{bmatrix}AB_1,\dots,AB_p\end{bmatrix}$。
Theorem 8 给出了矩阵乘法满足结合律的证明。其后说明了矩阵的幂是什么，以及 $A(BC)=(AB)C$说明linear combinations of linear combinations of the rows of C are again linear combinations of the rows of C，我认为这个诠释很有意思，Hoffman时刻在提示运算背后的思想。
最后就是要引入elementary matrix，与1.3节介绍的elementary row operations密切相关，elementary matrix的定义是从 $I$进行一次elementary row operations得到的矩阵，故其必是和 $I$同阶的方阵。Theorem 9 说明，如果用 $e$表示一个elementary row operation，则 $e(A)=e(I)A$，即行变换等于左乘对应初等矩阵。其推论显示： $A,B$两矩阵row equivalent的充要条件是 $B=PA$，其中 $P$是一堆初等矩阵乘积。

Exercises

1. Let

$A=\begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix},\quad B=\begin{bmatrix}3\\1\\-1\end{bmatrix},\quad C=\begin{bmatrix}1&-1\end{bmatrix}$

Compute $ABC$ and $CAB$.

Solution:
$ABC=\begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix} \begin{bmatrix}3\\1\\-1\end{bmatrix} \begin{bmatrix}1&-1\end{bmatrix} =\begin{bmatrix}4\\4\end{bmatrix} \begin{bmatrix}1&-1\end{bmatrix}=\begin{bmatrix}4&-4\\4&-4\end{bmatrix} \\ CAB=\begin{bmatrix}1&-1\end{bmatrix}\begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix}\begin{bmatrix}3\\1\\-1\end{bmatrix}=\begin{bmatrix}1&-3&0\end{bmatrix}\begin{bmatrix}3\\1\\-1\end{bmatrix}=0$

2. Let

$A=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix},\quad B=\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}$
Verify directly that $A(AB)=A^2B$
Solution:
$A(AB)=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\left(\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}\right)=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\begin{bmatrix}5&-1\\8&0\\10&-2\end{bmatrix}=\begin{bmatrix}7&-3\\20&-4\\25&-5\end{bmatrix} \\ A^2 B=\left(\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\right)\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}=\begin{bmatrix}2&-1&1\\5&-2&3\\6&-3&4\end{bmatrix}\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}=\begin{bmatrix}7&-3\\20&-4\\25&-5\end{bmatrix}$
thus $A(AB)=A^2 B$.

3.Find two different $2\times 2$ matrices $A$ such that $A^2=0$ but $A\neq 0$.

Solution: $A_1=\begin{bmatrix}0&1\\0&0\end{bmatrix}\quad,A_2=\begin{bmatrix}0&0\\1&0\end{bmatrix}$

4.For the matrix $A$ of Exercise 2, find elementary matrices $E_1,E_2,\dots,E_k$ such that

$E_k\dots E_2E_1A=I$
Solution:
$\begin{aligned}\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\xrightarrow{\text{add -2$\times$(1) to (2), add -3$\times$(1) to (3)}}&\begin{bmatrix}1&-1&1\\0&2&-1\\0&3&-2\end{bmatrix}\\ \xrightarrow{(2)\times(1/2)} &\begin{bmatrix}1&-1&1\\0&1&-1/2\\0&3&-2\end{bmatrix}\\ \xrightarrow{\text{add } -3\times(2) \text{ to }(3)}&\begin{bmatrix}1&-1&1\\0&1&-1/2\\0&0&-1/2\end{bmatrix}\\ \xrightarrow{(3)\times-2} &\begin{bmatrix}1&-1&1\\0&1&-1/2\\0&0&1\end{bmatrix}\\ \xrightarrow{\text{add } 1/2\times(3) \text{ to }(2),\text{ add } -1\times(3) \text{ to }(1)}&\begin{bmatrix}1&-1&0\\0&1&0\\0&0&1\end{bmatrix}\\\xrightarrow{\text{add } (2) \text{ to }(1)}&\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\end{aligned}$
so the corresponding elementary matrices are
$E_1=\begin{bmatrix}1&0&0\\-2&1&0\\0&0&1\end{bmatrix},E_2=\begin{bmatrix}1&0&0\\0&1&0\\-3&0&1\end{bmatrix},E_3=\begin{bmatrix}1&0&0\\0&1/2&0\\0&0&1\end{bmatrix},E_4=\begin{bmatrix}1&0&0\\0&1&0\\0&-3&1\end{bmatrix}\\E_5=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-2\end{bmatrix},E_6=\begin{bmatrix}1&0&0\\0&1&1/2\\0&0&1\end{bmatrix},E_7=\begin{bmatrix}1&0&-1\\0&1&0\\0&0&1\end{bmatrix},E_8=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$

5. Let

$A=\begin{bmatrix}1&-1\\2&2\\1&0\end{bmatrix},\quad B=\begin{bmatrix}3&1\\-4&4\end{bmatrix}$

Is there a matrix $C$ such that $CA=B$?

Solution: One possible choice of $C$ is $C=\begin{bmatrix}1&1&0\\2&3&-12\end{bmatrix}$

6. Let $A$ be an $m\times n$ matrix and $B$ and $n\times k$ matrix. Show that the columns of $C=AB$ are linear combinations of the columns of $A$. If $\alpha_1,\dots,\alpha_n$ are the columns of $A$ and $\gamma_1,\dots,\gamma_k$ are the columns of $C$, then

$\gamma_j=\sum_{r=1}^nB_{rj}\alpha_r$
Solution: We write $A=[\alpha_1,\dots,\alpha_n ]$, and $C=[\gamma_1,\dots,\gamma_k ]$, then as $C=AB$, for $1\leq i\leq m$, we have
$C_{ij}=\sum_{r=1}^nA_{ir} B_{rj}=\sum_{r=1}^nB_{rj} A_{ir},\quad 1\leq i\leq m$
thus
$\gamma_j=\begin{bmatrix}C_{1j}\\\vdots\\C_{mj}\end{bmatrix}=\begin{bmatrix}\sum_{r=1}^nB_{rj} A_{1r}\\\vdots\\\sum_{r=1}^nB_{rj} A_{mr}\end{bmatrix}=B_1j \begin{bmatrix}A_{11}\\\vdots\\A_{m1}\end{bmatrix}+\cdots+B_nj \begin{bmatrix}A_{1n}\\\vdots\\A_{mn}\end{bmatrix}=B_{1j}\alpha_1+\cdots+B_{nj} \alpha_n$

7.Let $A$ and $B$ be $2\times 2$ matrices such that $AB=I$. Prove that $BA=I$.

Solution: We have $ABA=IA=A$, thus $A(BA-I)=0$, assume $BA\neq I$, then the system $AX=0$ has non-trivial solutions, thus $A$ is row-equivalent to a row-reduced echelon matrix which is not $I$, so there’s $P$ which is product of elementary matrices such that
$PA=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
this means the second row of $P=PAB$ is $0$, a contradiction.

8. Let

$C=\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}$

be a $2\times 2$ matrix. We inquire when it is possible to find $2\times 2$ matrices $A$ and $B$ such that $C=AB-BA$. Prove that such matrices can be found if and only if $C_{11}+C_{22}=0$.

Solution: Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix},B=\begin{bmatrix}e&f\\g&h\end{bmatrix}$, then $AB=\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix},BA=\begin{bmatrix}ea+fc&eb+df\\ag+hc&gb+dh\end{bmatrix}$, thus
$AB-BA=\begin{bmatrix}bg-cf&af+bh-eb-df\\ce+dg-ag-hc&cf-bg\end{bmatrix}$
if $C=AB-BA$, then $C_{11}+C_{22}=bg-cf+cf-bg=0$.
Conversely, if $C_{11}+C_{22}=0$, then $C_{11}=-C_{22}$, thus solve the equation
$\begin{aligned}bg-cf&=C_{11}\\ af+bh-eb-df=(a-d)f+b(h-e)&=C_{12}\\ ce+dg-ag-hc=(d-a)g+c(e-h)&=C_{21}\end{aligned}$
It’s a system of 3 equations with 8 unknowns, fix $a=1,d=0,h=0,e=1$, the system becomes
$\begin{aligned}bg-cf&=C_{11}\\ f-b&=C_{12}\\ g-c&=C_{21}\end{aligned}$
thus $b(C_{21}+c)-c(C_{12}+b)=C_{11}$, or $bC_{21}=C_{11}+cC_{12}$, if $C_21\neq 0$, let $c=1$; if $C_{21}=0$, let $c=-C_{11}/C_{12}$ and $b=1$, all other unkowns can be solved.