1.3 Matrices and Elementary Row Operations

Hoffman这本书有个特点，初读文字有些拗口，英语没点阅读功底容易理解错误他的意思，但是如果能读进去，就能发现这本书最好的地方：视角高。如果说Rudin的分析是大巧不工的大师级，Hoffman的线代就是一个略有些唠叨的大师级。从1.3开始，就有些体会出这样的意思。这一节至少有如下两个拔高：矩阵拔高到函数维度，row-equivalence是一个等价关系。
这一节说Matrix和elementary row operations，我在学北大王萼芳那本高等代数时候也是先学的这一块（当然要在行列式之后），当时就觉得很难理解，把一堆数排成方块形有什么作用，对行进行变换的时候为何选那特定的三种变换。Hoffman是从上一节说的system of linear equations得出的matrix，实际是系数矩阵，但随后正式给的定义又是：matrix是从整数对 $(i,j)$到 $F$的一个函数，其entry就是这一函数在特定的 $(i,j)$上的取值；我们常见的矩阵的样子是矩阵最方便的一种表示。
接下来给出三种elementary row operations：某一行变为原来的 $c$倍 $(c\neq 0)$；变为自身加上另外一行的 $c$倍；调换两行。这三种都可以分别用一个函数 $e$精确的表示出来，这个 $e$是定义在一个固定的 $m$（矩阵的行数）上，不用管列是多少。为何选这三种？最重要的是这三种方式可以用同样的方式可逆，也就是Theorem 2的结论。由elementary row operations得到的矩阵和原来矩阵是row-equivalent的，并且揭示了row-equivalent满足self-reflexive,symmetric和transitivity，故而是一个等价关系。由于elementary row operations不会改变以该矩阵作为coefficient matrix的方程组系统的解空间，因此Theorem 3说明：row-equivalent的两个矩阵为coefficient matrix对应的方程组系统有相同的解空间。
最后，有一个row-reduced的定义，即满足一是每一个非零行从左往右先碰到的非零元素（leading non-zero entry）必须是1，二是含有“一是”里这样的1的列，其他元素都是0。Theorem 4说明任何一个matrix都可以被row-reduced。证明是纯说理性的，可能那个时代（20世纪70年代）画图不易，硬是像讲故事一样把证明说清楚了。

Exercises

1. Find all solutions to the system of equations

$(1-i)x_1-ix_2=0\\2x_1+(1-i)x_2=0$
Solution: We have
$\begin{cases}(1-i) x_1-ix_2=0\\2x_1+(1-i) x_2=0\end{cases} \Rightarrow \begin{cases}2x_1-(1+i)ix_2=0\\2x_1+(1-i) x_2=0\end{cases} ⇒ \begin{cases}2x_1+(1-i)x_2=0\\2x_1+(1-i) x_2=0\end{cases}$
so let $x_2=c$, we have $x_1=(i-1)c/2$, so all solutions to the system is
$x_1=\dfrac{(i-1)c}{2},\quad x_2=c,\quad ∀c\in C$

2. If

$A=\begin{bmatrix}3&-1&2\\2&1&1\\1&-3&0 \end{bmatrix}$

find all solutions of $AX=0$ by row-reducing $A$.

Solution: Since
$\begin{bmatrix}3&-1&2\\2&1&1\\1&-3&0\end{bmatrix}→\begin{bmatrix}0&8&2\\0&7&1\\1&-3&0\end{bmatrix}→\begin{bmatrix}0&1&1\\0&7&1\\1&-3&0\end{bmatrix}→\begin{bmatrix}0&1&1\\0&0&-6\\1&0&3\end{bmatrix}→\begin{bmatrix}0&1&1\\0&0&1\\1&0&3\end{bmatrix}→\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$
so $AX=0$ has only trivial solutions.

3. If

$A=\begin{bmatrix}6&-4&0\\4&-2&0\\-1&0&3 \end{bmatrix}$

find all solutions of $AX=2X$ and $AX=3X$.

Solution: It’s equivalent to solving $(A-2I)X=0$ and $(A-3I)X=0$, use row-reducing we have
$A-2I=\begin{bmatrix}4&-4&0\\4&-4&0\\-1&0&1\end{bmatrix}→\begin{bmatrix}1&-1&0\\0&0&0\\0&1&1\end{bmatrix}$
so let $x_2=c$, we have $x_1=c$ and $x_3=-c$, so all solutions to the system $(A-2I)X=0$ is $(c,c,-c),c\in C$.
$A-3I=\begin{bmatrix}3&-4&0\\4&-5&0\\-1&0&0\end{bmatrix}→\begin{bmatrix}0&0&0\\0&1&0\\1&0&0\end{bmatrix}$
so all solutions to the system $(A-3I)X=0$ is $(0,0,c),c\in C$.

4. Find a row-reduced matrix which is row-equivalent to

$A=\begin{bmatrix}i&-(1+i)&0\\1&-2&1\\1&2i&-1 \end{bmatrix}$
Solution:
$\begin{bmatrix}i&-(1+i)&0\\1&-2&1\\1&2i&-1\end{bmatrix}→\begin{bmatrix}0&-1+i&-i\\1&-2&1\\0&2i+2&-2\end{bmatrix}\\→\begin{bmatrix}0&-1+i&-i\\1&-2&1\\0&i+1&-1\end{bmatrix}→\begin{bmatrix}0&0&0\\1&-2&1\\0&1&\frac{i-1}{2}\end{bmatrix}→\begin{bmatrix}0&0&0\\1&0&i\\0&1&\frac{i-1}{2}\end{bmatrix}$

5. Prove that the following two matrices are not row-equivalent:

$\begin{bmatrix}2&0&0\\a&-1&0\\b&c&3 \end{bmatrix},\quad \begin{bmatrix}1&1&2\\-2&0&-1\\1&3&5 \end{bmatrix}$
Solution: The first matrix can be row reduced to
$\begin{bmatrix}2&0&0\\a&-1&0\\b&c&3\end{bmatrix}→\begin{bmatrix}1&0&0\\0&-1&0\\0&c&3\end{bmatrix}→\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
the second matrix can be row-reduced to
$\begin{bmatrix}1&1&2\\-2&0&-1\\1&3&5\end{bmatrix}→\begin{bmatrix}1&1&2\\0&2&3\\0&2&3\end{bmatrix}→\begin{bmatrix}1&1&2\\0&1&1/2\\0&0&0\end{bmatrix}→\begin{bmatrix}1&0&2\\0&1&1/2\\0&0&0\end{bmatrix}$
thus they are not row-equivalent.

6.Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ be a $2\times 2$ matrix with complex entries. Suppose that $A$ is row-reduced and also that $a+b+c+d=0$. Prove that there are exactly three such matrices.

Solution: The zero matrix $\begin{bmatrix}0&0\\0&0\end{bmatrix}$ is row-reduced and satisfy $a+b+c+d=0$, suppose there’re nonzero entries, then at least two entries shall be nonzero, otherwise contradicting $a+b+c+d=0$, consider the case when there’re exactly $2$ nonzero entries, then they can’t belong to different rows, since then the matrix would be $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ or $\begin{bmatrix}0&1\\1&0\end{bmatrix}$, under the restriction of row-reduced, but then contradicting $a+b+c+d=0$, in the case when they belong to the same row, we get the matrix $\begin{bmatrix}1&-1\\0&0\end{bmatrix}$ or $\begin{bmatrix}0&0\\1&-1\end{bmatrix}$ satisfy the condition.
Now consider the case when there’re exactly $3$ nonzero entries, then the matrix is of the form $\begin{bmatrix}1&b\\0&d\end{bmatrix}$ or $\begin{bmatrix}0&b\\1&d\end{bmatrix}$, but notice then either $b=0,d=1$ or $b=1,d=0$, under the restriction of row-reduced, this contradicting $a+b+c+d=0$.
Finally the case when all $4$ entries is nonzero contradicts the restriction of row-reduced.
Thus there’re exactly three such matrices: $\begin{bmatrix}0&0\\0&0\end{bmatrix}$, $\begin{bmatrix}1&-1\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\1&-1\end{bmatrix}$.

7.Prove that the interchange of two rows of a matrix can be accomplished by a finite sequence of elementary row operations of the other two types.

Solution: It’s enough to show the process for a two-row matrix:
$\begin{aligned}\begin{bmatrix}a_1&\cdots&a_n\\b_1&\cdots&b_n\end{bmatrix} \xrightarrow {\text{add row1 to row2}} \begin{bmatrix}a_1&\cdots&a_n\\b_1+a_1&\cdots&b_n+a_n\end{bmatrix} \\ \xrightarrow{\text{add (-1)*row2 to row1}} \begin{bmatrix}-b_1&\cdots&-b_n\\b_1+a_1&\cdots&b_n+a_n\end{bmatrix}\\ \xrightarrow{\text{add row1 to row2}} \begin{bmatrix}-b_1&\cdots&-b_n\\a_1&\cdots&a_n\end{bmatrix}\\ \xrightarrow{\text{multiply row1 by -1}} \begin{bmatrix}b_1&\cdots&b_n\\a_1&\cdots&a_n\end{bmatrix}\end{aligned}$

8. Consider the system of equations $AX=0$ where $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is a $2\times 2$ matrix over the field $F$. Prove the following:

( a ) If every entry of $A$ is $0$, then every pair $(x_1,x_2)$ is a solution of $AX=0$.

( b ) If $ad-bc\neq 0$, the system $AX=0$ has only the trivial solution $x_1=x_2=0$.

( c ) If $ad-bc= 0$, and some entry of $A$ is different from $0$, then there is a solution $(x_1^0,x_2^0)$ such that $(x_1,x_2)$ is a solution if and only if there is a scalar $y$ such that $x_1=yx_1^0,x_2=yx_2^0$.

Solution:
( a ) We have $0X=0$ for every $X$.
( b ) Since $a$ and $c$ can’t both be $0$, we suppose $a\neq 0$, use row reduction, we have
$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}→\begin{bmatrix}a&b\\ca&da\end{bmatrix}→\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix}→\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
and the conclusion follows.
( c ) If both $a$ and $c$ is $0$, then $A$ is row-equivalent to $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ or $\begin{bmatrix}0&0\\0&1\end{bmatrix}$. If not, then $A$ is row-equivalent to $\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix}$ or $\begin{bmatrix}0&bc-ad\\c&d\end{bmatrix}$, in all cases $A$ is row-equivalent to a matrix that has nonzero entries in only one row, thus the system $AX=0$ has the same solution with
$αx_1+βx_2=0,\quad (α≠0)∨(β≠0)$
let $(x_1^0,x_2^0)$ be a nonzero solution, then $αx_1^0=-βx_2^0$, thus either $α≠0$, which leads to $x_1^0=kx_2^0,k=-β/α$, or $β≠0$, which leads to $x_2^0=lx_1^0,l=-α/β$. If there’s any $(x_1,x_2)$ which satisfies $αx_1+βx_2=0$, then we must also have $x_1=kx_2$ or $x_2=lx_1$, and the conclusion follows.
In fact, $(x_1^0,x_2^0)$ is a basis for the solution space in this case, so any $(x_1,x_2)$ being a solution, we must have $(x_1,x_2 )=y(x_1^0,x_2^0)$.