本文为《Linear algebra and its applications》的读书笔记

If $A$A is an $m \times n$m×n matrix, each column of $A$A is a list of $m$m real numbers, which identifies a vector in $\mathbb R^m$Rm.
$A=[\ \ \boldsymbol a_1\ \ \ \boldsymbol a_2\ \ \ ...\ \ \ \boldsymbol a_n\ \ ]$A=[  a1​   a2​   ...   an​  ]

The diagonal entries (对角线元素) in an $m \times n$m×n matrix $A = [a_{ij}]$A=[aij​] are $a_{11}$a11​, $a_{22}$a22​, $a_{33}$a33​… and they form the main diagonal (主对角线) of $A$A. A diagonal matrix (对角矩阵) is a square $n \times n$n×n matrix whose nondiagonal entries are zero. An $m \times n$m×n matrix whose entries are all zero is a zero matrix (零矩阵) and is written as $0$0. The size of a zero matrix is usually clear from the context.

Sums and Scalar Multiples

The arithmetic for vectors described earlier has a natural extension to matrices.

The sum $A + B$A+B is the $m \times n$m×n matrix whose columns are the sums of the corresponding columns in $A$A and $B$B. The sum $A + B$A+B is defined only when $A$A and $B$B are the same size.

If $r$r is a scalar and $A$A is a matrix, then the scalar multiple $rA$rA is the matrix whose columns are $r$r times the corresponding columns in $A$A.

Matrix Multiplication

When a matrix $B$B multiplies a vector $\boldsymbol x$x, it transforms $\boldsymbol x$x into the vector $B\boldsymbol x$Bx. If this vector is then multiplied in turn by a matrix $A$A, the resulting vector is $A(B\boldsymbol x)$A(Bx). See Figure 2.

Thus $A(B\boldsymbol x)$A(Bx) is produced from $\boldsymbol x$x by a $composition$composition of mappings. Our goal is to represent this composite mapping as multiplication by a single matrix, denoted by $AB$AB, so that
$A(B\boldsymbol x)=(AB)\boldsymbol x$A(Bx)=(AB)x

If $A$A is $m \times n$m×n, $B$B is $n \times p$n×p, and $\boldsymbol x$x is in $\mathbb R^p$Rp. Then
$B\boldsymbol x=x_1\boldsymbol b_1+...+x_p\boldsymbol b_p$Bx=x1​b1​+...+xp​bp​By the linearity of multiplication by $A$A,
$A(B\boldsymbol x)=A(x_1\boldsymbol b_1)+...+A(x_p\boldsymbol b_p)=x_1A\boldsymbol b_1+...+x_pA\boldsymbol b_p$A(Bx)=A(x1​b1​)+...+A(xp​bp​)=x1​Ab1​+...+xp​Abp​The vector $A(B\boldsymbol x)$A(Bx) is a linear combination of the vectors $A\boldsymbol b_1$Ab1​,…,$A\boldsymbol b_p$Abp​, using the entries in $\boldsymbol x$x as weights. In matrix notation, this linear combination is written as
$A(B\boldsymbol x)=[\ \ A\boldsymbol b_1\ \ \ A\boldsymbol b_2\ \ \ ...\ \ \ A\boldsymbol b_p\ \ ]\boldsymbol x$A(Bx)=[  Ab1​   Ab2​   ...   Abp​  ]xThus multiplication by $[\ \ A\boldsymbol b_1\ \ \ A\boldsymbol b_2\ \ \ ...\ \ \ A\boldsymbol b_p\ \ ]$[  Ab1​   Ab2​   ...   Abp​  ] transforms $\boldsymbol x$x into $A(B\boldsymbol x)$A(Bx). We have found
the matrix we sought!

Multiplication of matrices corresponds to composition of linear transformations.

The definition of $AB$AB lends itself well to parallel processing on a computer. The columns of $B$B are assigned individually or in groups to different processors, which independently and hence simultaneously compute the corresponding columns of $AB$AB.

EXAMPLE 3
Compute $AB$AB, where $A =\begin{bmatrix}2&3\\1&-5\end{bmatrix}$A=[21​3−5​] and $B =\begin{bmatrix}4&3&6\\1&-2&3\end{bmatrix}$B=[41​3−2​63​].
SOLUTION

Each column of $AB$AB is a linear combination of the columns of $A$A using weights from the corresponding column of $B$B. ($AB$AB 的每一列都是$A$A 的各列的线性组合，以$B$B 的对应列的元素为权)

Obviously, the number of columns of $A$A must match the number of rows in $B$B in order for a linear combination such as $A\boldsymbol b_1$Ab1​ to be defined. Also, the definition of $AB$AB shows that $AB$AB has the same number of rows as $A$A and the same number of columnsas $B$B.

The definition of $AB$AB is important for theoretical work and applications, but the following rule provides a more efficient method for calculating the individual entries in $AB$AB when working small problems by hand.

Let $row_i(A)$rowi​(A) denote the $i$i th row of a matrix $A$A. Then
$row_i(AB)=row_i(A)\cdot B$rowi​(AB)=rowi​(A)⋅B

View vectors in $\mathbb R^n$Rn as $n \times 1$n×1 matrices. For $\boldsymbol u$u and $\boldsymbol v$v in $\mathbb R^n$Rn, the matrix product $\boldsymbol u^T \boldsymbol v$uTv is a $1 \times 1$1×1 matrix, called the scalar product (数量积), or inner product (内积), of $\boldsymbol u$u and $\boldsymbol v$v. It is usually written as a single real number without brackets. The matrix product $\boldsymbol u\boldsymbol v^T$uvT is an $n \times n$n×n matrix, called the outer product (外积) of $\boldsymbol u$u and $\boldsymbol v$v.

inner products ($\boldsymbol u^T \boldsymbol v$uTv and $\boldsymbol v^T \boldsymbol u$vTu) have the transpose symbol in the middle. Outer products ($\boldsymbol u\boldsymbol v^T$uvT and $\boldsymbol v\boldsymbol u^T$vuT) have the transpose symbol on the outside.

内积 $\boldsymbol u^T \boldsymbol v=\boldsymbol v^T \boldsymbol u$uTv=vTu
外积 $\boldsymbol u\boldsymbol v^T$uvT 和 $\boldsymbol v\boldsymbol u^T$vuT 互为对称矩阵

EXAMPLE 4
Suppose the last column of $AB$AB is entirely zero but $B$B itself has no column of zeros. What can you say about the columns of $A$A?
SOLUTION
Let $\boldsymbol b_p$bp​ be the last column of $B$B. By hypothesis, the last column of $AB$AB is zero. Thus, $A\boldsymbol b_p$Abp​ = 0. However, $\boldsymbol b_p$bp​ is not the zero vector. Thus, the equation $A\boldsymbol b_p = \boldsymbol 0$Abp​=0 is a linear dependence relation among the columns of $A$A, and so the columns of $A$A are linearly dependent.

Properties of Matrix Multiplication

Recall that $I_m$Im​ represents the $m \times m$m×m identity matrix and $I_m\boldsymbol x=\boldsymbol x$Im​x=x for all $\boldsymbol x$x in $\mathbb R^m$Rm.

PROOF (只证明性质(a)，其他性质的证明类似)
Property (a)
Property (a) follows from the fact that matrix multiplication corresponds to composition of linear transformations (which are functions), and it is known that the composition of functions is associative(结合律).

在离散数学中，函数的复合运算 $f\circ g$f∘g 被定义为关系乘积 $g*f$g∗f，可以证明关系乘积是满足结合律的，因此函数的复合运算也满足结合律

Here is another proof of (a) that rests on the “column definition” of the product of two matrices. Let
$C=[\ \ \boldsymbol c_1\ \ \ \boldsymbol c_2\ \ \ ...\ \ \ \boldsymbol c_p\ \ ]$C=[  c1​   c2​   ...   cp​  ]$BC=[\ \ B\boldsymbol c_1\ \ \ B\boldsymbol c_2\ \ \ ...\ \ \ B\boldsymbol c_p\ \ ]$BC=[  Bc1​   Bc2​   ...   Bcp​  ]$A(BC)=[\ \ A(B\boldsymbol c_1)\ \ \ A(B\boldsymbol c_2)\ \ \ ...\ \ \ A(B\boldsymbol c_p)\ \ ]$A(BC)=[  A(Bc1​)   A(Bc2​)   ...   A(Bcp​)  ]Recall that the definition of $AB$AB makes $A(B\boldsymbol x)= (AB)\boldsymbol x$A(Bx)=(AB)x for all $\boldsymbol x$x, so
$A(BC)= [(AB)\boldsymbol c_1\ \ \ ...\ \ \ (AB)\boldsymbol c_p]=(AB)C$A(BC)=[(AB)c1​   ...   (AB)cp​]=(AB)C

If $AB = BA$AB=BA, we say that $A$A and $B$B commute with one another.(可交换的)

WARNINGS:

1. In general, $AB \neq BA$AB​=BA.
2. The cancellation laws(消去律) do not hold for matrix multiplication. That is, if $AB = AC$AB=AC, then it is not true in general that $B = C$B=C.
3. If a product $AB$AB is the zero matrix, you cannot conclude in general that either $A = 0$A=0 or $B = 0$B=0.

Tip: When $B$B is square and $C$C has fewer columns than $A$A has rows, it is more efficient to compute $A(BC)$A(BC) than $(AB)C$(AB)C .

$Checkpoint$Checkpoint: Show that if $\boldsymbol y$y is a linear combination of the columns of $AB$AB, then $\boldsymbol y$y is a linear combination of the columns of $A$A.

$Answer\ to\ Checkpoint$Answer to Checkpoint: If $\boldsymbol y$y is a linear combination of the columns of $AB$AB, then there is a vector $\boldsymbol x$x such that $\boldsymbol y$y = $(AB)\boldsymbol x$(AB)x. By definition of matrix multiplication, $\boldsymbol y = A(B\boldsymbol x)$y=A(Bx). This expresses y as a linear combination of the columns of $A$A using the entries in the vector $B\boldsymbol x$Bx as weights.

Powers of a Matrix

If $A$A is an $n \times n$n×n matrix and if $k$k is a positive integer, then

If $k = 0$k=0; then $A^0\boldsymbol x$A0x should be $\boldsymbol x$x itself. Thus $A^0$A0 is interpreted as the identity matrix.

The Transpose of a Matrix

PROOF
Property (d)
The $(i, j)$(i,j)-entry of $(AB)^T$(AB)T is the $(j, i)$(j,i)-entry of $AB$AB, which is
$a_{j1}b_{1i}+...+a_{jn}b_{ni}$aj1​b1i​+...+ajn​bni​The entries in row $i$i of $B^T$BT are $b_{1i}, . . . , b_{ni}$b1i​,...,bni​, because they come from column $i$i of $B$B. Likewise, the entries in column $j$j of $A^T$AT are $a_{j1}, …, a_{jn}$aj1​,…,ajn​, because they come from row $j$j of $A$A. Thus the $(i, j)$(i,j)-entry in $B^TA^T$BTAT is $a_{j1}b_{1i}+...+a_{jn}b_{ni}$aj1​b1i​+...+ajn​bni​ , as above.

The generalization of Theorem 3(d) to products of more than two factors can be stated in words as follows: