Linear Algebra - Matrices

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Most of the content of this article comes from teacher Li Yongle's postgraduate entrance examination teaching materials and video lessons.

basic concept

• Matrix: by $m\times$mmcomposed of$n\; numbers$$m$ rownnA tablewith$n$ columns is called an$m\times n$ matrix, denoted as$A$。当$m=n$ , called$A$ is$n$阶矩阵。
  $$\left[\begin{array}{ccccc}{a}_{11}& a_{12}& \dots & a_{1n}& \\ a_{21}& a_{22}& \dots & a_2& \\ ⋮& ⋮& & ⋮& \\ a_m& a_m& \dots & a_{mn}& \end{array}\right]$$
• Homotype matrix: if $A$ and$B$ are all$m\times n$ matrix, then called$A$ and$B$ isthe homotype matrix.
• Equal matrix: let $A,B$ is the same matrix, if$a_{ij}=b_{ij}(\forall i=1,2,\dots ,m;j=1,2,\dots ,n)$ , then it is called$A$ and$B$ is equal, recorded as$A=B$。
• Zero Matrix: If all elements of a matrix are $0$ , then this matrix is ​​called a zero, denoted as$O$。
• Specifications: $$\left[\begin{array}{cccc}{a}_{11}& & & \\ & a_{22}& & \\ & & \ddots & \\ & & & a_{nn}\end{array}\right]$$
• Identity matrix: $$\left[\begin{array}{cccc}1& & & \\ & 1& & \\ & & \ddots & \\ & & & 1\end{array}\right]$$denoted as $E$。
• Details: $$\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ & a_{22}& \dots & a_2\\ & & \ddots & ⋮\\ & & & a_{nn}\end{array}\right]$$当$i>j$时，$a_{ij}=0$
• Details: $$\left[\begin{array}{cccc}{a}_{11}& & & \\ a_{21}& a_{22}& & \\ ⋮& ⋮& \ddots & \\ a_n& a_n& \dots & a_{nn}\end{array}\right]$$当$i<j$时，$a_{ij}=0$

Matrix operations

Addition of matrices

若$A=[a_{ij}],B=[b_{ij}]$ is the same matrix, then
$$A+B=[a_{ij}+b_{ij}]$$
Addition algorithm ($A,B,C$ same type):

• $A+B=B+A$
• $(A+B)+C=A+(B+C)$
• $A+O=A$
• $A+(-A)=O$

Multiply numbers and matrices

$$toA=\left[k{a}_{ij}\right]$$
Number multiplication algorithm:

• $k\left(mA\right)\_=m\left(kA\right)=(mk)A$
• $(k+m)A=toA+mA$
• $k(A+B)=toA+kB$
• $1A=A$
• $0A=O$

matrix multiplication

若$A=[a_{ij}{]}_{m\times s},B=[b_{ij}{]}_{s\times n}$，则$$A\times B=C=[c_{ij}{]}_{m\times n}$$其中$$c_{ij}=a_i{b}_1+a_{i2}{b}_2+\cdots +a_{is}{b}_{sj}=\sum _{k=1}^n{a}_{}{b}_{kj}$$
Multiplication algorithm:

• $A\left(BC\right)=(AB)\; Proof\; ofC$
  : Let$A=[a_{ij}{]}_{m\times s}$，$B=[b_{ij}{]}_{s\times t}$，$C=[c_{ij}{]}_{t\times n}$：
  $$\begin{gathered} A\left(BC\right)=D_{m\times n} \\ (AB)C=E_{m\times n} \end{gathered}$$
  D D iiin$D$row$i$ , jjThe elements of column$j$$A$ middleiiLine$i$$BC$中ddMultiply and add the corresponding elements of column$j$
  $$\sum _{k=1}^s{a}_{}(\sum _{p=1}^t{b}_{kp}{c}_{pj})=\sum _{k=1}^s\sum _{p=1}^t{a}_{}{b}_{kp}{c}_{pj}$$
  Empathy $E$ middleiirow$i$ , jjThe elements of column$j$
  $$\sum _{p=1}^t(\sum _{k=1}^s{a}_{}{b}_{kp})c_{pj}=\sum _{k=1}^s\sum _{p=1}^t{a}_{}{b}_{kp}{c}_{pj}$$
  Certificate completed.
• $A(B+C)=AB+AC$
• $\left(kA\right)\left(lB\right)=klAB$
• $AE=A,EA=A$
• $OA=O,The\_=O$

Notice:

• $AB\ne BA$
• $AB=O\nRightarrow A=O$ or$B=O$
• $AB=AC,A\ne O\nRightarrow B=C$
• Young $A,B$ is a diagonal matrix, then$AB=BA$
• ${\left[\begin{array}{ccc}{a}_1& & \\ & a_2& \\ & & a_3\end{array}\right]}^n=A=\left[\begin{array}{ccc}{a}_1^n& & \\ & a_2^n& \\ & & a_3^n\end{array}\right]$

Transpose of matrix

设$A=[a_{ij}{]}_{m\times n}$, General $The\; rows\; and\; columns\; of\; A$ are exchanged, and the obtained$n\times m$矩阵$[a_{ji}{]}_{n\times m}$called AAThe transpose matrix of$A$$A^{T.}$ _ Transpose algorithm:

• $(A+B{)}^T=A^T+B^T$
• $(kA{)}^T=to{A}^T$
• $(AB{)}^T=B^T{A}^T$
  证明：设$A=[a_{ij}{]}_{m\times s},B=[b_{ij}{]}_{s\times n}$，则有：
  $$\begin{gathered} AB=[c_{ij}{]}_{m\times n} \\ \Downarrow \\ (AB{)}^T=[c_{ji}{]}_{n\times m} \end{gathered}$$
  where
  $$c_{ji}=\sum _{k=1}^s{a}_{}{b}_{kj}$$
  令
  $$B^T{A}^T=[d_{ij}{]}_{n\times m}$$
  $B^T{A}^T$ 's$j$ line$The\; i$ column element is$B^T$ 's$Line\; j$ and$A^T$ of$The\; income\; in\; column\; i$ is also$B$ 's$Column\; j$ and$A$ ii_The result of row$i$
  $$d_{ji}=\sum _{k=1}^s{b}_{kj}{a}_{}=c_{ji}$$
  Certificate completed.
• $(A^T{)}^T=A$

若$A^T=A$ , aka$A$ isa symmetric matrix, if$A^T=-A$ , abbreviated as$A$ isan antisymmetric matrix.

Matrices and Systems of Equations

There is a system of linear equations in two variables:
$$\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1(1)\\ a_{2}x+b_{2}y=c_2(2)\end{array}$$
Available matrix representations are:
$$\begin{gathered} \left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right] \\ \Downarrow \\ AX=C \end{gathered}$$
in$A$ is called the coefficient matrix,$X$ is called the unknown matrix,$C$ is called a constant term matrix.

Square Matrix and Determinant

任$A=\left[a_{ij}\right]$ fornnThe determinant formed by a square matrix of order$n$$The\; determinant\; of\; A$ , denoted as$$|A|$$ Note:

• Only square matrices have determinants.
• $A=O$ sum$|A|=0$ doesn't matter.

Properties of the determinant of a square matrix:

• $|A^T|=|A|$
• $|kA|=k^n|A|$
• $|AB|=|A||B|Proof$
  : Let$A$、$B$ is$n$ -order matrix (with$n=2$）为例：
  $$A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]B=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]$$
  form the fourth order determinant$D$：
  $$D=|A||B|=\left|\begin{array}{cc}A& O\\ -E& B\end{array}\right|=\left|\begin{array}{cccc}{a}_{11}& a_{12}& 0& 0\\ a_{21}& a_{22}& 0& 0\\ -1& 0& b_{11}& b_{12}\\ 0& -1& b_{21}& b_{22}\end{array}\right|=\left|\begin{array}{cccc}{a}_{11}& a_{12}& a_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}& a_{22}& a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\\ -1& 0& 0& 0\\ 0& -1& 0& 0\end{array}\right|=\left|\begin{array}{cc}A& AB\\ -E& O\end{array}\right|=-1^{(2\times 2)}|-E||AB|=|AB|$$

Adjoint matrix

任$A=\left[a_{ij}\right]$ is$Square\; matrix\; of\; order\; n$ , determinant$|A|$ each element$a_{ij}$The algebraic remainder $A_{ij}$The following square matrix formed is called $A$的伴随矩阵。$$A^{\ast }=\left[\begin{array}{cccc}{A}_{11}& A_{21}& \dots & A_n\\ A_{12}& A_{22}& \dots & A_n\\ ⋮& ⋮& & ⋮\\ A_{1n}& A_2& \dots & A_{nn}\end{array}\right]$$Properties of the adjoint matrix:

• $A{A}^{\ast }=A^{\ast }A=|A|E$
  proof: Let$A$ is$n$ -order matrix (with$n=2$）为例：
  $$A{A}^{\ast }=\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\left|\begin{array}{cc}{A}_{11}& A_{21}\\ A_{12}& A_{22}\end{array}\right|=\left|\begin{array}{cc}{a}_{11}{A}_{11}+a_{12}{A}_{12}& a_{11}{A}_{21}+a_{12}{A}_{22}\\ a_{21}{A}_{11}+a_{22}{A}_{12}& a_{21}{A}_{21}+a_{22}{A}_{22}\end{array}\right|=\left|\begin{array}{cc}|A|& O\\ O& |A|\end{array}\right|=|A|\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=|A|E$$
• The adjoint matrix of the second-order matrix: the main diagonal is swapped, and the sub-diagonal is changed in sign.

invertible matrix

for $n$ -order square matrix$A$ , if there exists$n$ order square matrix$B$，使$$AB=BA=E$$ is called matrix$A$ is invertible, matrix$B$ isAAThe inverse matrix of$A.$If matrix$A$ is reversible, then$The\; inverse\; matrix\; of\; A$ is unique, denoted as$$A^{The\; properties\; of\; -1}$$ invertible matrix are as follows$${}^{:}$$

• If $A$ is reversible, then$A^{-1}$ is also reversible, and$(A^{-1}{)}^{-1}=A.$ _
  Proof: Decree$$B=A^{-1}$$
  :$$BA=AB=E$$
  so$B$ reversible and$$B^{-1}=(A^{-1}{)}^{-1}=A$$
• If $A$ is reversible, and$k\ne 0$ ,$kA$ is reversible, and$(kA{)}^{-1}=\frac{1}{k}{A}^{-1}$。
• Fruit $A,B$ is reversible, then$AB$ is also reversible, and$(AB{)}^{-1}=B^{-1}{A}^{-1}$。
• If $A$ is reversible, then$A^T$ is also invertible, and$(A^T{)}^{-1}=(A^{-1}{)}^T$。
• $A$可逆$\iff |A|\ne 0$ .
  Proof: Because
  $$|A{A}^{-1}|=|A||A^{-1}|=|E|=1$$
  so$|A|\ne 0$
• $A,B$ is for$Square\; matrix\; of\; order\; n$ , if$AB=E$ ,$A^{-1}=B$。
• If $A$ reversible,$A^{-1}=\frac{1}{|A|}{A}^{\ast }$。
• If $A$ reversible,$|A^{-1}|=\frac{1}{|A|}$。
• Let $A=\left[\begin{array}{ccc}{a}_1& & \\ & a_2& \\ & & a_3\end{array}\right]$is a diagonal matrix, then $A^{-1}=\left[\begin{array}{ccc}\frac{1}{a_1}& & \\ & \frac{1}{a_2}& \\ & & \frac{1}{a_3}\end{array}\right]$。

Partitioned Matrix

Appropriate block processing of the matrix can be efficiently calculated, and the following algorithm is available after block:

• $\left[\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right]+\left[\begin{array}{cc}{B}_1& B_2\\ B_3& B_4\end{array}\right]=\left[\begin{array}{cc}{A}_1+B_1& A_2+B_2\\ A_3+B_3& A_4+B_4\end{array}\right]$
• $\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{cc}X& Y\\ Z& W\end{array}\right]=\left[\begin{array}{cc}AX+BZ& AY+BW\\ CX+DZ& CY+DW\end{array}\right]$
• ${\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]}^T=\left[\begin{array}{cc}{A}^T& C^T\\ B^T& D^T\end{array}\right]$

A $A,B$ are m , nm, nrespectively$m,n$ order square matrix, then:

• ${\left[\begin{array}{cc}A& O\\ O& B\end{array}\right]}^n=\left[\begin{array}{cc}{A}^n& O\\ O& B^n\end{array}\right]$

A $A,B$ are m , nm, nrespectively$m,n$ -order reversible matrix, then:

• ${\left[\begin{array}{cc}A& O\\ O& B\end{array}\right]}^{-1}=\left[\begin{array}{cc}{A}^{-1}& O\\ O& B^{-1}\end{array}\right]$
• ${\left[\begin{array}{cc}O& A\\ B& O\end{array}\right]}^{-1}=\left[\begin{array}{cc}O& C^{-1}\\ B^{-1}& O\end{array}\right]$

Young $A$是$m\times n-$ matrix,$B$ is$n\times s$ matrix and$AB=C$ , then for$B$，$C$ ratio:
$$\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_2\\ ⋮& ⋮& & ⋮\\ a_m& a_m& \dots & a_{mn}\end{array}\right]\left[\begin{array}{cccc}{b}_1& b_2& \dots & b_n\end{array}\right]=\left[\begin{array}{cccc}{c}_1& c_2& \dots & c_n\end{array}\right]$$
即$$\{\begin{array}{l}{a}_{11}{b}_1+a_{12}{b}_2+\cdots +a_{1n}{b}_n=c_1\\ a_{21}{b}_1+a_{22}{b}_2+\cdots +a_2{b}_n=c_2\\ \dots \\ a_n{b}_1+a_n{b}_2+\cdots +a_{nn}{b}_n=c_n\end{array}$$

Elementary Transformation of Matrix

To solve the following system of linear equations:
$$\{\begin{array}{l}2x{\_}_1-x_2+3x{\_}_3=1\\ 4x{\_}_1+2x{\_}_2+5x_3=4\\ 2x{\_}_1+2x{\_}_3=6\end{array}$$
According to the method of addition and subtraction :

• Multiply both sides of the equation by a non-zero number.
• kk of an equation$k$ times is added to another equation.
• Swap the positions of the two equations.

Simplify the system of equations to:
$$\{\begin{array}{l}2x{\_}_1-x_2+3x{\_}_3=1\\ x_2-x_3=5\\ x_3=-6\end{array}$$
The solution of the equation can be obtained. The essence of addition, subtraction and elimination is to change the unknown coefficients and constant items, so the unknown coefficients and constant items can be written as a matrix: [ 2 − 1 3 1 4 2 5 4 2 0 2
$$\left[\begin{array}{cccc}2& -1& 3& 1\\ 4& 2& 5& 4\\ 2& 0& 2& 6\end{array}\right]$$
This matrix is ​​called the augmented matrix of the system of linear equations . And the following three transformations are called the elementary row (column) transformation of the matrix, collectively referred to as the elementary transformation of the matrix:

• Multiply a row (column) of a matrix by a nonzero constant.
• kk of a row (column)$k$ times to another row (column).
• Swaps the positions of two rows (columns) in the matrix.

Now perform elementary row transformation on the matrix from top to bottom to simplify the matrix to a ladder type (this process is also called forward elimination):
$$\left[\begin{array}{cccc}2& -1& 3& 1\\ 0& 1& -1& 5\\ 0& 0& 1& -6\end{array}\right]$$
Then add the simplified matrix to the unknown from bottom to top to get the solution of the linear equation (this process is also called reverse solution).

elementary matrix

The matrix obtained by an elementary transformation from the identity matrix is ​​called an elementary matrix . The properties of elementary matrices are as follows:

• Elementary Matrix $P$ is left multiplied by$A$ obtained$PA$ is to$A$ do it once with$P$ The same elementary row transformation.
• Elementary Matrix $P$ right multiplied by$A$ obtained$AP$ is to$A$ do it once with$P$ The same elementary column transformation.
• ${\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& k& 1\end{array}\right]}^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& -k& 1\end{array}\right]$
• ${\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]}^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$
• ${\left[\begin{array}{ccc}1& 0& 0\\ 0& k& 0\\ 0& 0& 1\end{array}\right]}^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{k}& 0\\ 0& 0& 1\end{array}\right]$

equivalence matrix

If matrix $A$ undergoes a finite number of elementary transformations to become a matrix$B$ , we call the matrix$A$ and$B$ is equivalent, recorded as$A\cong B.$ _ The properties of matrix equivalence are as follows:

• Reflexivity: $A\cong A$
• Symmetry: If $A\cong B$，则$B\cong A$
• Transitivity: if $A\cong B,B\cong C$，则$A\cong C$

row echelon matrix

AA $A$ is a$m\times The\; matrix\; of\; n$ , if it satisfies:

• If the matrix has zero rows, the zero rows are all at the bottom of the matrix.
• The elements below the column where the pivot element of the matrix of each non-zero row (that is, the first non-zero element on the far left in a row) are all $0$。

AA $A$ is a row echelon matrix.

row minimal matrix

AA $A$ is a$m\times The\; matrix\; of\; n$ , if it satisfies:

• $A$ is a row echelon matrix.
• Pivots of non-zero rows are all $1$ , and the other elements in the column where the pivot is located are all$0$。

AA $A$ is a row minimal matrix.

Application of Elementary Transformation in Solving Matrix

• Find the inverse of a matrix by elementary row transformation: $A$ matrix is ​​invertible$\iff$$A$ can be expressed as the product of several elementary matrices. That is,
  $$P_n\dots P_2{P}_1=A$$
  then
  $$(P_n\dots P_2{P}_1{)}^{-1}A=Ebecause$$
  ( P n … P 2 P 1 ) − 1 = P 1 − 1 P 2 − 1 … P n − 1 = Q$(P_n\dots P_2{P}_1{)}^{-1}=P_1^{-1}{P}_2^{-1}\dots P_n^{-1}=Q_1{Q}_2\dots Q_n$（$Q$ is still an elementary matrix), so the original formula can be written as:
  $$Q_1{Q}_2\dots Q_{n}A=E$$
  then
  $$Q_1{Q}_2\dots Q_{n}E=A^{-1}$$
  so:$The\; A$ matrix can be transformed into an identity matrix after several times of row transformation, and the identity matrix can be transformed into$The\; inverse\; matrix\; of\; A$ , namely:
  $$(A|E)\to \cdots \to (E|A^{-1})$$
• Solve the matrix equation by elementary row transformation: if $AX=B$，You Fruit$A$ reversible, n 么$$X=A^{-1}B$$又$$PA=EnPB$$ = A − 1 B = X PB=A^{-1}B=X$$PB=A^{-1}B=X$$ because$$P(A|B)=(E|X)$$

Rank of the matrix

at $m\times$Matrix AAof order$n$$In\; A$ , choose$k$ row and$k$ columns ($k\le n,k\le m$ ), k 2 k^2at the intersections of these rows and columns$k^2$ elements according to their original matrix$The\; sequence\; of\; A$ can form a$Determinant\; of\; order\; k$ , call it matrixAAA kkof$A$$k-$ order sub-formula. If matrixAAkkexists in$A$$The\; k-$ order subform is not$0$，$k+1st-$ order subformula (if any) are all zeros, it is called$k$ is matrixAAThe rankof$A$ , recorded as$r(A)=k$ , the rank of the zero matrix is ​​specified as$0$ . The properties of rank are as follows:

• Young $A$ is$n$ order square matrix, then
  • $r\left(A\right)=n\iff |A|\ne 0\iff Areversible$
  • $r(A)<n\iff |A|=0\iff Irreversible\_$
• The rank of the matrix is ​​unchanged after the elementary transformation.
• $0\le r(A_{m\times n})\le min(m,n)$
• $r\left(A^T\right)=r\left(A\right)$
• $r(A+B)\le r\left(A\right)+r\left(B\right)$
  Extract:
  $$\left[\begin{array}{cc}E& O\\ O& O\end{array}\right]+\left[\begin{array}{cc}O& O\\ O& E\end{array}\right]$$
• $r\left(kA\right)=r(A)(k\ne 0)$
• $r(AB)\le min(r(A),r(B))$
  proof: Let$r\left(A\right)=r$ , then there are invertible matrices$P,Q$ input$$PAQ={\left[\begin{array}{cc}{E}_r& O\\ O& O\end{array}\right]}_{m\times n}$$则$$PAB=\left[\begin{array}{cc}{E}_r& O\\ O& O\end{array}\right]Q^{-1}B=\left[\begin{array}{cc}{E}_r& O\\ O& O\end{array}\right]\left[\begin{array}{c}{B}_{r\times s}\\ B_{(n-r)\times s}\end{array}\right]=\left[\begin{array}{c}{B}_{r\times s}\\ O\end{array}\right]$$即$$r\left(AB\right)=r\left(PAB\right)=r(\left[\begin{array}{c}{B}_{r\times s}\\ O\end{array}\right])=r(B_{r\times s})\le r\le r(A)$$由此可得$$r\left(AB\right)=r((AB{)}^T)=r(B^T{A}^T)\le r(B^T)=r(B)$$故$r\left(AB\right)\le min\left(r\right(A),r(B))$
• If the matrix $P,Q$ is reversible, then$r\left(PAQ\right)=r\left(A\right)$
• $r\left(\left[\begin{array}{cc}A& O\\ O& B\end{array}\right]\right)=r\left(A\right)+r(B)$
• $max(r(A),r(B))\le r(A|B)\le r(A)+r(B)$
  proof:$r\left(A\right)=r,r(B)=t$ ,$$\begin{gathered} P{A}^T=A_1(row\; ladder,rnon-zero\; rows) \\ Q{B}^T=B_1(row\; ladder,tnon-zero\; rows) \end{gathered}$$ then$$\left[\begin{array}{cc}P& O\\ O& Q\end{array}\right]\left[\begin{array}{c}{A}^T\\ B^T\end{array}\right]=\left[\begin{array}{c}{A}_1\\ B_1\end{array}\right]$$则$$r(A|B)=r\left[\begin{array}{c}{A}^T\\ B^T\end{array}\right]=\left[\begin{array}{c}{A}_1\\ B_1\end{array}\right]\le r+t\le r(A)+r(B)$$
• $n$ -element linear equation system$AX=Determination\; of\; B$ solution: its augmented matrix is$C$ Then its solution is as follows:

Condition	illustrate
No solution	$r\left(A\right)+1=r(C)$
unique solution	$r\left(A\right)=r\left(C\right)=n$
infinite solution	$r(A)=r(C)<n)$

• $n$元齐次方程组$AX=O$有非零解$\iff r(A)<n$
• 矩阵方程$AX=B$有解$\iff r(A)=r(A,B)$
  证明：设$A-m\times n$，$X-n\times t$，$B-m\times t$，将$B$矩阵按列分块:
  $$B=[{\beta }_1,{\beta }_2,\dots ,{\beta }_n]$$
  $AX=B$有解$\iff$$AX={\beta }_j(j=1,2,\dots ,t)$都有解，设$r(A)=r$，则$r(A,{\beta }_j)=r$，化为行最简$\iff r(PA,P{\beta }_j)=r$，那么$P{\beta }_j$的后$m-r$行就为$0\iff r(PA,PB)=r\iff r(A,B)=r=r(A)$。