Follow teacher Li to learn online generation - matrix (continuously updated)

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Basic concepts

• matrix ： from $m\times n$ A number of $m$ That's ok $n$ The table of columns is called a $m\times n$ Matrix , When $m=n$ when , be called $n$ Order matrix , Write it down as $A$.
• Homomorphic matrices ： If $A$ and $B$ All are $m\times n$ Order matrix , So called $A$ and $B$ It's a homomorphic matrix .
• Equality matrix ： set up $A,B$ It's a homomorphic matrix , If $a_{ij}=b_{ij}(\forall i=1,2,\dots ,m;j=1,2,\dots ,n)$, said $A$ and $B$ equal , Write it down as $A=B$.
• Zero matrix ： If all the elements of a matrix are $0$, Then call this matrix zero matrix , Write it down as $O$.
• Diagonal matrix ： $$\left[\begin{array}{cccc}{a}_{11}& & & \\ & a_{22}& & \\ & & \ddots & \\ & & & a_{nn}\end{array}\right]$$
• Unit matrix ：$$\left[\begin{array}{cccc}1& & & \\ & 1& & \\ & & \ddots & \\ & & & 1\end{array}\right]$$ Write it down as $E$.
• Upper triangular matrix ：$$\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ & a_{22}& \dots & a_{2n}\\ & & \ddots & ⋮\\ & & & a_{nn}\end{array}\right]$$ When $i>j$ when ,$a_{ij}=0$
• Lower triangular matrix ：$$\left[\begin{array}{cccc}{a}_{11}& & & \\ a_{21}& a_{22}& & \\ ⋮& ⋮& \ddots & \\ a_{n1}& a_{n2}& \dots & a_{nn}\end{array}\right]$$ When $i<j$ when ,$a_{ij}=0$
• Symmetric matrix ： if $A^T=A$, said $A$ Is symmetric matrix .
• Antisymmetric matrix ： if $A^T=-A$, said $A$ Antisymmetric matrix .

Matrix operation

• Addition of matrix ：$A=[a_{ij}],B=[b_{ij}]$ Are all $m\times n$ Order matrix , that $A+B=[a_{ij}+b_{ij}]$.
• Addition algorithm ： if $A,B,C$ Same type , be $$\begin{gathered} A+B=B+A \\ (A+B)+C=A+(B+C) \\ A+O=A \\ A+(-A)=O \end{gathered}$$
• Number multiplied by matrix ：$kA=[k{a}_{ij}]$.
• Number multiplication algorithm ：$$\begin{gathered} k(mA)=m(kA)=(mk)A \\ (k+m)A=kA+mA \\ k(A+B)=kA+kB \\ 1A=A \\ 0A=O \end{gathered}$$
• Multiplication of matrices ： if $A=[a_{ij}{]}_{m\times s},B=[b_{ij}{]}_{s\times n}$, be $$A\times B=C=[c_{ij}{]}_{m\times n}$$ among $$c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+\cdots +a_{is}{b}_{sj}=\sum _{k=1}^n{a}_{ik}{b}_{kj}$$
• Multiplication algorithm ：$$\begin{gathered} A(BC)=(AB)C \\ A(B+C)=AB+AC \\ (kA)(lB)=klAB \\ AE=A,EA=A \\ OA=O,AO=O \end{gathered}$$ Be careful ：$$\begin{gathered} AB\ne BA,ifA,BIt\text{'}s\; a\; diagonal\; matrix\; ,\; beAB=BA \\ AB=O\nRightarrow A=OorB=O \\ AB=AC,A\ne 0\nRightarrow B=C \\ {\left[\begin{array}{ccc}{a}_1& & \\ & a_2& \\ & & a_3\end{array}\right]}^n=A=\left[\begin{array}{ccc}{a}_1^n& & \\ & a_2^n& \\ & & a_3^n\end{array}\right] \end{gathered}$$
• Transposition ： set up $A=[a_{ij}{]}_{m\times n}$, take $A$ The rows and columns are interchanged , Got $n\times m$ Matrix $[a_{ji}{]}_{n\times m}$ be called $A$ The transpose matrix of , Write it down as $A^T$.
• Transpose algorithm ：$$\begin{gathered} (A+B{)}^T=A^T+B^T \\ (kA{)}^T=k{A}^T \\ (AB{)}^T=B^T{A}^T \\ (A^T{)}^T=A \end{gathered}$$

Matrices and determinants

• The determinant of a square ： set up $A=[a_{ij}]$ by $n$ Order matrix , The determinant formed by keeping all its elements in place is called a matrix $A$ The determinant of , Write it down as $$|A|$$ Be careful ： Only square matrix has determinant ;$A=O$ and $|A|=0$ It doesn't matter. .
• The formula of the determinant of the square matrix ：$$\begin{gathered} |A^T|=|A| \\ |kA|=k^n|A| \\ |AB|=|A||B| \\ |A^2|=|A{|}^2 \end{gathered}$$

Adjoint matrix

• Adjoint matrix ： set up $A=[a_{ij}]$ yes $n$ Order matrix , determinant $|A|$ Each element of $a_{ij}$ The algebraic covalent of $A_{ij}$ The following matrix $$A^{\ast }=\left[\begin{array}{cccc}{A}_{11}& A_{21}& \dots & A_{n1}\\ A_{12}& A_{22}& \dots & A_{n2}\\ ⋮& ⋮& & ⋮\\ A_{1n}& A_{2n}& \dots & A_{nn}\end{array}\right]$$ It's called a matrix $A$ The adjoint matrix of .
• $A{A}^{\ast }=A^{\ast }A=|A|E$
• Adjoint matrix of second order matrix ： The main diagonals are interchangeable , The sub diagonal changes sign .

Invertible matrix

• Invertible matrix ： about $n$ Order matrix $A$, If $\exists n$ Order matrix $B$, send $$AB=BA=E$$ It's called matrix $A$ It's reversible , matrix $B$ yes $A$ The inverse matrix .
• If matrix $A$ It's reversible , that $A$ The inverse matrix of is unique , Write it down as $A^{-1}$.
• If $A$ reversible , be $A^{-1}$ It's also reversible , And $(A^{-1}{)}^{-1}=A$.
• If $A$ reversible , And $k\ne 0$, be $kA$ reversible , And $(kA{)}^{-1}=\frac{1}{k}{A}^{-1}$.
• If $A,B$ reversible , be $AB$ It's reversible , And $(AB{)}^{-1}=B^{-1}{A}^{-1}$.
• If $A$ reversible , be $A^T$ It's reversible , And $(A^T{)}^{-1}=(A^{-1}{)}^T$.
• $A$ reversible $\iff |A|\ne 0$.
• $A,B$ yes $n$ Order matrix , If $AB=E$, be $A^{-1}=B$.
• If $A$ reversible , be $A^{-1}=\frac{1}{|A|}{A}^{\ast }$.
• If $A$ reversible , be $|A^{-1}|=\frac{1}{|A|}$.
• If $A=\left[\begin{array}{ccc}{a}_1& & \\ & a_2& \\ & & a_3\end{array}\right]$ It's a diagonal matrix , be $A^{-1}=\left[\begin{array}{ccc}\frac{1}{a_1}& & \\ & \frac{1}{a_2}& \\ & & \frac{1}{a_3}\end{array}\right]$.

Block matrix

Block the matrix properly , It can be calculated effectively , After blocking, there are the following algorithms ：

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

set up $A,B$ Respectively $m,n$ Order matrix , be ：

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

set up $A,B$ Respectively $m,n$ Second order invertible matrices , be ：

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

if $A$ yes $m\times n$ matrix ,$B$ yes $n\times s$ Matrix and $AB=C$, On the other hand $B,C$ There are ：$$\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{m1}& a_{m2}& \dots & a_{mn}\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right]$$ namely $$\{\begin{array}{l}{a}_{11}{b}_1+a_{12}{b}_2+\cdots +a_{1n}{b}_n=c_1\\ a_{21}{b}_1+a_{22}{b}_2+\cdots +a_{2n}{b}_n=c_2\\ \dots \\ a_{n1}{b}_1+a_{n2}{b}_2+\cdots +a_{nn}{b}_n=c_n\end{array}$$

Elementary transformation of matrix

The elementary transformation of a matrix is the sum of elementary row transformation and elementary column transformation ：

• Multiply a row of the matrix with a non-zero constant （ Column ）.
• Put a line （ Column ） Of $k$ Multiply to another line （ Column ）.
• Exchange two rows in the matrix （ Column ） The location of .

Elementary row transformation and linear equations

Now solve the following 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 Addition and subtraction Law ：

• Multiply both sides of the equation by a non-zero number .
• Put the... Of an equation $k$ Add times to another equation .
• Exchange the positions of two equations .

Simplify the 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 the change of unknown coefficient and constant term , Therefore, the unknown coefficient and constant term can be written into a matrix ：
$$\left[\begin{array}{cccc}2& -1& 3& 1\\ 4& 2& 5& 4\\ 2& 0& 2& 6\end{array}\right]$$
This matrix is called the matrix of linear equations Augmented matrix . The line performs elementary row transformation on the matrix to simplify the matrix （ Positive elimination ） to ：
$$\left[\begin{array}{cccc}2& -1& 3& 1\\ 0& 1& -1& 5\\ 0& 0& 1& -6\end{array}\right]$$
Then reverse the solution , The solution of the linear equation can be obtained .

Elementary matrix

The matrix obtained from the identity matrix through an elementary row transformation is called Elementary matrix .

• Elementary matrix $P$ Left multiplication $A$ What you get $PA$ That's right $A$ Do it once with $P$ The same elementary line transformation .
• Elementary matrix $P$ Take the right $A$ What you get $AP$ That's right $A$ 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 $A$ After a finite number of elementary changes, it is transformed into a matrix $B$, It's called a matrix $A$ And $B$ Equivalent , Write it down as $A\cong B$. Matrix equivalence satisfies ：

• Reflexivity ：$A\cong A$
• symmetry ： if $A\cong B$, be $B\cong A$
• Transitivity ： if $A\cong B,B\cong C$, be $A\cong C$

Row ladder matrix

set up $A$ It's a $m\times n$ Matrix , If meet ：

• If the matrix has zero rows , Then the zero row is at the bottom of the matrix .
• The principal element of every non-zero row matrix （ That is, the first non-zero yuan on the leftmost side of a line ） The elements below the column are $0$.

said $A$ For row ladder matrix .

Row simplest matrix

set up $A$ It's a $m\times n$ Matrix , If meet ：

• $A$ It's a row ladder matrix
• The principal elements of non-zero elements are $1$, And other elements in the column where the primary element is located are $0$.

said $A$ Is the row simplest matrix .

Application of elementary transformation in matrix solution

• Solve the inverse matrix of the matrix by elementary row transformation ：$A$ The necessary and sufficient condition for the invertibility of a matrix is $A$ It can be expressed as the product of several elementary matrices . namely $$P_n\dots P_2{P}_1=A$$ that $$(Pn\dots P_2{P}_1{)}^{-1}A=E$$ because $(Pn\dots P_2{P}_1{)}^{-1}=P_n^{-1}\dots P_2^{-1}{P}_1^{-1}=Q_n\dots Q_2{Q}_1$（$Q$ Still elementary matrix ）, So the original formula can be written as ：$$Q_n\dots Q_2{Q}_{1}A=E$$ that $$Q_n\dots Q_2{Q}_{1}E=A^{-1}$$ therefore $$(A|E)\to \cdots \to (E|A^{-1})$$
• Solve the matrix equation by elementary row transformation ： if $Ax=B$, If $A$ reversible , that $$x=A^{-1}B$$ also $$PA=E$$ that $$PB=A^{-1}B=x$$ therefore $$P(A|B)=(E|x)$$

The rank of a matrix

• $k$ Order subformula ： stay $m\times n$ Matrix of order $A$ in , Take whatever you like $k$ Line with $k$ Column （$k\le n,k\le m$）, Located at the intersection of these rows and columns $k^2$ Elements in the original matrix $A$ The order of can form a $k$ Step determinant , Call it a matrix $A$ One of the $k$ Order subformula .

• The rank of a matrix ： If the matrix $A$ in $r$ The order formula is not $0$,$r+1$ Order subformula （ If there is ） It's all zero , It's called matrix $A$ The rank of is $r$, Write it down as $r(A)=r$, The rank of the zero matrix is specified as $0$.

• if $A$ yes $n$ Order matrix , that $$\begin{gathered} r(A)=n\iff |A|\ne 0\iff Areversible \\ r(A)<n\iff |A|=0\iff AIrreversible \end{gathered}$$

• After elementary transformation, the rank of the matrix is invariant .

• If matrix $P,Q$ reversible , Yes $PAQ=B$, that $r(PAQ)=r(B)$.

• $0\le r(A_{m\times n})\le min(m,n)$

• $r(A^T)=r(A)$

• $r(A+B)\le r(A)+r(B)$

• $r(kA)=r(A)(k\ne 0)$

• $r(AB)\le min(r(A),r(B))$. prove ： set up $r(A)=r$, Then there is any invertible matrix $P,Q$ bring $$PAQ={\left[\begin{array}{cc}{E}_r& O\\ O& O\end{array}\right]}_{m\times n}$$ be $$PAB=\left[\begin{array}{cc}{E}_r& O\\ O& O\end{array}\right]Q^{-1}B=\left[\begin{array}{cc}{E}_r& O\\ O& O\end{array}\right]\left[\begin{array}{c}{B}_{r\times s}\\ B_{(n-r)\times s}\end{array}\right]=\left[\begin{array}{c}{B}_{r\times s}\\ O\end{array}\right]$$ namely $$r(AB)=r(PAB)=r(\left[\begin{array}{c}{B}_{r\times s}\\ O\end{array}\right])=r(B_{r\times s})\le r\le r(A)$$ It can be obtained. $$r(AB)=r((AB{)}^T)=r(B^T{A}^T)\le r(B^T)=r(B)$$ so $r(AB)\le min(r(A),r(B))$

• $r(\left[\begin{array}{cc}A& O\\ O& B\end{array}\right])=r(A)+r(B)$

• $max(r(A),r(B))\le r(A|B)\le r(A)+r(B)$. prove ： set up $r(A)=r,r(B)=t$,$$\begin{gathered} P{A}^T=A_1(Step\; by\; step\; ,rA\; nonzero\; line) \\ Q{B}^T=B_1(Step\; by\; step\; ,tA\; nonzero\; line) \end{gathered}$$ that $$\left[\begin{array}{cc}P& O\\ O& Q\end{array}\right]\left[\begin{array}{c}{A}^T\\ B^T\end{array}\right]=\left[\begin{array}{c}{A}_1\\ B_1\end{array}\right]$$ be $$r(A|B)=r\left[\begin{array}{c}{A}^T\\ B^T\end{array}\right]=\left[\begin{array}{c}{A}_1\\ B_1\end{array}\right]\le r+t\le r(A)+r(B)$$

• $n$ A system of elementary linear equations $Ax=B$ Determination of solution ： Its augmented matrix is $C$ Then the solution is as follows ：

situation	explain
unsolvable	$r(A)+1=r(C)$
Unique solution	$r(A)=r(C)=n$
Infinite solutions	$r(A)=r(C)<n)$

• $n$ Element homogeneous equations $Ax=0$ There is a nonzero solution $\iff r(A)<n$
• matrix equation $Ax=B$ Have solution $\iff r(A)=r(A,B)$