Advanced Mathematics: Matrices

1. Matrix concept

$${\left(\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_m& \cdots & a_{mn}\end{array}\right)}_{m\times n}$$
The matrix has m rows and n columns

The matrices are equal only if all the elements in the matrix are correspondingly equal.

Second, the operation of the matrix

Sum of matrices: $C=A+B$ (the added matrix must have equal number of rows and columns)
$$A_{m\times n}=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_2\\ ⋮& ⋮& \ddots & ⋮\\ a_m& a_m& \cdots & a_{mn}\end{array}\right]B_{m\times n}=\left[\begin{array}{cccc}{b}_{11}& b_{12}& \cdots & b_{1n}\\ b_{21}& b_{22}& \cdots & b_2\\ ⋮& ⋮& \ddots & ⋮\\ b_m& b_m& \cdots & b_{mn}\end{array}\right]$$
$$c_{m\times n}=A_{m\times n}+B_{m\times n}=\left[\begin{array}{cccc}{a}_{11}+b_{11}& a_{12}+b_{12}& \cdots & a_{1n}+b_{1n}\\ a_{21}+b_{21}& a_{22}+b_{22}& \cdots & a_2+b_2\\ ⋮& ⋮& \ddots & ⋮\\ a_m+b_m& a_m+b_m& \cdots & a_{mn}+b_{mn}\end{array}\right]$$

Associative law: $A+(B+C)=(A+B)+C$

Commutative law: $A+B=B+A$

Matrix multiplication:
$$A=(a_{}{)}_{s\times n},B=(b_{kj}{)}_{n\times m},thenC=AB=(c_{ij}{)}_{s\times m}$$
$$c_{ij}=a_i{b}_i+a_2{b}_{i2}+...+a_{in}{b}_{nj}=\sum _{k=1}^n{a}_{}{b}_{kj}$$
In one sentence: the product of matrix A and B, the element of row i and column j of matrix A is equal to the product of the first row of the first matrix A and the corresponding element of the first column of the second matrix B

Mistakes:

1. $AB\ne BA$
2. $AB=AC$ does not necessarily$\Rightarrow B=C$

Unity matrix: n×n matrix with all elements on the main diagonal being 1 and all other elements being 0
$$\left[\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 1\end{array}\right]$$
It is called the n-order identity matrix, denoted as $E_n$

Form:
$$toA=\left[\begin{array}{cccc}k{a}_{11}& k{a}_{12}& \cdots & k{a}_{1n}\\ k{a}_{21}& k{a}_{22}& \cdots & k{a}_2\\ ⋮& ⋮& \ddots & ⋮\\ k{a}_m& k{a}_m& \cdots & k{a}_{mn}\end{array}\right]$$
, each element in A is multiplied by k

转置（transform）：
$$A_{m\times n}={\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_2\\ ⋮& ⋮& \ddots & ⋮\\ a_m& a_m& \cdots & a_{mn}\end{array}\right]}_{m\times n}$$
$$A^{\top }={\left[\begin{array}{cccc}{a}_{11}& a_{21}& \cdots & a_n\\ a_{12}& a_{22}& \cdots & a_n\\ ⋮& ⋮& \ddots & ⋮\\ a_1& a_2& \cdots & a_{nm}\end{array}\right]}_{n\times m}$$
The transpose of matrix A is to exchange the rows and columns of A

The law of matrix transposition:
$$\begin{gathered} (A^{\top }{)}^{\top }=A \\ (A+B{)}^{\top }=A^{\top }+B^{\top } \\ (AB{)}^{\top }=B^{\top }{A}^{\top } \\ (kA{)}^{\top }=to{A}^{\top } \end{gathered}$$

3. Determinant and rank of matrix product

Theorem 1: Suppose A and B are two n × nn\times n on the number field P$n\times n$ matrix, then

$\left|AB\right|=\left|A\right||B|$ , that is, the determinant of the matrix product is equal to the product of the determinants of its factors

Theorem 6: n × nn\times n on the number field P$n\times The\; n-$ matrix A is called non-degenerate if$\left|A\right|\ne 0$ ; otherwise called degenerate.

Theorem 2: Let A be m × nm\times n on the number field P$m\times n$ matrix, B is m × sm\times son the number field P$m\times s$ matrix, so we have:

rank(AB) $\le min\left[Chichi\right(A),rank(B)]$ , that is, the rank of the product does not exceed the rank of each factor

Proof: Just prove that rank $(AB)\le rank\left(A\right)$ and$Rank\left(AB\right)\le 秩(B)$
$$A_{m\times n}=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_2\\ ⋮& ⋮& \ddots & ⋮\\ a_m& a_m& \cdots & a_{mn}\end{array}\right]B_{n\times s}=\left[\begin{array}{cccc}{b}_{11}& b_{12}& \cdots & b_1\\ b_{21}& b_{22}& \cdots & b_{2s}\\ ⋮& ⋮& \ddots & ⋮\\ b_n& b_n& \cdots & b_{ns}\end{array}\right]=\left[\begin{array}{c}{B}_1\\ B_2\\ ⋮\\ B_n\end{array}\right]$$

Among them, $B_1,B_2...B_n$Represents the row vector of matrix B, $C_1,C_2...C_m$Display rectangle AB line direction
$$C_i=a_i{B}_1+a_{i2}{B}_2+...+a_{in}{B}_n,i=1,\mathrm{2...}n$$
is the row vector of matrix AB can be expressed linearly by the row vector of matrix B, so

The rank of matrix AB cannot exceed the rank of matrix A (from the conclusion "If vector group 1 can be expressed linearly by vector group 2, then the rank of vector group 1 cannot exceed the rank of vector group 2")

Similarly, if A is written as a column vector, the rank of matrix AB cannot exceed the rank of B

To sum up: $Rank\; (AB)\le min[Chichi\; (A),rank(B)]$

Inference: If $A=A_1{A}_2...A_n$那么，
$$r(A)\le min[r(A_1),r(A_2)...r(A_n)]$$

Fourth, the inverse of the matrix (inv)

Definition 7: A square matrix A of order n is called reversible, if there is a square matrix B of order n such that $AB=BA=E$ , E is the identity matrix

Where matrix B is the inverse matrix of matrix A, recorded as $B=A^{-1}$

Two calculation methods of the inverse matrix:

1）伴随矩阵：
$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_2\\ ⋮& ⋮& \ddots & ⋮\\ a_n& a_n& \cdots & a_{nn}\end{array}\right]$$
Let A divide the square root
$$A^{\ast }=\left[\begin{array}{cccc}{A}_{11}& A_{12}& \cdots & A_{1n}\\ A_{21}& A_{22}& \cdots & A_2\\ ⋮& ⋮& \ddots & ⋮\\ A_n& A_n& \cdots & A_{nn}\end{array}\right]$$
($A_{ij}$AA ∗ = A ∗ A = d E AA^{*}=A^{*}A= dE
$$A{A}^{\ast }=A^{\ast }A=dE$$
$$\therefore A^{-1}=\frac{1}{|A|}{A}^{\ast }(provided\; that|A|\ne 0)$$

2) Elementary matrix:

(A,E) column transformation and corresponding row transformation to get $(E,A^{-1})$

For example: Find
$$A=\left(\begin{array}{ccc}0& 1& 2\\ 1& 1& 4\\ 2& -1& 0\end{array}\right)$$
Integer
$$(A,E)=\left(\begin{array}{cccccc}0& 1& 2& 1& 0& 0\\ 1& 1& 4& 0& 1& 0\\ 2& -1& 0& 0& 0& 1\end{array}\right)$$
After a series of elementary row and column transformations,
$$(E,A^{-1})=\left(\begin{array}{cccccc}1& 0& 0& 2& -1& 1\\ 0& 1& 0& 4& -2& 1\\ 0& 0& 1& -\frac{3}{2}& 1& -\frac{1}{2}\end{array}\right)$$
$$A^{-1}=\left(\begin{array}{ccc}2& -1& 1\\ 4& -2& 1\\ -\frac{3}{2}& 1& -\frac{1}{2}\end{array}\right)$$

Corollary: If the matrices A and B are invertible, then $A^{-1}$ and AB are also invertible, and
$$(A^T{)}^{-1}=(A^{-1}{)}^T$$
$$(AB{)}^{-1}=B^{-1}{A}^{-1}$$
$$(kA{)}^{-1}=\frac{1}{k}{A}^{-1}(k\ne 0)$$

5. Elementary Matrix

Definition 10: The matrix obtained by an elementary transformation from the identity matrix E is called an elementary matrix

Left multiplication row transformation, right multiplication column transformation

Definition 11: If B can be obtained from A through a series of elementary transformations, then A and B are said to be equivalent

6. Problem-solving skills about matrices

1) $Order\; (A+B)\le Rank(A)\; +\; Rank(B)$

2) $Rank\; (AB)\le min\; (chichi\; (A),\; chichi\; (B))$

3) $\left|kA\right|＝k^n|A|(A\; is\; a\; matrix\; of\; order\; n)$

4) A is a third-order matrix, $｜A｜＝２$
$$\left|{\left(２Ａ\right)}^{－１}－３{Ａ}^{＊}\right|=|\frac{1}{2}{A}^{-1}-3A^{\ast }|=|\frac{1}{2}{A}^{-1}-3|A|A^{-1}|=|-\frac{11}{2}{A}^{-1}|=(-\frac{11}{2}{)}^3\frac{1}{|A|}$$
5) Let A be $n\times n$ matrix, if for any n-dimensional vector
$$X=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)$$
All have $AX=0,\; thenA=0$

Proof: Take
$$X=\left(\begin{array}{c}1\\ 0\\ ⋮\\ 0\end{array}\right)$$
or
$$\left(\begin{array}{c}0\\ 1\\ ⋮\\ 0\end{array}\right)$$
Or
$$\left(\begin{array}{c}0\\ 0\\ ⋮\\ 1\end{array}\right)$$
Taken into the calculation, each element in A is 0, that is, A is a zero matrix

6) Let B be $r\times r$ matrix, C is$r\times n$ matrix, and rank$（C）=r$ , prove that:

1. If BC= 0, then B=0

2. If BC=C, then B=E

Proof: 1) $\because r(C)=r$

Let's take the maximum linear independent group of C as $C_1,C_2...C_r$
$$BC=0\Rightarrow B(C_1,C_2...C_r)=0$$
又$(C_1,C_2...C_r)\ne 0$ , reversible

So, B=0

2) Same as 1) $BC=C\Rightarrow (B-E)(C_1,C_2...C_r)=0$

又$(C_1,C_2...C_r)\ne 0$ , reversible

So, B=E

Welcome everyone to make corrections, there are some