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1. Matrix concept
( a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a m 1 ⋯ a m n ) m × n \begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix} _{m\times n}
a11⋮am 1⋯⋱⋯a1n⋮amn
m×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 + BC=A+BC=A+B (the added matrix must have equal number of rows and columns)
A m × n = [ a 11 a 12 ⋯ a 1 na 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ am 1 am 2 ⋯ amn ] B m × n = [ b 11 b 12 ⋯ b 1 nb 21 b 22 ⋯ b 2 n ⋮ ⋮ ⋱ ⋮ bm 1 bm 2 ⋯ bmn ] A_{m\times n}= \begin{bmatrix} a_{11}& a_ {12}& \cdots & a_{1n} \\ a_{21}& a_{22}& \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1}& a_{m2}& \cdots & a_{mn} \end{bmatrix} B_{m\times n}= \begin{bmatrix} b_{11}& b_{12}& \cdots & b_{1n} \\ b_ {21}&b_{22}& \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{m1}& b_{m2}& \cdots & b_{mn} \end{ bmatrix}Am×n=
a11a21⋮am 1a12a22⋮am 2⋯⋯⋱⋯a1na2 n⋮amn
Bm×n=
b11b21⋮bm 1b12b22⋮bm 2⋯⋯⋱⋯b1nb2 n⋮bmn
c m × n = A m × n + B m × n = [ a 11 + b 11 a 12 + b 12 ⋯ a 1 n + b 1 n a 21 + b 21 a 22 + b 22 ⋯ a 2 n + b 2 n ⋮ ⋮ ⋱ ⋮ a m 1 + b m 1 a m 2 + b m 2 ⋯ a m n + b m n ] c_{m\times n}=A_{m\times n}+B_{m\times n}= \begin{bmatrix} a_{11}+b_{11}& a_{12}+b_{12}& \cdots & a_{1n}+b_{1n} \\ a_{21}+b_{21}& a_{22}+b_{22}& \cdots & a_{2n}+b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1}+b_{m1}& a_{m2}+b_{m2}& \cdots & a_{mn}+b_{mn} \end{bmatrix} cm×n=Am×n+Bm×n=
a11+b11a21+b21⋮am 1+bm 1a12+b12a22+b22⋮am 2+bm 2⋯⋯⋱⋯a1n+b1na2 n+b2 n⋮amn+bmn
Associative law: A + ( B + C ) = ( A + B ) + C A+(B+C)=(A+B)+CA+(B+C)=(A+B)+C
Commutative law: A + B = B + A A+B=B+AA+B=B+A
Matrix multiplication:
A = ( aik ) s × n , B = ( bkj ) n × m , then C = AB = ( cij ) s × m A=(a_{ik})_{s\times n},B =(b_{kj})_{n\times m} , then C=AB=(c_{ij})_{s\times m}A=(ai)s×n,B=(bkj)n×m,then C=AB=(cij)s×m
c i j = a i 1 b i 1 + a 2 i b i 2 + . . . + a i n b n j = ∑ k = 1 n a i k b k j c_{ij}=a_{i1}b_{i1}+a_{2i}b_{i2}+...+a_{in}b_{nj}=\sum_{k=1}^{n}{a_{ik}b_{kj}} cij=ai 1bi 1+a2 ibi2+...+ainbnj=k=1∑naibkj
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:
- AB ≠ NOT AB\neAB=BA
- A B = A C AB=AC AB=A C does not necessarily⇒ B = C \Rightarrow B=C⇒B=C
Unity matrix: n×n matrix with all elements on the main diagonal being 1 and all other elements being 0
[ 1 0 ⋯ 0 0 1 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 1 ] \begin{bmatrix} 1& 0& \cdots & 0 \\ 0& 1& \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0& 0& \cdots & 1 \end{bmatrix}
10⋮001⋮0⋯⋯⋱⋯00⋮1
It is called the n-order identity matrix, denoted as E n E_{n}En
Form:
k A = [ for 11 for 12 ⋯ for 1 but 21 for 22 ⋯ for 2 n ⋮ ⋮ ⋱ ⋮ kam 1 kam 2 ⋯ kamn ] kA= \begin{bmatrix} for_{11}& for_{12}&; \cdots & to_{1n} \\ to_{21}& to_{22}& \cdots & to_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ to_{m1}& to_{m2} &\cdots&to_{mn}\end{bmatrix}to A=
k a11k a21⋮k am 1k a12k a22⋮k am 2⋯⋯⋱⋯k a1nk a2 n⋮k amn
, each element in A is multiplied by k
转置(transform):
A m × n = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a m 1 a m 2 ⋯ a m n ] m × n A_{m\times n}= \begin{bmatrix} a_{11}& a_{12}& \cdots & a_{1n} \\ a_{21}& a_{22}& \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1}& a_{m2}& \cdots & a_{mn} \end{bmatrix} _{m\times n} Am×n=
a11a21⋮am 1a12a22⋮am 2⋯⋯⋱⋯a1na2 n⋮amn
m×n
A ⊤ = [ a 11 a 21 ⋯ a n 1 a 12 a 22 ⋯ a n 2 ⋮ ⋮ ⋱ ⋮ a 1 m a 2 m ⋯ a n m ] n × m A^{\top}= \begin{bmatrix} a_{11}& a_{21}& \cdots & a_{n1} \\ a_{12}& a_{22}& \cdots & a_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1m}& a_{2m}& \cdots & a_{nm} \end{bmatrix} _{n\times m} A⊤=
a11a12⋮a1 ma21a22⋮a2 m⋯⋯⋱⋯an 1an 2⋮anm
n×m
The transpose of matrix A is to exchange the rows and columns of A
The law of matrix transposition:
( A ⊤ ) ⊤ = A ( A + B ) ⊤ = A ⊤ + B ⊤ ( AB ) ⊤ = B ⊤ A ⊤ ( k A ) ⊤ = k A ⊤ (A^{\top} )^{\top}=A\\ (A+B)^{\top}=A^{\top}+B^{\top}\\ (AB)^{\top}=B^{\top }A^{\top}\\ (kA)^{\top}=kA^{\top}(A⊤)⊤=A(A+B)⊤=A⊤+B⊤(AB)⊤=B⊤A⊤( k A )⊤=to A⊤
3. Determinant and rank of matrix product
Theorem 1: Suppose A and B are two n × nn\times n on the number field Pn×n matrix, then
∣ A B ∣ = ∣ A ∣ ∣ B ∣ \left| AB \right|=\left| A \right| \left| B \right| ∣AB∣=∣A∣∣ 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 Pn×The n- matrix A is called non-degenerate if∣ A ∣ ≠ 0 \left| A \right|\ne0∣A∣=0 ; otherwise called degenerate.
Theorem 2: Let A be m × nm\times n on the number field Pm×n matrix, B is m × sm\times son the number field Pm×s matrix, so we have:
rank(AB) ≤ min [ rank( A ) , rank( B ) ] \leq min[ rank(A),rank(B)]≤min [ Chichi ( A ) ,rank ( B )] , that is, the rank of the product does not exceed the rank of each factor
Proof: Just prove that rank ( AB ) ≤ rank( A ) (AB)≤ rank(A)(AB)≤rank ( A ) andrank( AB ) ≤ rank( B ) rank(AB)≤ rank(B)Rank ( A B )≤秩(B)
A m × n = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a m 1 a m 2 ⋯ a m n ] B n × s = [ b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋱ ⋮ b n 1 b n 2 ⋯ b n s ] = [ B 1 B 2 ⋮ B n ] A_{m\times n}= \begin{bmatrix} a_{11}& a_{12}& \cdots & a_{1n} \\ a_{21}& a_{22}& \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1}& a_{m2}& \cdots & a_{mn} \end{bmatrix} B_{n\times s}= \begin{bmatrix} b_{11}& b_{12}& \cdots & b_{1s} \\ b_{21}&b_{22}& \cdots & b_{2s} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1}& b_{n2}& \cdots & b_{ns} \end{bmatrix} = \begin{bmatrix} B_{1}\\ B_{2} \\ \vdots \\ B_{n} \end{bmatrix} Am×n=
a11a21⋮am 1a12a22⋮am 2⋯⋯⋱⋯a1na2 n⋮amn
Bn×s=
b11b21⋮bn 1b12b22⋮bn 2⋯⋯⋱⋯b1 sb2s⋮bns
=
B1B2⋮Bn
Among them, B 1 , B 2 . . . B n B_{1},B_{2}...B_{n}B1,B2...BnRepresents the row vector of matrix B, C 1 , C 2 . . . C m C_{1},C_{2}...C_{m}C1,C2...CmDisplay rectangle AB line direction
C i = ai 1 B 1 + ai 2 B 2 + . a_{i2}B_{2}+...+a_{in}B_{n},i=1,2...nCi=ai 1B1+ai2B2+...+ainBn,i=1,2... 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 ) ≤ min [ rank( A ), rank( B ) ] rank(AB)\leq min[ rank(A), rank(B)]Rank ( A B )≤min [ Chichi ( A ) ,rank( B ) ]
Inference: If A = A 1 A 2 . . . A n A=A_{1}A_{2}...A_{n}A=A1A2...An那么,
r ( A ) ≤ m i n [ r ( A 1 ) , r ( A 2 ) . . . r ( A n ) ] r(A)\leq min[r(A_{1}),r(A_{2})...r(A_{n})] r(A)≤min[r(A1),r(A2)...r(An)]
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 AB=BA=EAB=BA=E , E is the identity matrix
Where matrix B is the inverse matrix of matrix A, recorded as B = A − 1 B=A^{-1}B=A−1
Two calculation methods of the inverse matrix:
1)伴随矩阵:
A = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ] A=\begin{bmatrix} a_{11}& a_{12}& \cdots & a_{1n} \\ a_{21}& a_{22}& \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1}& a_{n2}& \cdots & a_{nn} \end{bmatrix} A=
a11a21⋮an 1a12a22⋮an 2⋯⋯⋱⋯a1na2 n⋮ann
Let A divide the square root
A ∗ = [ A 11 A 12 ⋯ A 1 n A 21 A 22 ⋯ A 2 n ⋮ ⋮ ⋱ ⋮ A n 1 A n 2 ⋯ A nn ] A^{*}=\begin{bmatrix } A_{11}& A_{12}&\cdots&A_{1n}\A_{21}&A_{22}&\cdots&A_{2n}\\\vdots&\vdots&\ddots&\vdots \\A_{n1}&A_{n2}&\cdots&A_{nn}\end{bmatrix}A∗=
A11A21⋮An 1A12A22⋮An 2⋯⋯⋱⋯A1nA2 n⋮Ann
( A i j A_{ij} AijAA ∗ = A ∗ A = d E AA^{*}=A^{*}A= dE
AA∗=A∗A=d E
∴ A − 1 = 1 ∣ A ∣ A ∗ (provided that ∣ A ∣ ≠ 0 ) \therefore A^{-1}=\frac{1}{|A|}A^{*} (provided that| A|≠0)∴A−1=∣A∣1A∗ (provided that∣A∣=0)
2) Elementary matrix:
(A,E) column transformation and corresponding row transformation to get ( E , A − 1 ) (E,A^{-1})(E,A−1)
For example: Find
A = ( 0 1 2 1 1 4 2 − 1 0 ) A=\begin{pmatrix} 0 & 1 & 2\\ 1 & 1 & 4\\ 2 & -1 &0 \end{pmatrix}A=
01211−1240
Integer
( A , E ) = ( 0 1 2 1 0 0 1 1 4 0 1 0 2 − 1 0 0 0 1 ) (A,E)=\begin{pmatrix} 0 & 1 & 2&1&0&0\\ 1 & 1&4&0&1&0\\ 2&-1&0&0&0&1\end{pmatrix}(A,E)=
01211−1240100010001
After a series of elementary row and column transformations,
( E , A − 1 ) = ( 1 0 0 2 − 1 1 0 1 0 4 − 2 1 0 0 1 − 3 2 1 − 1 2 ) (E,A^{-1 })=\begin{pmatrix} 1 & 0 & 0&2&-1&1\\ 0 & 1 & 0&4&-2&1\\ 0 & 0 &1&-\frac{3}{2} &1&-\frac{1}{2} \ end{pmatrix}(E,A−1)=
10001000124−23−1−2111−21
A − 1 = ( 2 − 1 1 4 − 2 1 − 3 2 1 − 1 2 ) A^{-1}=\begin{pmatrix} 2&-1&1\\ 4&-2&1\\ -\frac{3}{2} &1&-\frac{1}{2} \end{pmatrix} A−1=
24−23−1−2111−21
Corollary: If the matrices A and B are invertible, then A − 1 A^{-1}A− 1 and AB are also invertible, and
( AT ) − 1 = ( A − 1 ) T (A^{T})^{-1}=(A^{-1})^{T}(AT)−1=(A−1)T
( AB ) − 1 = B − 1 A − 1 (AB)^{-1}=B^{-1}A^{-1}(AB)−1=B−1A− 1
( k A ) − 1 = 1 k A − 1 ( k ≠ 0 ) (kA)^{-1}=\frac{1}{k}A^{-1}(k\ne 0)( k A )−1=k1A−1(k=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) Chichi ( A + B ) ≤ Chichi ( A ) + Chichi ( B ) Chichi (A + B) ≤ Chichi (A) + Chichi (B)Order ( A + B )≤ Rank( A ) + Rank( B )
2) Chichi (AB) ≤ min (Chichi (A), Chichi (B)) Chichi (AB) ≤ min (Chichi (A), Chichi (B))Rank ( A B )≤ min (chichi ( A ), chichi ( B ))
3) ∣ k A ∣ = kn ∣ A ∣ (A is a matrix of order n) \left | kA \right | =k^{n}|A| (A is a matrix of order n)∣kA∣=kn ∣A∣(A is a matrix of order n)
4) A is a third-order matrix, | A |=2 |A|=2|A|=2
∣ ( 2A ) -1 -3 A * ∣ = ∣ 1 2 A − 1 − 3 A ∗ ∣ = ∣ 1 2 A − 1 − 3 ∣ A ∣ A − 1 ∣ = ∣ − 11 2 A − 1 ∣ = ( − 11 2 ) 3 1 ∣ A ∣ \left| \left( 2A \right)^{-1} -3A^{*}\right|=|\frac{1}{2}A^{-1}-3A^{*}|=|\frac{1}{2}A^{-1}-3|A|A^{-1}|=|-\frac{11}{2}A^{-1}|=(-\frac{11}{2})^{3}\frac{1}{|A|}
( 2A)-1-3A*
=∣21A−1−3 A∗∣=∣21A−1−3∣A∣A−1∣=∣−211A−1∣=(−211)3∣A∣1
5) Let A be n × nn\times nn×n matrix, if for any n-dimensional vector
X = ( x 1 x 2 ⋮ xn ) X =\begin{pmatrix}x_{1} \\x_{2} \\\vdots \\x_{n} \end{ pmatrix}X=
x1x2⋮xn
All have AX = 0, then A = 0 AX=0, then A=0AX=0 , then A=0
Proof: Take
X = ( 1 0 ⋮ 0 ) X =\begin{pmatrix}1 \\0 \\\vdots \\0 \end{pmatrix}X=
10⋮0
or
( 0 1 ⋮ 0 ) \begin{pmatrix}0 \\1 \\\vdots \\0 \end{pmatrix}
01⋮0
Or
( 0 0 ⋮ 1 ) \begin{pmatrix}0 \\0 \\\vdots \\1 \end{pmatrix}
00⋮1
Taken into the calculation, each element in A is 0, that is, A is a zero matrix
6) Let B be r × rr\times rr×r matrix, C isr × nr\times nr×n matrix, and rank( C ) = r (C) = r(C)=r , prove that:
1. If BC= 0, then B=0
2. If BC=C, then B=E
Proof: 1) ∵ r ( C ) = r \because r(C)=r∵r(C)=r
Let's take the maximum linear independent group of C as C 1 , C 2 . . . C r C_{1},C_{2}...C_{r}C1,C2...Cr
B C = 0 ⇒ B ( C 1 , C 2 . . . C r ) = 0 BC = 0\Rightarrow B(C_{1},C_{2}...C_{r})=0 BC=0⇒B(C1,C2...Cr)=0
又( C 1 , C 2 . . . C r ) ≠ 0 (C_{1},C_{2}...C_{r})\ne 0(C1,C2...Cr)=0 , reversible
So, B=0
2) Same as 1) BC = C ⇒ ( B − E ) ( C 1 , C 2 . . . C r ) = 0 BC=C\Rightarrow (BE)(C_{1},C_{2}.. .C_{r})=0BC=C⇒(B−E)(C1,C2...Cr)=0
又( C 1 , C 2 . . . C r ) ≠ 0 (C_{1},C_{2}...C_{r})\ne 0(C1,C2...Cr)=0 , reversible
So, B=E
Welcome everyone to make corrections, there are some