[PubMed Mathematics] Linear Algebra Chapter 4 - Linear Equations (2, General Solutions of Linear Equations | Theoretical Extension)


introduction

Continuing from the previous article, continue to learn the content of linear equations, starting from the general solution of the equations.


4. General solutions of linear equations

4.1 Homogeneous linear equations

(1) Basic solution system - Let r ( A ) = r < nr(A)=r<nr(A)=r<n , thenAX = 0 \pmb{AX=0}AX=0 The maximum linearly independent group of the solution vector group formed by all solutions is called the equation systemAX = 0 \pmb{AX=0}AX=A basic solution system of 0 . The number of linearly independent solution vectors contained in the basic solution system is( n − r ) (nr)(nr ) pieces.

Because r ( A ) = nr(A)=nr(A)=What about n ? because ifr ( A ) = nr(A)=nr(A)=If it is n , then the homogeneous equation has only zero solutions, and there is nothing to discuss.

When finding the basic solution system of a homogeneous linear equation system, the coefficient matrix is ​​transformed into steps by elementary row transformation (elementary row transformation of the coefficient matrix is ​​equivalent to the same solution transformation of the equation system), and the first non-zero element of each row is located The unknowns corresponding to the columns are constrained variables, and the remaining variables are free variables, so that the basic solution system can be determined (it is best to turn the first non-zero element of each row to 1, and turn the rest of the elements in the column to zero).

For example, suppose the equation system AX = 0 \pmb{AX=0}AX=0 coefficient matrixA \pmb{A}After elementary row transformation, A can be reduced to the following form:
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thenr ( A ) = 3 < 5 r(A)=3<5r(A)=3<5 , the equation systemAX = 0 \pmb{AX=0}AX=The basic solution system of 0 contains n − r = 5 − 3 = 2 nr=5-3=2nr=53=2 solution vectors, wherex 1 , x 2 , x 3 x_1,x_2,x_3x1,x2,x3is the constraint variable, x 4 , x 5 x_4,x_5x4,x5is a free variable, ( x 4 , x 5 ) (x_4,x_5)(x4,x5) take( 1 , 0 ) (1,0)(1,0 ) and( 0 , 1 ) (0,1)(0,1 ) , then the basic solution system is: ξ 1 = ( − 2 , 1 , − 3 , 1 , 0 ) T , ξ 2 = ( 3 , − 4 , 2 , 0 , 1 ) T . \xi_1=(-2 ,1,-3,1,0)^T,\xi_2=(3,-4,2,0,1)^T.X1=(2,1,3,1,0)T,X2=(3,4,2,0,1)T . (2) General solution——Supposeξ 1 , ξ 2 , … , ξ n − r \xi_1,\xi_2,\dots,\xi_{nr}X1,X2,,XnrFor the homogeneous linear equation system AX = 0 \pmb{AX=0}AX=0 , calledk 1 ξ 1 + k 2 ξ 2 + ⋯ + kn − r ξ n − r k_1\xi_1+k_2\xi_2+\dots+k_{nr}\xi_{nr}k1X1+k2X2++knrXnrFor the homogeneous linear equation system AX = 0 \pmb{AX=0}AX=0 , wherek 1 , k 2 , … , kn − r k_1,k_2,\dots,k_{nr}k1,k2,,knrfor any constant.

4.2 Inhomogeneous linear equations

r ( A ) = r ( A ‾ ) < n r(A)=r(\overline{A})<n r(A)=r(A)<n,且ξ 1 , ξ 2 , … , ξ n − r \xi_1,\xi_2,\dots,\xi_{nr}X1,X2,,Xnrfor AX = b \pmb{AX=b}AX=b 's derived equation systemAX = 0 \pmb{AX=0}AX=A basic solution system of 0 , η 0 \pmb{\eta_0}the0for AX = b \pmb{AX=b}AX=A solution of b , then AX = b \pmb{AX=b}AX=The general solution of b is k 1 ξ 1 + k 2 ξ 2 + ⋯ + kn − r ξ n − r + η 0 , k_1\xi_1+k_2\xi_2+\dots+k_{nr}\xi_{nr}+\eta_0,k1X1+k2X2++knrXnr+the0, wherek 1 , k 2 , … , kn − r k_1,k_2,\dots,k_{nr}k1,k2,,knrfor any constant.

1. Homogeneous linear equation system AX = 0 \pmb{AX=0}AX=The basic solution system of 0 is not unique, but the number of linearly independent solution vectors is unique.
2,r ( A ) = r ( A ‾ ) < nr(A)=r(\overline{A})<nr(A)=r(A)<When n , the non-homogeneous linear equation systemAX = b \pmb{AX=b}AX=b The number of maximum linearly independent vectors of all solution vectors is( n − r + 1 ) (n-r+1)(nr+1 )亪。
3,任η 1 , η 2 , … , η n − r + 1 \eta_1,\eta_2,\dots,\eta_{n-r+1}the1,the2,,thenr+1For non-homogeneous linear equations AX = b \pmb{AX=b}AX=A maximal linear independent group of b , its general solution can also be expressed ask 1 η 1 + k 2 η 2 + ⋯ + kn − r + 1 η n − r + 1 k_1\eta_1+k_2\ eta_2+\dots+k_{n-r+1}\eta_{n-r+1}k1the1+k2the2++knr+1thenr+1, where k 1 , k 2 , … , kn − r + 1 k_1,k_2,\dots,k_{n-r+1}k1,k2,,knr+1is any constant, and k 1 + k 2 + ⋯ + kn − r + 1 = 1. k_1+k_2+\dots+k_{n-r+1}=1.k1+k2++knr+1=1.


5. Theoretical extension of the solution of equations

Theorem 1 - Let AAA m × n m\times n m×n- matrix,BBB isn × sn\times sn×s matrix, ifAB = O AB=OAB=O , thenBBThe column vector system of B is the equation systemAX = 0 AX=0AX=0 solution.
Proof:LetB = ( β 1 , β 2 , … , β s ) B=(\beta_1,\beta_2,\dots,\beta_s)B=( b1,b2,,bs),则AB = ( A β 1 , A β 2 , … , A β s ) AB=(A\beta_1,A\beta_2,\dots,A\beta_s)AB=(Aβ1,Aβ2,,Aβs),若AB = O AB=OAB=O,则A β 1 = 0 , A β 2 = 0 … , A β s = 0 A\beta_1=0,A\beta_2=0\dots,A\beta_s=0Aβ1=0,Aβ2=0,Aβs=0 , the original proposition is proved.

Theorem 2 - Let the equation system AX = 0 \pmb{AX=0}AX=0 andBX = 0 \pmb{BX=0}BX=0 is the same solution equation system, thenr ( A ) = r ( B ) r(A)=r(B)r(A)=r ( B ) , and vice versa.

Theorem 3 - Let the equation system AX = 0 \pmb{AX=0}AX=The solution of 0 is BX = 0 \pmb{BX=0}BX=0Solution , forr ( A ) ≥ r ( B ) r(A) \geq r(B).r(A)r(B).

1. Let the equation system AX = 0 \pmb{AX=0}AX=The solution of 0 is BX = 0 \pmb{BX=0}BX=0 , but not all, thenr ( A ) > r ( B ) . r(A) > r(B).r(A)>r ( B ) .
2, let the equation systemAX = 0 \pmb{AX=0}AX=The solution of 0 is BX = 0 \pmb{BX=0}BX=0 , andr ( A ) = r ( B ) r(A) = r(B)r(A)=r ( B ) , then the two equations have the same solution.

Theorem 4 —— Let AX = b , BX = c \pmb{AX=b},\pmb{BX=c}AX=b,BX=c , then the linear equation system( A , B ) TX = ( b , c ) T (A,B)^TX=(b,c)^T(A,B)TX=(b,c)The solution of T is the common solution of the two equations.

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