Linear Algebra lecture8 note

时间:2023-03-09 08:40:01
Linear Algebra lecture8 note

Compute solution of AX=b (X=Xp+Xn)

rank r

r=m solutions exist

r=n solutions unique

 


example:

Linear Algebra lecture8 note

若想方程有解,b1,b2,b3需要满足什么条件? 观察矩阵可知,第三行是前两行的和,所以b1+b2=b3

Linear Algebra lecture8 note

Solvability Condition on b:

Ax=b is solvable when b is in C (A)

If a combination of Rows of A gives zero row, then the same combination of entries of b must give 0

假设Linear Algebra lecture8 note,则上述矩阵变为:

Linear Algebra lecture8 note

To find complete solution to AX=b:

1.Xp (particular): set all free variables to zero, solve AX=b for pivot variable

此例中,X2=0,X4=0

Linear Algebra lecture8 note

Linear Algebra lecture8 note

2.Xn(nullspace) 上一节已经解出

Linear Algebra lecture8 note

3.X(complete)=Xp+Xn

Linear Algebra lecture8 note

以上操作可解释为:

Linear Algebra lecture8 note

 


m by n matrix A of rank r(r<=m,r<=n)

Full column of rank(r=n):

所有列均有主元; no free variables;  N(A)=zero vector; solution to AX=b is X=Xp which means if solution exists then the solution is unique(0 or 1 solution)

这种情况实际就是,除zero组合之外,列之间的线性组合无法产生零列

Full row of rank(r=m):

所有行均有主元; no zero rows; can solve AX=b for every b; left with n-r(n-m) free variables

Full rank(r=m=n):

N(A)=zero vector; R(行最简形)=I(单位矩阵)

 

summary:

矩阵的秩决定了方程组解的数目

Linear Algebra lecture8 note