已知矩阵A,求可逆阵P,使得(P^-1)AP为对角阵A= [2,0,00,1,-10,-1,1]
已知矩阵A,求可逆阵P,使得(P^-1)AP为对角阵A= [2,0,00,1,-10,-1,1]
已知矩阵A,求可逆阵P,使得(P^-1)AP为对角阵
A= [2,0,0
0,1,-1
0,-1,1]
已知矩阵A,求可逆阵P,使得(P^-1)AP为对角阵A= [2,0,00,1,-10,-1,1]
|A-λE| = -λ(2-λ)^2
所以A的特征值为0,2,2
解得 AX=0 的基础解系:a1=(0,1,1)'
解得 (A-2E)X=0 的基础解系:a2=(1,0,0)',a3=(0,1,-1)'
令P=(a1,a2,a3)=
0 1 0
1 0 1
1 0 -1
则P可逆,且P^-1AP = diag(0,2,2).
P = [ 1,0,0
0,1,1
0, 0,1]
P^(-1) = P^t = [ 1,0,0
0,1,0
0,1,1]
P^(-1)A = [2,0,0
...
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P = [ 1,0,0
0,1,1
0, 0,1]
P^(-1) = P^t = [ 1,0,0
0,1,0
0,1,1]
P^(-1)A = [2,0,0
0,1,-1
0,0,0]
P^(-1)AP = [2,0,0
0,2,-2
0,0,0]
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