A=(3 -2 0 -1 3 -1 -5 7 -1)求A的特征值，判断A是否相似于对角矩阵若相似求可逆矩阵P使(P^-1)AP为对角矩阵
A=(3 -2 0 -1 3 -1 -5 7 -1)
有1个网友回答
解: |A-λE|=
3-λ -2 0
-1 3-λ -1
-5 7 -1-λ
c1+c2+c3
1-λ -2 0
1-λ 3-λ -1
1-λ 7 -1-λ
r2-r1,r3-r1
1-λ -2 0
0 5-λ -1
0 9 -1-λ
= (1-λ)[(λ-5)(λ+1)+9]
= (1-λ)(λ^2-4λ+4)
= (1-λ)(λ-2)^2
所以A的特征值为1,2,2.
因为 A-2E =
1 -2 0
-1 1 -1
-5 7 -3
-->
r2+r1,r3+5r1
1 -2 0
0 -1 -1
0 -3 -3
r3-3r2
1 -2 0
0 -1 -1
0 0 0
所以 r(A-2E)=2, A的属于二重特征值2的线性无关的特征向量有 3-2=1 个
故A不能对角化.
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