Question

linear algebra~~~线性代数】~~这个矩阵的特征值和特征向量怎么求？有无通俗方法，解决此问题？

|A-λE| = 1-λ 2 3以下是网上的解法，我看不大懂！仅供您的参考_______________________________________2 1-λ 33 3 6-λr1-r2-1-λ 1+λ 02 1-λ 33 3 6-λc2+c1-1-λ 0 02 3-λ 33 6 6-λ= (-1-λ)[(3-λ)(6-λ)-18]= (-1-λ)[λ^2-9λ]= λ(9-λ)(1+λ)所以A的特征值为 0,9,-1AX = 0 的基础解系为 (1,1,-1)'所以,A的属于特征值0的全部特征向量为:c1(1,1,-1)',c1为非零常数.(A-9E)X = 0 的基础解系为 (1,1,2)'所以,A的属于特征值9的全部特征向量为:c2(1,1,2)',c2为非零常数.(A+E)X = 0 的基础解系为 (1,-1,0)'所以,A的属于特征值-1的全部特征向量为:c3(1,-1,0)',c3为非零常数

Answer 1

如图

追问

I have readed some related PPT-S
about eigen-values and eigen-vectors for all one day!

I have grasped it!

And the same time, I must thank you very much!
THANKS A LOT1

追答

You are welcome.And,I will try my best to explain well.In that way, the questioner would not cost time on PPTS.