λE - A = E - A =
[-1 1 -2]
[ 0 0 0]
[ 2 -2 4]
行初等变换为
[1 -1 2]
[ 0 0 0]
[ 0 0 0]
方程组(λE - A)x = 0 化为 x1 - x2 + 2x3 = 0
即 x1 = x2 - 2x3
取 x2 = 1, x3 = 0, 得基础解系(1, 1, 0)^T;
取 x2 = 0, x3 = 1, 得基础解系(-2, 0, 1)^T;
即得 2 个特征向量。
追问有点看不懂,×1中的×代表着什么?
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/63d9f2d3572c11df32060ec66a2762d0f603c204?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
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原题和过程
λ3的基础解系