...矩阵B=(kE+A)2,其中k为实数,E为单位矩阵,求对角矩阵Λ,使得B与Λ...

发布网友 发布时间：2024-03-04 18:53

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热心网友 时间：2024-03-07 14:25

由于：.λE?A.=.λ?1    0?10   λ?2    0?1    0λ?1.=λ（λ-2）2，
可得A的特征值为：λ1=λ2=2，λ3=0，
设λ为A的特征值，x≠0为相应的特征向量，
即：Ax=λx，Bx=（kE+A）2x=（k+λ）2x，
即B有特征值：（k+λ）2，（k+λ）2，k2，
又因为B为实对称矩阵，从而可以对角化，由此可得：
A＝(k+λ)2   (k+λ)2   (k+λ)2，
且：B～A
B=P（kE+D）2PT=P(k+λ)2   (k+λ)2   (k+λ)2已赞过已踩过你对这个回答的评价是？评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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热心网友 时间：2024-03-07 14:17

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