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kang1534

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默默的就好

[求助] 一个矩阵题 求大神

1、已知n阶矩阵A≠0(零矩阵),A^2=0,
证明:(1) 求A的特征值;
  (2)求A的Jordan标准形
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wurongjun

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【答案】应助回帖

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kang1534: 金币+3, ★★★很有帮助 2013-10-29 11:49:59
(1):
       特征值 : r^2=0,所以r=0;
(2):A^2=0;A~=O,所以A中含有二阶若当块!
例如:四阶阵有以下情况
A =
     0     1     0     0
     0     0     0     0
     0     0     0     1
     0     0     0     0
A =
     0     1     0     0
     0     0     0     0
     0     0     0     0
     0     0     0     0
A =
     0     0     0     0
     0     0     1     0
     0     0     0     0
     0     0     0     0
A =
     0     0     0     0
     0     0     0     0
     0     0     0     1
     0     0     0     0
善恶到头终有报,人间正道是沧桑.
2楼2013-10-28 22:09:53
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yinhu1988

木虫 (著名写手)

【答案】应助回帖

★ ★
感谢参与,应助指数 +1
kang1534: 金币+2, ★★★很有帮助 2013-10-29 11:50:05
Ax=lambda x
A^2x=lambda Ax=lambda^2 x =0
x is nonzero vector, lambda^2=0, so lambda=0.
君子务本,静水流深
3楼2013-10-29 03:31:09
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