Re: [理工] [線代]-判斷相似
※ 引述《GI9 ( )》之銘言：
: Let A and B be two nxn matrices over C . Suppose that the 2nx2n matrix
: [A 0] [B 0]
: [0 A] is similar to [0 B]
: Show that A is similar to B
我有問題...
M=[A 0] N=[B 0]
[0 A] [0 B]
A特徵值:入a1 入a2... B特徵植:入b1 入b2...
M特徵值:入a1 入a1 入a2 入a2... N特徵值:入b1 入b1 入b2 入b2...
A B不相似
M N光是特徵值就不一樣了
怎麼可能相似 ~_~?
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◆ From: 140.113.139.124
※ 編輯: itsforte 來自: 140.113.139.124 (12/17 18:08)
推
12/17 18:23, , 1F
12/17 18:23, 1F
→
12/17 18:35, , 2F
12/17 18:35, 2F
推
12/17 21:01, , 3F
12/17 21:01, 3F
推
12/18 01:05, , 4F
12/18 01:05, 4F
→
12/18 13:18, , 5F
12/18 13:18, 5F
assume A= p J p^(-1), B= q K q^(-1), which A is NOT similar to B
[A 0] = [p J p^(-1) 0 ] = [p 0] [J 0] [p^(-1) 0 ]
[0 A] [ 0 p J p^(-1)] [0 p] [0 J] [0 p^(-1)]
= P [J 0] P^(-1)
[0 J]
[B 0] = [q K q^(-1) 0 ] = [q 0] [K 0] [q^(-1) 0 ]
[0 B] [ 0 q K q^(-1)] [0 q] [0 K] [0 q^(-1)]
= Q [K 0] Q^(-1)
[0 K]
=> [A 0] is not similar to [B 0]
[0 A] [0 B]
<=> [A 0] is similar to [B 0] => A is similar to B
[0 A] [0 B]
※ 編輯: itsforte 來自: 140.113.139.124 (12/18 13:49)
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