Re: [線代] 若兩個矩陣相似(similar)則秩(rank)一樣?

看板Math作者JohnMash (Paul)時間15年前 (2011/01/25 13:40)推噓2(2推 0噓 1→)

留言3則, 2人參與討論串4/4 (看更多)

:※ 引述《wyob (Go Dolphins)》之銘言： :藉這篇問一下，那A會和A^t相似嗎 :因為我想要解以下這題不知道路線對不對 :prove or disprove A(A^t)和(A^t)A有相同的非負eigenvalue :-- :※ 發信站: 批踢踢實業坊(ptt.cc) :◆ From: 134.208.85.174

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sc841104

01/24 23:19,

01/24 23:19

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wyob

01/24 23:20,

01/24 23:20

推

sc841104

01/24 23:24,

01/24 23:24

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sc841104

01/24 23:25,

01/24 23:25

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sc841104

01/24 23:26,

01/24 23:26

D is Jordan block matrix, and then?

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wyob

01/25 00:03,

01/25 00:03

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Vulpix

01/25 00:36,

01/25 00:36

Suppose a NONZERO vector x and a POSITIVE numberλsatisfying A.A^t.x=λ x...........................................(1) then A^t.A.A^t.x=λ A^t.x....................................(2) that is, if x is an eigenvector of A.A^t with eigenvalue λ. By (1), A^t.x CANNOT be the null vector. By(2), A^t.x is an eigenvector of A^t.A with eigenvalue λ. Suppose a NONZERO vector x satisfying A.A^t.x=0 (i.e. λ=0) then det(A)=0 there is a NONZERO vector y such that A.y=0 then A^t.A.y=0 Done. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 112.104.88.209 ※ 編輯: JohnMash 來自: 112.104.88.209 (01/25 13:44)

推

wyob

01/25 23:00, , 1^F

01/25 23:00, 1^F

For example, A=[1 0 0] [0 1 0] A.A^t=[1 0] [0 1] A^t.A=[1 0 0] [0 1 0] [0 0 0] ※ 編輯: JohnMash 來自: 112.104.98.144 (01/26 00:02)

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ofd168

01/26 22:42, , 2^F

01/26 22:42, 2^F

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ofd168

01/26 22:42, , 3^F