[理工] 離散 對角化 105成大

看板Grad-ProbAsk作者 (水能載舟亦能洗澡)時間8年前 (2017/08/29 15:48), 編輯推噓3(3推 0噓 8→)
留言11則, 3人參與, 最新討論串1/1
http://i.imgur.com/F5KiXt1.jpg
想請問第三題該如何證明 想了很久都想不出如何利用rank=3的條件 麻煩各位大大幫忙解惑 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 1.175.150.93 ※ 文章網址: https://www.ptt.cc/bbs/Grad-ProbAsk/M.1503992912.A.D22.html
推
08/29 16:33, , 1F
det(A-λI)=0, since A is singular, det(A)=0, we have
08/29 16:33, 1F
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08/29 16:33, , 2F
λ=0 is a eigenvalue of A. dim{N(A-λI),λ=0} = dim{N
08/29 16:33, 2F
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08/29 16:33, , 3F
(A)} = nullity = n-r = 4-3 = 1 = GM(λ=0)<=AM(λ=0). s
08/29 16:33, 3F
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08/29 16:33, , 4F
ince A is 4x4, and λ≠0 has multiplicity 3 , so AM(λ
08/29 16:33, 4F
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08/29 16:33, , 5F
=0) = 4-3 = 1.
08/29 16:33, 5F
推
08/29 16:36, , 6F
GM(λ=0)應該是1 你算錯了 不過AM>=GM 要對角化每個λ一定
08/29 16:36, 6F
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08/29 16:37, , 7F
要AM=GM(各eigenspace的基底總和=4) 題目沒說清楚 不過從
08/29 16:37, 7F
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08/29 16:37, , 8F
答案來看應該是指AM=GM的情況
08/29 16:37, 8F
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08/29 21:30, , 9F
所以這題的λ不為0的eigenvalue不用去驗證am=gm囉？ 從
08/29 21:30, 9F
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08/29 21:30, , 10F
題目好像沒辦法驗證
08/29 21:30, 10F
推
11/17 18:51, , 11F
你那本題目好像打錯了！lambda = 0
11/17 18:51, 11F
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