Re: [線代] 矩陣 AB=I => BA = I 嗎

看板Math作者 (Ian)時間10年前 (2015/03/01 02:23), 編輯推噓1(1推 0噓 7→)
留言8則, 2人參與, 最新討論串11/12 (看更多)
※ 引述《kyoiku (生死間有大恐怖)》之銘言： : 反方陣的定義 : AB = BA = I，則 B 稱為 A 的反矩陣 : 如果只有 AB = I 那是否必然 BA = I 呢 : 如何證明？ Assume AB=I. To prove BA=I, it is suffice to show that B is a linear combination of I,A,A^2,A^3,...A^k,... Proof: Consider the infinitely long sequence of matrices I,A,A^2,A^3,.... They are all in a vector space of dim n^2. Hence they are linearly dependant: so there are some coefficients c_0, c_1,c_2,...,c_s such that c_0*I+c_1*A+c_2*A^2+..+c_s*A^s = 0. By multiplying B from the left, we may assume c_0 is non-zero, says c_0=-1. Thus, we have I=c_1*A+c_2*A^2+...+c_s*A^s. Mutiply B from the left again, we get B=c_1*I+c_2*A+....+c_s*A^{s-1}. -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.165.196.85 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1425147783.A.366.html
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03/01 13:09, , 1F
Sorry,it seemed that you proved necessary
03/01 13:09, 1F
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03/01 13:10, , 2F
condition instead of sufficient one; Namely, you
03/01 13:10, 2F
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03/01 13:10, , 3F
assumed BA=I, and then showed B is a linear
03/01 13:10, 3F
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03/01 13:11, , 4F
combination of A^k for k non-negative integers.
03/01 13:11, 4F
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03/01 13:21, , 5F
thx...change all left to right..than its fine
03/01 13:21, 5F
推
03/01 20:55, , 6F
Yes,it is a good proof which aims to show A
03/01 20:55, 6F
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03/01 20:56, , 7F
commutes with B, so it is sufficient to show
03/01 20:56, 7F
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03/01 20:57, , 8F
that B is a polynomial of A.
03/01 20:57, 8F
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