Matrices

2011-10-12 10:30 pm

For an n x n square matrix A, if it is given that A^3 = I, is it always true that :
(a) A must be equal to I ?
(b) det A must be equal to 1 ?
(c) In general, if A^n = I, then (a) and (b) are always true ?
Please prove your result.

更新1:

To : Forretrio,自由自在 and doraemonpaul. Appreciate your helps. Forretrio provided answer to (a) although may be incomplete, it is acceptable to me. For (b), 自由自在 showed that det A can be +1 or - 1, but is this the only possible answers? Is this answers also apply to A^n = I ? Please explain, thks.

回答 (3)

doraemonpaul

2011-10-13 5:28 pm

✔ 最佳答案

Forretrio的回答還有不少問題：
1. a部舉出了太刁鑽的反例
2. 當complex matrix冇到

2011-10-13 09:28:37 補充：

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahooknowledge/matrixequation/matrixequproperties9.jpg

參考資料：
my maths knowledge + http://en.wikipedia.org/wiki/Root_of_unity + example from http://tw.knowledge.yahoo.com/question/question?qid=1010020803771

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用戶9112

2011-10-13 4:12 am

A^n = I => det(A) = 1 is false
for example A = row 1 is (1 , 0) row 2 is (0, -1)
det(A)= -1 and A^2 = I

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Forretrio

2011-10-13 3:19 am

a) No. One counter example is enough to disproof the claim:

圖片參考：http://*****/3b3mvhw

b) Yes.
Let A^3 = I
det (A^3) = (det A)^3 = det I = 1
det A = 1.
c) For (a) is not always true as for A^n = I,
(A-I)(I+A+A^2+...+A^(n-1)) = zero matrix
There exist a solution A for I+A+A^2+,,,+A^(n-1) = 0 in which A =/= I.
For (b) it's still true as
det (A^n) = (det A)^n = 1
det A = 1
as long as the matrix is real.

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收錄日期: 2021-04-23 23:23:17
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