推 plover:不錯呀 ~(￣▽￣)~(＿△＿)~(￣▽￣)~(＿△＿)~(ꄠ140.112.218.142 08/23
※ 編輯: hips 來自: 140.112.7.59 (08/23 14:28)
※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: If V is a finite dimensional vector space, T: V ---> V is a
: linear transformation such that T^3 - 3T^2 + 3T - I = O, where
: O: V ---> V, O(v) = 0 for all v in V.
: (1) Show that there is a v≠0 in V such that T(v) = v.
Since (T - I)^3 (v) = 0 for all v , T - I is not injective
so T has eigenvalue 1. Thus (1) is valid.
: (2) Show that T is invertible.
Let T(v) = T(k) , then T^3(v) - 3T^2(v) + 3T(v) - v = 0
= T^3(k) - 3T^2(k) + 3T(k) - k , so v = k ,i.e. T is injective
,by 題目,T is surjective, so T is invertible.
: 1 1 0
: (3) Let A = ( 0 1 1 ). Show that A is invertible, and express
: 0 0 1
: A^{-1} as a polynomial in A with real coefficients.
第一部分最快的方法就是 det(A)≠0 , 但是其中的原理我不了解@@
故採用之前的結果. T^3 - 3T^2 + 3T - I = O ,把T用L_A代會符合
根據(2), A 可逆. L_A (left multiplication).
A^3 - 3A^2 + 3A = I = AA^{-1} , so
A^{-1}(A^3 - 3A^2 + 3A) = A^2 -3A +3I = A^{-1}
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有錯請指正,我不敢保證是對的orz
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