Re: [線代] eigenvalue一題...
: 麻煩各位了O_Q
Clearly, T(T(A)) = A = I(A), I is the identity map.
Let f(x) = x^2-1, f(T) = 0.
Hence,the mini.poly p of T , p|x^2-1 = (x-1)(x+1).
So, p = x^2-1 or x-1 or x+1.
It's easy to see that p ≠ x-1 or x + 1
(otherwise, T = I or T = -I)
Consequencely,
The eigenvalue of T is 1 and -1.
p = (x-1)(x+1), so T is diagonalizable.
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※ 編輯: VFresh 來自: 111.251.175.194 (03/02 01:55)
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