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※ 引述《Taeyeon20 (ygyhjddsfdsfdsfdsgvsdv)》之銘言：
: Let A be a 2x2 skew-Hermitian matrix. Prove that if A^n=I for some n
: positive integer, then A^4=I.
: 不知道如何證 希望有人教我 謝謝
A = ai b+ci
-b+ci di
where a,b,c,d are real
characteristic eq.
x^2 - (a+d)i x + (-ad+b^2+c^2) = 0
denote i x = u
then
u^2 + (a+d) u + (ad-b^2-c^2) = 0
Δ = (a+d)^2 - 4(ad-b^2-c^2) = (a-d)^2 + 4b^2 + 4c^2
(i) Δ=0, then a=d, b=0, c=0
and A^n = I, then a=d=1,-1, then A^4=I
(ii) Δ>0, u has two distinct real roots, α,β
x = -iα, -iβ
because x^n - 1 is divided by characteristic polynomial
then {α,β} = {1, -1}
then x^2 + 1 = 0, A^2 + I = 0
hence, A^4 = I
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