2) Let B be an n x n skew-Hermitian matrix (B* = -B). Prove that if B^(n) = I , then A⁴ = I.
3) Let A, B be two n x n diagonalizable real matrices. Prove that if A, B have the same eigenvalues, then they commute, i.e. AB = BA.
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Answer: An nxn Hermitian matrix has a basis v(1),....,v(n) of eigenvectors, which have eigenvalues a(1)...a(n) which are real: A v(k) = a(k) v(k). Assuming n>1. If we know that Aⁿ = I, then for every k: Aⁿ v(k)=a(k)ⁿ v(k) = I v(k). Therefore a(k)ⁿ=1. The only real roots of the equation xⁿ=1 are x=+1 (if n is odd) and x=+1 or -1 if n is even. We therefore conclude that a(k)²=1 for every (k)....