Let $A$ be a Hermitian $n\times n$ matrix over ${\Bbb C}$ such that $A^m = I$ for some natural number $m$.
Prove:
- $A^2=I$.
- If $m$ is odd, then $A=I$
Well, for the first question I did this: Since $A$ is Hermitian then $A$ is diagonalizable. $A$'s eigenvalues are $\{-1,1\}$. The minimal polynomial that reset $A$ and $M(x) = (x+1)^z (x-1)^k $ for $k,z$ natural numbers. So the options for the minimal polynomial are:
- $(x-1)$
- $(x+1)$
- $(x-1)(x+1)$
- $(x-1)^2 $
Eventually I got $A$ can be: $ A=I \quad \mathrm{or}\quad A=-I $
Is that even right?
For the second part of the question I have no clue.