A is a Hermitian $n\times n$ matrix over ${\Bbb C}$ . $$ A^m = I $$ for $m$ a natural number.
Prove $$A^2=I $$
Suppose $m$ is odd. Prove $$ A=I$$
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.