Prove If $A_{n\times n}$ is hermitian over $\mathbb C$ and $A^m=I, m\in \mathbb N$, then $A^2=I$
In other words, we need to show that $A$ is unitary, since its hermitian $A=A^*$, then $A^2=A^*A=AA^*=I$.
I don't know how to tackle this though, maybe somehow use that the polynomial $x^m-1$ annihilate $A$ and both it and $x^2-1$ and have a common root?
Combine that:
The matrix is diagonalizable. Hermitian Matrices are Diagonalizable
The eigenvalues satisfy $\lambda^m = 1$.
The eigenvalues of a Hermitian matrix are real.
To get that the matrix is similar to a diagonal matrix with entries $\pm 1$, for which the claim holds, and your result follows directly.
Or, closer to your attempt. The minimal polynomial $P$ of $A$ divides $X^m -1 =\prod_{j= 0}^{m-1} (X - \zeta^j)$ where $\zeta$ is a primitive $m$-th root of unity. And the roots of $P$ are the eigenvalues of $A$. These eigenvalues are all real (as $A$ is hermitian). Yet the only real roots of $X^m -1$ are $1$, and possibly $-1$ (depending on the parity of $m$). Therefore $P$ divides $ (X-1)(X+1)= X^2 -1$ and $A^2 - I= 0$.
Hint:
Using the spectral decomposition theorem, there exists a unitary matrix $U \in M_n(\Bbb C)$, such that:
$$U^* A U = \text{diag}(\lambda_1,...,\lambda_n) \ \ \ \ (\star)$$
With $\lambda_i$'s being the eigenvalues of $A$. Raising both sides to the $m$th power, then comparing, we obtain: $\lambda_i ^m = 1$, for all $i$.
However, we know that the eigenvalues of a Hermitian matrix are real. So, for all $i$, we have $\lambda_i ^2 = 1$. Going back to $(\star)$, we get the result.
We can play an old trick that does not require the use of eigenvalues or diagonalisation. Suppose $A$ is Hermitian and $A^m=I$ for some natural number $m$.
When $m=2k$, let $x$ be an arbitrary vector and $y=(A^2-I)x$. Then \begin{align} 0 &= x^\ast(A^m-I)y\\ &=x^\ast(A^2-I)\left[(A^2)^{k-1} + (A^2)^{k-2} + \cdots + (A^2) + I\right]y\\ &=y^\ast\left[(A^2)^{k-1} + (A^2)^{k-2} + \cdots + (A^2) + I\right]y\\ &=\|A^{k-1}y\|^2 + \|A^{k-2}y\|^2 + \cdots + \|Ay\|^2 + \|y\|^2\tag{1} \end{align} where $\|\cdot\|$ denotes the Euclidean norm. Now, what does $(1)$ tell you?
If $m=2k+1$ instead, let $y=(A-I)x$. Then \begin{align} 0 &= x^\ast(A^m-I)y\\ &=x^\ast(A-I)\left(A^{2k} + A^{2k-1} + \cdots + A + I\right)y\\ &=y^\ast\left(A^{2k} + A^{2k-1} + \cdots + A + I\right)y\\ &=y^\ast\left(\sum_{j=0}^kA^{2j} + \sum_{j=1}^k A^{2j-1}\right)y\\ &=y^\ast\left(\sum_{j=0}^kA^{2j} + \sum_{j=1}^k A^{2j+2k}\right)y. \end{align} In this case, can you prove something stronger?
