A problem related to hermitian and unitary matrix Let $A$ be a $n\times n$ matrix which is both hermitian and unitary.Then which of the followings are true?


*

*$A^2=I$

*$A$ is real

*characteristic polynomial and minimal polynomial are same.

*eigenvalues are 0,1,-1


I think (1) true, implies minimal polynomial of $A$ is $x^2-1$.But characteristic polynomial is of degree $n$.So(3)and (4) are false.I have no idea about (2).please help.Thanks
 A: $A$ is unitary if and only if $A$ is invertible and
$A^\dagger = A^{-1}; \tag{1}$
likewise, $A$ is Hermitian if and only if
$A^\dagger = A; \tag{2}$
combining (1) and (2) we find
$A^{-1} = A; \tag{3}$
multiplying through by $A$ yields
$A^2 = I; \tag{4}$
thus item (1) binds.
Item (2) is false.  A family of counterexamples is provided by
$2 n \times 2n$ matrices of the form
$A = \begin{bmatrix} 0 & -iI_n \\
iI_n & 0 \end{bmatrix}, \tag{5}$
where $I_n$ is the $n \times n$ identity matrix; it is easy to see that such $A$ satisfy equation (2); furthermore,
$A^{\dagger}A = A^2 = i^2 \begin{bmatrix} 0 & -I_n \\
I_n & 0 \end{bmatrix}^2$
$= (-1)(-I_{2n}) = I_{2n}, \tag{6}$
showing both that such $A$ are invertible and that equation (1) binds; however such $A$ are clearly not real.
Item (3) is false, since for $\text{size}(A) \ge 3$, the characteristic polynomial is of degree $\text{size}(A) > 2$.
Item (4) is also false, since (4) implies every eigenvalue $\mu$ of $A$ satisfies $\mu^2 = 1$, as shown by the usual argument given by Bungo in his comment; thus $0$ cannot be an eigenvalue of $A$.
