Let $A$ be an $n \times n$ matrix which is both Hermitian and unitary

Question

Let $A$ be an $n \times n$ matrix which is both Hermitian and unitary.Then

a) $A^2=I$

b) $A$ is real

c) the eigenvalues of $A$ are $0$, $1$, $-1$

d) The minimal and characteristic polynomials are same.

Answer 1

(a) True. $A^*=A$ and $A^*=A^{-1}$ implies $A=A^{-1}$ so $A^2=I$.

(b) False. Choose $A=\begin{bmatrix}{0}&{i}\\{-i}&{0}\end{bmatrix}.$

(c) False. $A^2=I$ implies $p(\lambda)=\lambda^2-1$ is an annihilator polynomial of $A$, hence its minimal polynomial divides to $(\lambda-1) (\lambda+1)$. The only possible eigenvalues are $\lambda=1$ or $\lambda=-1.$

(d) False. Choose $A=\begin{bmatrix}{1}&{0}\\{0}&{1}\end{bmatrix}.$