A matrix with imaginary entries satisfies the property?

Question

Let $A$ be $n\times n$ complex matrix. If $A$ is both Hermitian and Unitary, then show that absolute value of all eigenvalues are $1$.

I just know what is Hermitian and Unitary matrix, and that's all. I can't go further.

$$A^{\mathrm{H}}A = AA^{\mathrm{H}}=I_{n}$$ is unitary and,

$$A^{\mathrm{H}}=A$$ means A is Hermitian.

Answer 1

You can see the magnitude of eigenvalues of a unitary matrix by thinking about the eigenvectors. Let $v, \lambda$ be any eigenvector, eigenvalue pair of $A$, so we have $Av=v\lambda$. We can write $A^HAv = I v$ by the nature of $A$ being unitary, as the matrices cancel each other out.

By using the hermetian symmetry we can write $A^H Av = AAv = Av\lambda = v \lambda^2$.

This gives the property of $\lambda^2 = 1$ which has only two possible solutions.

In fact, it is somewhat enough talking about unitary properties to conclude your $|\lambda| = 1$ claim. The hermitian symmetry reduces the possible values to numbers on the real line.

Answer 2

If $Ax = \lambda x$ then $$x=Ix =A^HAx = A^H(\lambda x) = A(\lambda x) = \lambda^2 x$$ so clearly $\lambda^2 = 1$. This implies $|\lambda| = 1$.