Unitary matrix has eigenvalues of 1 or -1

Question

$A\in M_n(C)$

\begin{pmatrix} 1 & 0&0 \\ 0& i & 0\\ 0& 0 & i \end{pmatrix}

$AA^*=A^*A=I_3$ so the matrix is unitary but the eigenvalues are $1, i$.

the eigenvalues not supposed to be just $1$ or $-1$? am I wrong?

Answer 1

You are wrong. A complex number $\omega$ is an eigenvalue of some unitary matrix if and only if $|\omega|=1$.

Answer 2

If $\lambda_1,...,\lambda_n \in \mathbb C$ with$|\lambda_j|=1$ for $j=1,...,n$, then

$$A= diag(\lambda_1,...,\lambda_n )$$

is unitary and the eigenvalues of $A$ are $\lambda_1,...,\lambda_n $.

Answer 3

The matrix \begin{bmatrix} \cos t & -\sin t & 0 \\ \sin t & \cos t & 0 \\ 0 & 0 & 1 \end{bmatrix} is easily seen to be unitary (actually real orthogonal) for every real $t$. Its eigenvalues are $1$, $\cos t+i\sin t$ and $\cos t-i\sin t$.

A real eigenvalue of a unitary matrix can only be $1$ or $-1$; but generally unitary matrices have complex eigenvalues all with modulo $1$. Indeed, if $\lambda$ is an eigenvalue of the unitary $U$, with eigenvector $v$, we have $$ v^Hv=v^HU^HUv=(Uv)^H(Uv)=(\lambda v)^H(\lambda v)=\bar{\lambda}\lambda v^Hv =|\lambda|^2(v^Hv) $$ forcing $|\lambda|^2=1$ and so $|\lambda|=1$. Nothing more can be said, because it's quite easy to make a unitary matrix having as eigenvalues whatever complex numbers of modulo $1$ you want.