For every $\lambda \in C$ with $|\lambda| = 1$, there exists a unitarian matrix $B$ with eigenvalue $\lambda$ How can I prove: > _For every $\lambda \in C$, with $|\lambda| = 1$, there exist unitarian matrix $B$ with eigenvalue $\lambda$._ I tried to find a counterexample and I was not succeeded. I believe I need to prove that statement. Any help is highly appreciated.

First off, it is well known any eigenvalue$~\lambda$ of a unitary matrix$~U$ has $|\lambda|=1$; this is so because of $v\in\Bbb C^n$ is a corresponding eigenvector then $0\
eq v^*v=v^*U^*Uv=(\lambda v)^*(\lambda v)=\lambda\overline\lambda\,(v^*v)$, so $1=\lambda\overline\lambda=|\lambda|^2$.

To conversely show that for each $\lambda\in\Bbb C$ with $|\lambda|=1$ there exists a unitary matrix with eigenvalue$~\lambda$, it suffices to consider $1\times 1$ matrices; these are unitary if and only if their unique entry $a$ has $|a|=1$. If you insist on having a square matrix of a given size$~n$, then you can, provided that $n>0$, take a diagonal matrix whose diagonal entries all have norm$~1$ (so that it is unitary) and at least one of them equals$~\lambda$; for instance take $\lambda I_n$. For $n=0$ you cannot find such a matrix, because the $0\times0$ matrix has no eigenvalues at all.