question about schur canonical form If $A$ is an $n\times n$-matrix and it has $n$ orthonormal eigenvectors, 
is it true that $U^*AU$ is diagonal? $U$ is an unitary matrix and $U^*$ is the conjugate transpose of $U$.
If it is true, how to prove it?
Thanks!

Unitary diagonalization is really the same thing as normal diagonalization, except the eigenvectors form an orthonormal basis instead of a regular basis. 
Recall that you diagonalize by forming an eigenbasis $\mathcal{B}$ of some matrix $A$. If the matrix $P$ has columns $\mathcal{B}$ then $P^{-1}AP$ will be diagonal. Now if the basis you took was orthonormal, then it follows that $P$ is not only invertible, but unitary. You have unitarily diagonalized $A$.
Of course, this is not always possible. Not all matrices are diagonalizable, and of those that are, still not all are unitarily diagonalizable. But you are given that your matrix has an orthonormal set of $n$ eigenvectors which by the above discussion is equivalent to the matrix being unitarily diagonalizable. The condition of being unitarily diagonalizable is equivalent to that of being normal in the sense of $AA^* = A^*A$.
Now Schur decomposition is something different, albeit related. Over $\mathcal{C}$ there always exists a unitary matrix $U$ for which $U^*AU = T$ is upper triangular. This is what is typically called the Schur decomposition or Schur triangularization. I assume this is what you mean by "Schur canonical form". It turns out that if your matrix $A$ is normal, then the unitary matrix which triangularizes $A$ also diagonalizes $A$. I suspect this is what your question is really asking. In fact this is the standard proof given that a normal matrix is unitarily diagonalizable. I will sketch the proof below.
Theorem: Let $A$ be an $n\times n$ complex matrix. Then $A$ is unitarily diagonalizable if and only if $A$ is normal.
Proof: I will sketch a proof for the backwards direction. Suppose that $A$ is normal. Then $AA^* = A^*A$. Let $A$ be triangularized by the unitary matrix $U$. We then have $T=U^*AU$. Now consider
$$T^*T = U^*A^*AU  = U^*AA^*U = TT^*$$
If you compare the diagonal entries of $T^*T$ and $TT^*$ you will see that the equality forces the off-diagonal entries to be zero.
