Normal matrix proof

Question

I've just been shown the Spectral Theorem proof, which states $A$ is a normal matrix iff there exist a unitary matrix $U$ and a diagonal matrix $\Lambda$ such that $A = U\Lambda U^*$. Now, I've been told the following statement directly follows from the spectral theorem, but I din't understand how? The statement is; $A \in \mathbb{C}^{nxn}$ is normal iff it has n orthogonal eigenvectors? I have no idea how it directly follows, though?

Answer 1

$U$ being unitary means that $U^{-1} = U^* = \overline {U^T}$. That is, $$ U^*U = I. \tag{1} $$

Writing this out: Let $u_1,\ldots,u_n$ be the columns of $U$. Then for all $k, l$ it holds that $$ 0 = \overline u_k^T u_l = <u_k, u_l>, \text{ if $k \ne l$} \\ 1 = \overline u_k^T u_l = <u_k, u_l>, \text{ if $k = l$}. $$ This means that the $n$ columns of $U$ are orthogonal (even orthonormal).

Since the columns of $U$ are the eigenvectors of $A$, the claim follows. (The columns of $U$ are the eigenvectors of $A$ because $A = U\Lambda U^*$ is equivalent to $AU = U\Lambda$)