Orthogonally Diagonalize a matrix with nonreal eigenvectors I am given the matrix A and asked to orthogonally diagonalize it. 
$$A=\begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix} $$
While doing this I got $\lambda = 0,2.$
Then I found the eigenvectors corresponding to the eigenvalues to be $$\begin{pmatrix} i \\ 1 \\ \end{pmatrix} $$ and $$\begin{pmatrix} -i \\ 1 \\ \end{pmatrix}, $$ respectively. While trying to divide each eigenvector by its norm I run into a problem, take $V_1$ for example $\|V_1\| = 0$ so I obviously can't divide by $ V_1$ by its norm to make it orthogonal. My question is, is $A$ even orthogonally diagonizable at all? Or if the $\|V_1\| = 0$ does this mean that it is already orthogonal and I can just use my eigenvectors as is to generate the matrix $U$ such that $A=UDU^*?$

It happens that $\|(a,b)\|=\sqrt{|a|^2+|b|^2}$. Therefore, the norm of both vectors that you mentioned is $\sqrt2$.
