Orthogonally Diagonalize a matrix with nonreal eigenvectors

Question

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 $A=UDU^*?$

Answer 1

The norm of a vector $v$ is

$$ \|v\|^2 = (v^*)^Tv = \sum_k v_k^* v_k $$

where $^*$ denotes the complex conjugate. So in your case

$$ \|v_1\|^2 = \pmatrix{i & 1}^* \pmatrix{i \\ 1} = \pmatrix{-i & 1} \pmatrix{i \\ 1} = 1 + 1 = 2 $$

Answer 2

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