Unitary diagonalization of a normal matrix

Question

I will gladly appreciate explanation on how to do so on this matrix:

$$ \begin{pmatrix} i & 0 \\ 0 & i \\ \end{pmatrix} $$

I got as far as calculating the eigenvalues and came up with $λ = i$. when trying to find the eigenvectors I came up with the $0$ matrix.

what am I doing wrong?

Much appreciation and thanks in advance.

Answer 1

If $A=\begin{pmatrix}{i}&{0}\\{0}&{i}\end{pmatrix}$, then $I_2^*AI_2=A=D\text{ (diagonal)}$ and $I_2=\begin{pmatrix}{1}&{0}\\{0}&{1}\end{pmatrix}$ is an unitary matrix.

Answer 2

You got zero matrix that means every vector is in the eigenspace of $i$. So choose any two linearly independent vector and that will be your basis. Also its already diagonal so standard basis are the eigenvectors(if the matrix is with respect to standard basis).

Answer 3

Since $Ov=O$ is satisfied for all non-zero vectors $v$ so take any two arbitrary but linearly independent of them like $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$.