Define $$T\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} = \begin{pmatrix} 2b & ia+c\\ -3d & i(a-b)\\ \end{pmatrix}. $$
Prove that $\operatorname{Mat}(2, 2, \mathbb{C}$) has no basis consisting of eigenvectors of $T$ which are orthonormal with respect to the Hilbert-Schmidt inner product, where the Hilbert-Schmidt inner product is defined as $$\langle\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} \begin{pmatrix} w & x\\ y & z\\ \end{pmatrix} \rangle =(a\overline{w}+b\overline{x})+(c\overline{y}+d\overline{z})$$
I've tried everything from finding the zeros the inner product of the general form of $T^2$ to manipulating the characteristic polynomial and Jordan form. All I have got so far is very messy numerical solutions. Is there an elegant approach to this problem?
The inner product space $(M_2(\mathbb{C}), \left < \cdot, \cdot \right>_{HS})$ is isometric to the inner product space $(\mathbb{C}^4, \left< \cdot, \cdot \right>_{\text{standard}})$ using the map $\varphi \colon M_2(\mathbb{C}) \rightarrow \mathbb{C}^4$ given by
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}. $$
You can use this isometry $\varphi$ to translate the question to a more familiar context. Namely, $T$ will be orthogonally diagonalizable if and only if the operator $S = \varphi \circ T \circ \varphi^{-1} \colon \mathbb{C}^4 \rightarrow \mathbb{C}^4$ will be orthogonally diagonalizble. More explicitly, we have
$$ S \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} 2b \\ ia + c \\ -3d \\ i(a - b) \end{pmatrix} = \begin{pmatrix} 0 & 2 & 0 & 0 \\ i & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \\ i & -i & 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = A \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} $$
which shows that the operator $S$ is represented with repsect to the standard (orthonormal) basis of $\mathbb{C}^4$ by the matrix $A$. The only thing left is to show that $A$ is not normal ($AA^{*} \neq A^{*}A$) and so $A$ (and $S$) is not orthogonally diagonalizable.