Consider $V = \mathbb C^{n \times n}$ as an inner product space with respect to the dot product. Let $A\in V$ be a diagonalize orthonormal matrix. Show that the linear operator $T : V \to V$ defined by $T(B)=AB$ is orthonormal diagonalizable.
We were told a hint that we can show that if $v \in \mathbb C^{n \times 1}$ is an eigenvector of $A,$ we can conclude that for a matrix $B$ such that $v$ is the column and the rest are zeroes, there exists $T(B)=\lambda B.$
I concluded that, but how do I proceed from here?
For each eigenvector $v_i$ of $A$, you can form $n$ eigenvector $B_{i,1},\ldots,B_{i,n}$ of $T$ as follows:
$$B_{i,1}=(v_i\, 0\,\cdots 0),\; B_{i,2}=(0\,v_i\,\cdots\,0)\cdots B_{i,n}=(0\,\cdots\,0\,v_i).$$ Note that we have written $v_i$ and $0$ in the column form, so the $B_{i,j}$ are $n\times n$ matrices. Moreover, if $Av_i=\lambda v_i$ then $TB_{i,j}=\lambda B_{i,j}$ for all $j=1,\ldots,n$.
Now let $i$ vary from $1$ to $n$, to get $n^2$ eigenvectors of $T$, namely the $B_{i,j}$'s for $i,j=1,\ldots,n$.
Next, you need to check that the $B_{i,j}$'s are linearly independent. This is not difficult, and follows from the linear independence of the $v_i$'s. Since $\dim V=n^2$, it tells you that $T$ is diagonalizable.
For the orthonormality, we have $\left\langle B_{i,j},B_{k,\ell}\right\rangle=\mathrm{tr}(B_{i,j}B_{k,\ell}^{T})=\delta_{i,k}\delta_{j,\ell}$. So the set of $B_{i,j}$ is orthonormal.