TS = ST for all S, then T has to be a multiple of identity

Question

Some one asked me a question- if TS = ST for all S, then prove that T is diagonalizable. We convinced ourselves that T has to be a diagonal matrix. For the 2x2 case however, I showed that the diagonals have to be equal as well. This is because (writing TS=ST out and comparing (1,2) components) -

$t_{1,1} s_{1,2}+t_{1,2}s_{2,2} = s_{1,1}t_{1,2}+s_{1,2}t_{2,2}$

Now if $t_{1,2} = 0$ (which can be shown by comparing the other entries of the two matrices), then $t_{1,1} = t_{2,2}$. So for 2x2, we showed that T has to be a constant multiple of the identity. How to show this for nxn?

Answer 1

This is true (and well-known). To prove it take $S=E_{ij}$, the matrix with all zeros save for a $1$ in the $(i,j)$-position, for all $i$ and $j$.

Answer 2

Just to provide a counter example: if $S = T$ but is not diagonalizable. $ST = TS$ providing a counterexample to your proposition.

However, if $S$ and $T$ have the same set of eigenvectors.

$ST = TS$