Diagonalizable maps commute? What went wrong?

Question

This seems weird: If $T: V \rightarrow V$ is diagonalizable linear operator on finite dimensional vector space $V$, and $S:V \rightarrow V$ is another linear operator, then take basis $B$ for which $[T]_B$ is diagonal matrix. Do we not have

$[TS]_B= [T]_B [S]_B = [S]_B [T]_B =[ST]_B$ hence $ST=TS$? What went wrong?

Answer 1

Why would a diagonal matrix commutes with another matrix? Take for example:

$$\begin{pmatrix}1&0\\0&2\end{pmatrix},\begin{pmatrix}0 & 1\\0&0\end{pmatrix}.$$

In other words, your computation went wrong here: $$[T]_B[S]_B=[S]_B[T]_B.$$

Answer 2

If you're not simply mistaken, then you're probably trying to remember one of:

• Scalar operators — operators that are scalar multiples of the identity — commute with every matrix.
• Simultaneously diagonalizable operators — a pair of operators that are both diagonal relative to a single basis — commute.

Answer 3

The problem is that diagonal matrices only commute with every other matrix when all of the diagonal elements are equal, that is, when they are multiples of the identity. $T$ is only given as diagonalizable, so we don’t know whether or not all of its eigenvalues are equal.