Is there a counter-example where the matrices $A$ and $B$ commutes, $A$ have distinct eigenvalues but $B$ is not diagonalisable?

Question

Suppose that $A$ and $B$ are matrices in $M_n(\mathbb{C})$ with $AB=BA$. We know that if $A$ have $n$ distinct eigenvalues, then every eigenvalue of $A$ is also an eigenvalue for $B$, so $B$ is diagonalisable. What I wonder is if $A$ have distinct eigenvalues but maybe some of them are repeated, do $B$ still have to be diagonalisable, if not what is the counter-example.

Answer 1

Since you want $A$ to have at least $2$ different eigenvalues, the identity matrix will not work. Actually your requirement is not reasonable, as the following block matrix example shows:

Let $A=\begin{pmatrix}0&0&0\\0&1&0\\0&0&1\end{pmatrix}$ and $B=\begin{pmatrix}0&0\\0&B'\end{pmatrix}$ with $B'$ any non-diagonalizable $2 \times 2$-matrix.

As you see, I just took the identity matrix and any non-diagonalizable matrix. Then I inserted a zero-block at the upper left to fulfill your requirement of two different eigenvalues...