= Solution
For $n=2$, take
$$
A=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad B=I_2.
$$
They are <commuting operators> because every <matrix> commutes with the <identity matrix>. However $A$ has only the <eigenvalue> zero, and $\ker A=\operatorname{span}\{(1,0)^T\}$ has <dimension> one. It has no <eigenbasis>, so it cannot appear in any <simultaneous diagonalization>. Equivalently, a <diagonal matrix> similar to this nonzero <nilpotent matrix> would have to be zero, an impossibility. For any prescribed $n\ge2$, use $A=E_{12}$ and $B=I_n$; its zero <eigenspace> has <dimension> $n-1$. For $n=1$ no such example exists.
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