If a commutator has an eigenvalue $0$, do the two operators share an eigenvector? Let $A,B$ be two diagonalizable linear operators, such that their commutator $\left[A,B\right] = AB-BA$ has an eigenvalue $\lambda = 0$. Does this mean $A$ and $B$ share an eigenvector? 


Of course, the vice-versa is correct. If they do share an eigenvector, the commutator necessarily has an eigenvalue of $0$. I'm not sure though if this is a sufficient condition, or only a necessary one.
Clearly we can pick a basis so that one of the operators is diagonal already. Here's a small counterexample of this sort:
$$ A = \begin{pmatrix} \lambda &amp; 0 &amp; 0 \\ 0 &amp; \lambda &amp; 0 \\ 0 &amp; 0 &amp; \mu \end{pmatrix}  \qquad B = \begin{pmatrix} 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 0 \end{pmatrix}. $$
The eigenvalues of $B$ are $\pm \sqrt{2}$ and $0$, and all eigenvectors are of the form $(1,a,\pm 1)$, none of which live in one of the eigenspaces of $A$. But $AB-BA$ has first row and first column all zero, so $(1,0,0)$ is an eigenvector with eigenvalue zero.
Here is another counterexample. Let $n&gt;2$, the underlying field is real and

$A=\operatorname{diag}(n,\,n-1,\,\ldots,\,1)$,
$T$ be the skew-symmetric Toeplitz matrix whose first row is $(0,\,1,\,\frac12,\,\ldots,\,\frac1{n-1})$,
$C$ be any skew-symmetric matrix with a nonzero off-diagonal entry on every column when $n$ is odd, or any hollow, singular symmetric matrix with a nonzero off-diagonal entry on every column when $n$ is even. E.g. when $n$ is even, one may take $C=\pmatrix{S&amp;S\\ S&amp;S}$ where $S$ is any hollow symmetric matrix with a nonzero off-diagonal entry on every column,
$B=T\circ C$, the Hadamard product of $T$ and $C$.

Note that $C$ is singular and $B$ is is symmetric when $n$ is odd or skew-symmetric when $n$ is even. Moreover, $(AB-BA)_{ij}=(a_i-a_j)t_{ij}c_{ij}=c_{ij}$. Hence $A,B$ are diagonalisable and $AB-BA=C$ is singular. However, by construction, every column of $B$ possesses a nonzero off-diagonal entry. Hence $A$ and $B$ do not share any eigenvector.