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.
 A: Clearly we can pick a basis so that one of the operators is diagonal already. Here's a small counterexample of this sort:
$$ A = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu \end{pmatrix}  \qquad B = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}. $$
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.
A: Take $A,B$ to be $4 \times 4$ matrices such that the first column of $A$ is $e_2$, the last column of $A$ is $e_3$, the second column of $B$ is $e_3$, and the first column of $B$ is $e_4$. Then 
$$(AB-BA)e_1=ABe_1-BAe_1=Ae_4-Be_2=e_3-e_3=0.$$
(The point of the choices was to create two "parallel paths" from $1$ to $3$, namely $1 \to 4 \to 3$ and $1 \to 2 \to 3$.)
Notice though that the choices we made here did not create an invariant subspace yet: $A$ sends the span of $\{ e_1,e_4 \}$ to the span of $\{ e_2,e_3 \}$ and $B$ sends the span of $\{ e_1,e_2 \}$ to the span of $\{ e_3,e_4 \}$. So these choices do not uniquely specify any eigenvectors of either matrix. Thus taking the other two unspecified columns of $A$ and those of $B$ at random will typically give all different eigenvectors.
A: Here is another counterexample. Let $n>2$, 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 $C=\pmatrix{S&S\\ S&S}$ 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.
