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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.

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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.

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  • $\begingroup$ This is just the first example I came up with: there are bound to be many more of this sort. In the $2\times2$ case one can check there are no such examples. $\endgroup$ – Chappers Jul 15 '17 at 21:27
  • $\begingroup$ Is there some simple intuition behind this example? In my example the point is to set up $AB$ and $BA$ to send a given vector to another given vector without fixing an invariant subspace for either (so that no eigenvectors are determined by my setup). You seem to have done something slightly different. $\endgroup$ – Ian Jul 16 '17 at 2:37
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    $\begingroup$ @Ian I calculated $\det{[A,B]}$ for diagonal $A$ and a general $B$ to find out what the condition for having a zero eigenvalue was. It turned out it could be factored to $(d_1-d_2)(d_2-d_3)(d_3-d_1)(b_{12}b_{23}b_{31}-b_{13}b_{32}b_{21})$. So we can force a zero eigenvalue by setting two diagonal elements equal, and then any $B$ with no eigenvectors in the eigenspace of the multiple eigenvalue of $A$ will work. $\endgroup$ – Chappers Jul 16 '17 at 10:15
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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.

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Here is another counterexample. Let $n>2$ 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 singular matrix with a zero diagonal and also a nonzero off-diagonal entry in its every column; e.g. when $n$ is odd, one may take any skew-symmetric matrix with nonzero off-diagonal entries; when $n$ is even, one may take $C=\pmatrix{J-I&J-I\\ J-I&J-I}$, where $J$ is the all-one matrix of size $n/2$.
  • $B=T\circ C$, the Hadamard product of $T$ and $C$.

Then $AB-BA=C$ is singular, but by construction, every column of $B$ possesses a nonzero off-diagonal entry. Hence $A$ and $B$ do not share any eigenvector.

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