# $AB - BA = \lambda B$ has a nontrivial solution, iff $\lambda = \lambda_i - \lambda_j$? [duplicate]

I have the following question in hand.

If $\lambda_1,\cdots,\lambda_n$ are the eigenvalues of a given matrix $A \in M_n$, then prove that the matrix equation $AB - BA = \lambda B$ has a nontrivial solution $B \neq 0 \in M_n$, if and only if $\lambda = \lambda_i - \lambda_j$ for some $i,j$.

## marked as duplicate by Carsten S, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 2 '18 at 12:57

• In the suggested solution of this link, an assumption is that $E$ is diagonalizable. – user550103 May 26 '18 at 13:39
• No, it says that we may assume - without loss of generality- that $E$ is diagonal. If $E$ is not diagonalizable, we can use Jordan blocks. – Dietrich Burde May 26 '18 at 14:58
• Seeing that n lambdas are given, it is fair to assume that the matrix is supposed to be diagonalizable. Otherwise one needs to assume an algebraically closed field. – Carsten S May 26 '18 at 16:50

Consider the operator $B\mapsto [A,B]$. What are its eigenvalues?

• I think $AB$ and $BA$ should have the same spectrum (or same characteristic polynomial). I guess only ordering of the eigenvalues is different. – user550103 May 26 '18 at 11:35
• @user550103 The question is a different one here, though. Not about the eigenvalues of the commutator, but of the map that takes the bracket with $A$. – Pedro Tamaroff May 26 '18 at 11:45
• I must admit that I feel I have knowledge gap here :(. I am not familiar with this mapping as such. If you can give some reference, then I would be very grateful to you. – user550103 May 26 '18 at 11:52
• Would you mind to review my below attempt? I have tried bit differently (utilizing "the toolset" I have so far). – user550103 May 26 '18 at 14:17

Here is my attempt. Does this make sense to you experts?

If we vectorize such that \begin{align} AB - BA &= \lambda B \\ &\Downarrow \\ \mbox{vec}\left(AB - BA \right) &= \mbox{vec}(\lambda B) \\ \mbox{vec}\left(ABI - IBA \right) &= \mbox{vec}(\lambda B) \\ \left(\left(I \otimes A\right) - \left(A^{\rm T} \otimes I\right)\right)\mbox{vec}(B) &= \lambda \mbox{vec}(B) \\ \end{align}

So, according to Theorem 13.16, the eigenvalues of the Kronecker sum $\left(\left(I \otimes A\right) - \left(A^{\rm T} \otimes I\right)\right)$ would be $\lambda_i - \lambda_j$. Hence, the solution should be non-trivial if and only if $\lambda = \lambda_i - \lambda_j$.

• This seems to be a valid reformulation. If you think about it a bit you will see that the proof of the theorem is not that different from the proof in the answer to the other question. – Carsten S May 26 '18 at 16:57
• This is basically how Bourbaki proves it (Groupes et algebres de Lie, ch. VII par. 2 no. 2 exemple 3), so you should be good. – Torsten Schoeneberg May 28 '18 at 5:51