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Let $A$ be an $n \times n$ complex matrix and let $T(M) = AM-MA$. I need to determine the eigenvalues of $T$ in terms of those of $A$.

This was an exercise from Artin, and I was not being able to think how to proceed. I couldnot find any proper answer to this on Stackexchange even though I saw this questioned before.

Can you help me? Seems like a good problem. It was listed under the section "Using coninuity" in the chapter "Applications of Linear Operators".

  • Are you sure you don't mean $T(M) = AM - MA$? Just curious! Cheers! – Robert Lewis Jun 05 '15 at 18:13
  • I think what Robert is saying seems to be correct. – Anurag A Jun 05 '15 at 18:33
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    It's helpful to have the notation $[A, B] = AB - BA$, so that you can write the operator you're interested in as $[A, -]$. For starters, can you solve this problem when $A$ is diagonal? – Qiaochu Yuan Jun 05 '15 at 18:38
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    Yes, its $T(M)$, sorry. @Qiaochu, I was unable to do it, when $A$ was diagonal. I was getting $T(M)=((a_{ii} - a_{jj})m_{ij})$ but putting it to be equal to $tM$ gives only $t = 0$ from diagonal entries. –  Jun 05 '15 at 18:51
  • Unless $m_{ii}=0$ for all $i$. In which case, I think we need a common value from $(a_{ii}-a_{jj})$. –  Jun 05 '15 at 18:55
  • Well, I think I have an answer: Are the eigenvalues $(a_{ii}-a_{jj})= \lambda_i-\lambda_j$ for all $i,j$, by taking appropriate $M$? Now how do I take this to the general case? $A$ may not be diagonalizable. –  Jun 05 '15 at 19:00
  • I know that diagonalizable matrices are dense in the set of all complex matrices. So somehow if I prove this for diagonalizable matrices $A$, I will be finished by continuity. But how does assuming $A$ diagonal help here? In other words, if this holds for diagonal matrices, how do I conclude this for diagonalizable matrices? –  Jun 05 '15 at 19:06
  • think about the properties of the trace function – Justine Jun 05 '15 at 19:29
  • Well, the trace of any matrix in the image of $T$ is $0$. –  Jun 05 '15 at 19:33
  • @Qiaochu, I got it. Thanks. –  Jun 05 '15 at 20:05
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    @Paramjit: the assignment $A \mapsto [A, -]$ respects conjugation. You also do not need to use a continuity argument: instead of diagonalizing $A$ you can upper-triangularize it and the argument doesn't change much, although you need to think in terms of generalized eigenvectors rather than eigenvectors. – Qiaochu Yuan Jun 05 '15 at 20:39
  • Okay, I will think about that approach. But it was unclear why on the right side conjugation of $[A,_]$ means anything. I mean this would be a map on the space of complex matrices and conjugation by a matrix doesnot mean anything. Am I missing something? –  Jun 05 '15 at 22:01
  • Possible duplicate for this – hamid kamali Jun 06 '15 at 10:46
  • Is this question what you're looking for? – Armadillo Jim Jul 20 '15 at 23:31

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