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Maybe it's simple but I can't see the solution of this problem (Russell Merris, Multilinear Algebra, CRC Press, 1997, chapter 6, p.202, exercise 4):

Let $\lambda_1,\ldots,\lambda_p$ be the eigenvalues of $A\in\mathbb C_{p,p}$ (multiplicities included), and $\omega_1,\ldots,\omega_q$ be the eigenvalues of $B\in\mathbb C_{q,q}$ respectively. Find the eigenvalues of

a. $A \otimes B - B \otimes A$.

b. $A \otimes B + B \otimes A$.

From chapter 5, I know the eigenvalues of $A\otimes B$ and $A\otimes I_q + I_p \otimes B$:

  1. The eigenvalues of $A \otimes B$ are $\lambda_i \cdot \omega_j$, $1 \leq i \leq p$, $1 \leq j \leq q$

  2. The eigenvalues of $A\otimes I_q + I_p \otimes B$ are $\lambda_i + \omega_j$, $1 \leq i \leq p$, $1 \leq j \leq q$

These facts may give us a decomposition of $A\otimes B \pm B \otimes A$.

This may be very simple but I need a hint.

I also made some Matlab calculations with integer matrices, and I get non-integer/non-real eigenvalues... maybe square roots are involved...

Thanks!

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  • $\begingroup$ What are the dimensions of the matrices $A$ and $B$? Are they the same? What about the dimension of $I$? $\endgroup$
    – Alex Silva
    Jan 7 '15 at 15:29
  • $\begingroup$ Edited the dimensions $\endgroup$
    – Leafar
    Jan 7 '15 at 15:36
  • $\begingroup$ I understand what you are saying. You are right! This is not possible find the desired eigenvalues. I have deleted the answer. $\endgroup$
    – Alex Silva
    Jan 7 '15 at 16:12
  • $\begingroup$ @AlexSilva Thanks for helping ;) $\endgroup$
    – Leafar
    Jan 7 '15 at 16:15
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    $\begingroup$ I think the author has simply left out some additional conditions. From his wording, he seems to imply that the eigenvalues of $A\otimes B\pm B\otimes A$ depend solely on the eigenvalues of $A$ and $B$, but this is not true. E.g. when $A=\pmatrix{1&1\\ 1&1}$ and $B=\pmatrix{0&1\\ 0&0}$, the spectra of $A\otimes B+B \otimes A$ and $A\otimes B-B \otimes A$ are respectively $\{-1,1,0,0\}$ and $\{-i,i,0,0\}$. However, if we replace $B$ by the zero matrix (that has the same spectrum as the original $B$), all eigenvalues of $A\otimes B\pm B \otimes A$ are zero. $\endgroup$
    – user1551
    Jan 10 '15 at 16:25
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I don't have Merris' book, but something seems to be forgotten in the question - in the present form it cannot be answered.

Consider, for instance, diagonal matrices $A$ and $B$ explicitly given by $$A=\operatorname{diag}\left\{\lambda_1,\lambda_2,\lambda_3\right\},\qquad B=\operatorname{diag}\left\{\omega_1,\omega_2\right\},$$ so that \begin{align} &A\otimes B+B\otimes A=\\ =\,&\operatorname{diag}\left\{2\lambda_1\omega_1,\lambda_2\omega_1+\lambda_1\omega_2, (\lambda_2+\lambda_3)\omega_1,(\lambda_1+\lambda_2)\omega_2,\lambda_3\omega_1+\lambda_2\omega_2,2\lambda_3\omega_2\right\}. \end{align} The spectrum of the last matrix is not invariant e.g. with respect to the exchange $\omega_1\leftrightarrow\omega_2$.

Furthermore, even setting $p=q$ as in the book referenced in the comment does not save the situation: again consider diagonal $2\times 2$ matrices $A,B$ and notice that the spectrum of $A\otimes B+ B\otimes A$, given by $\{2\lambda_1\omega_1,\lambda_1\omega_2+\lambda_2\omega_1,\lambda_1\omega_2+\lambda_2\omega_1,2\lambda_2\omega_2\}$ is not invariant w.r.t. the exchange of eigenvalues of one of them.

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