We define $T(X) = AX - XB$ for fixed $A,B$. We allow $X$ to be any matrix in $M_n(F)$.

Write down all the eigenvalues of $T$ in terms of the eigenvalues of $A$ and $B$.

I think I saw another question here which said that for $T(X) = AX - XA$, if $u$ is an eigenvector for $A$ and $v$ an eigenvector for $A^T$, then $uv^T$ is an eigenvector for $T$, but I don't know how to prove this nor do I know if it generalizes if we replace one instance of $A$ with $B$.

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    $\begingroup$ If you use the letter x, outside of MathJax, rather than \times within MathJax, then it will look like $n\text{ x }n$ instead of $n\times n.$ I edited accordingly. $\qquad$ $\endgroup$ – Michael Hardy Oct 4 '18 at 3:41
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    $\begingroup$ If $A$ and $B$ are both diagonalizable over $F$, then P. Quinton gave a complete answer. But when one of $A$ and $B$ is not, this problem becomes quite a tricky business. $\endgroup$ – user593746 Oct 4 '18 at 13:39
  • $\begingroup$ At least, when $A=B$ is not diagonalizable, $T$ is not diagonalizable. You will have to deal with generalized eigenvalues of $T$, and I am not sure how to do so. But if $F=\mathbb{R}$ or $F=\mathbb{C}$, you can employ continuity to show that the generalized eigenvalues of $T$ are of the form $\lambda_i-\mu_j$, for $i=1,2,\ldots,n$ and $j=1,2,\ldots,n$, given that $\lambda_1,\lambda_2,\ldots,\lambda_n$ and $\mu_1,\mu_2,\ldots,\mu_n$ are generalized eigenvalues of $A$ and $B$, respectively. $\endgroup$ – user593746 Oct 4 '18 at 13:39
  • $\begingroup$ I'm not familiar with arguments involving continuity when it comes to matrices. Is there another approach if we work in the complex field? Perhaps one which involves calculating $T$ on a nice basis? $\endgroup$ – Saad Oct 4 '18 at 16:26

Suppose that $\lambda_A$ and $u_A$ is an eigenvalue/eigenvector pair of $A$ and $\lambda_B$ and $u_B$ is an eigenvalue/eigenvector pair of $B^T$, then

\begin{align*} T(u_A u_B^T) &= A u_A u_B^T - u_A u_B^T B\\ &=(A u_A) u_B^T - u_A (B^T u_B)^T\\ &=\lambda_A u_A u_B^T - \lambda_B u_A u_B^T\\ &=(\lambda_A - \lambda_B) u_A u_B^T \end{align*}

which means that $\lambda_A-\lambda_B$ and $u_A u_B^T$ form an eigenvalue/eigenvector pair of $T$.

Now if the matrices $A$ and $B^T$ have exactly $n$ eigenvalues/eigenvectors pairs which means that we can have $n^2$ eigenvalues/eigenvectors pairs for $T$ this way (see below), but $T$ cannot have more of those since $T:M_n(F)\rightarrow M_n(F)$ is a transformation from an $n^2$ dimensional space to an $n^2$ dimensional space.

Let $u_{A_1}$ and $u_{A_2}$ be two linearly independents normalized eigenvectors of $A$ and let $u_{B_1}$ and $u_{B_2}$ be two eigenvectors of $B^T$ (they could be the same), then suppose that $u_{A_1} u_{B_1}^T=u_{A_2} u_{B_2}^T$, so we have \begin{align*} u_{A_1} = (u_{A_1} u_{B_1}^T) u_{B_1} = (u_{A_2} u_{B_2}^T) u_{B_1}=(u_{B_2}^T u_{B_1}) u_{A_2} \end{align*} and so $u_{A_1}$ and $u_{A_2}$ are collinear, since they are normalized they are equal which proves that $u_{A_1} u_{B_1}^T\neq u_{A_2} u_{B_2}^T$.

The same argument can be given for $u_{B_1}\neq u_{B_2}$ and $u_{A_1}$ and $u_{A_2}$ arbitrary.

This means that all the eigenvectors of this form are distinct.

  • $\begingroup$ Thanks. How would we show that all the eigenvalues of $T$ are of this form? Is it not possible for us to have an eigenvalue of $T$ which can be expressed as the difference of two distinct pairs of eigenvalues of $A$ & $B$? For example, we could have $\lambda_{A1}-\lambda_{B1}$ = $\lambda_{A2}-\lambda_{B2}$ and then we don't get $n^{2}$ eigenvalues for $T$, so it is possible for some to not be in the form mentioned. $\endgroup$ – Saad Oct 4 '18 at 6:59
  • $\begingroup$ in this case you will get $u_{A_1} u_{B_1}^T \neq u_{A_2} u_{B_2}^T$, which means that the eigenvalue will just have some multiplicity. $\endgroup$ – P. Quinton Oct 4 '18 at 7:06
  • $\begingroup$ Sorry, I still don't follow why there can be no other eigenvalues. The way I see it, there are at most $n^2$ eigenvalues of the form mentioned, so it is possible that there are, say only $n^{2} - 1$ eigenvalues which are the difference of an eigenvalue of $A$ and $B$, so there could be another eigenvalue which isn't of our form. How do we show there isn't? $\endgroup$ – Saad Oct 4 '18 at 7:09
  • $\begingroup$ Good point, see my edit, it proves that if $A$ and $B^T$ have each $n$ distinct eigenvectors then $T$ have $n^2$ eigenvectors. This fact depend a bit on what field you are using I think but in $\mathbb C$ it should be fine $\endgroup$ – P. Quinton Oct 4 '18 at 8:06
  • $\begingroup$ 'distinct' does not mean much for eigenvectors, it could be that $u_{A_1}=-u_{A_2}$, linear independence is the better notion $\endgroup$ – daw Oct 4 '18 at 13:49

The converse of the claim is not true, i.e., not every eigenvector of $T$ is of the from $v_Av_B^T$. To see this, take $A=B=\pmatrix{0&1\\0&0}$. Then $$ T\pmatrix{x_{11} & x_{12}\\x_{21} & x_{22} } = \pmatrix{x_{21}& x_{11}-x_{22}\\0&x_{21}}. $$ It follows that $\lambda=0$ is an eigenvalue with eigenspace spanned by $$ \pmatrix{0&1\\0&0}, \quad \pmatrix{1&0\\0&1}, $$ where the latter eigenvector is not of the particular form (and the eigenspace does not contain $\pmatrix{1&0\\0&0}$)

  • $\begingroup$ The eigenvalue which $T$ is supposed to contain is the first element of our eigenspace because we wanted eigenvectors where the first element is an eigenvector for $A$, and the second one is the transpose of an eigenvector for $B$. Also, this does disprove the converse, but the overall claim is still true as far as the eigenvalues go $\endgroup$ – Saad Oct 4 '18 at 16:21

If you want only the eigenvalues over $\bar{F}$, the algebraic closure of $F$, then it's not difficult. Here, we stack the matrices row by row. cf.


There are invertible $P,Q$ s.t. $P^{-1}AP=S=[s_{i,j}],Q^{-1}B^TQ=T=[t_{i,j}]$ are triangular. Then $(P\otimes Q)^{-1}(A\otimes I-I\otimes B^T)(P\otimes Q)=(P^{-1}AP)\otimes I-I\otimes (Q^{-1}B^TQ)$

$=S\otimes I-I\otimes T=[s_{i,j}I]-[\delta_{i,j}T]$. The last two matrices are triangular and the diagonal of the matrix result is


Finally, the required spectrum is $(\lambda_i-\mu_j)_{i,j}$. Moreover, the result is valid when $A\in M_n,B\in M_m$ ($m\not= n$). For example, for $n=2,m=3$, the diagonal is:



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