# Finding the eigenvalues of a linear transformation which takes inputs from the set of all $n\times n$ matrices.

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 $$B$$.

• 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. $\qquad$ – Michael Hardy Oct 4 '18 at 3:41
• 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. – user593746 Oct 4 '18 at 13:39
• 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. – user593746 Oct 4 '18 at 13:39
• 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? – 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.

• 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. – Saad Oct 4 '18 at 6:59
• 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. – P. Quinton Oct 4 '18 at 7:06
• 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? – Saad Oct 4 '18 at 7:09
• 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 – P. Quinton Oct 4 '18 at 8:06
• 'distinct' does not mean much for eigenvectors, it could be that $u_{A_1}=-u_{A_2}$, linear independence is the better notion – 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 $$\pmatrix{1&0\\0&0}$$)

• 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 – 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.

https://en.wikipedia.org/wiki/Kronecker_product

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

$$diag(s_{1,1}I,\cdots,s_{n,n}I)-diag(T,\cdots,T)$$.

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:

$$diag(s_{1,1},s_{1,1},s_{1,1},s_{2,2},s_{2,2},s_{2,2})-diag(t_{1,1},t_{2,2},t_{3,3},t_{1,1},t_{2,2},t_{3,3})$$.