# Invertibility of Linear Transformation

Let $A$ and $B$ be two $n\times n$ matrices that have no eigenvalues in common. Show that the transformation $T$ that maps the $n\times n$ matrices, $M_n$, to the $M_n$ and is defined by the formula $$T(S):=AS-SB$$ is invertible.

• I think the eigenvalues of $T$ are of the form $\lambda_i-\mu_j$ where $\lambda_i$ and $\mu_j$ are the eigenvalues of $A$ and $B$ respectively. I forget how to show this statement, but assuming it is true, it ensures that $T$ has no zero eigenvalues, i.e., it is invertible. – angryavian Nov 30 '14 at 3:13
• @angryavian Any hints on how to show this? – Dia McThrees Nov 30 '14 at 3:20

If $AS = SB$ and $u$ is an eigenvector of $B$ for eigenvalue $\lambda$, then $A S u = S B u = \lambda S u$. Since $A$ and $B$ have no eigenvalues in common, we must have $S u = 0$. That would settle matters in the case where $B$ is diagonalizable.

More generally, you have to consider generalized eigenvectors. Prove by induction on $k$ that $S v = 0$ for any $v$ such that $(B - \lambda I)^k v = 0$.

• How do we know that $AS=SB$? – Dia McThrees Nov 30 '14 at 15:36
• $T$ is a linear map from a finite dimensional vector space to itself. So $T$ is invertible iff $ker\ T = 0$. So if we can show that for any $S \in ker\ T$ we have $S = 0,$ then $T$ is invertible. So we pick an arbitrary $S \in ker\ T.$ Then we have $TS = 0,$ or equivalently $AS = SB.$ Now continue with the answer of Robert Israel. – jflipp Nov 30 '14 at 22:14
• @jflipp Proof by induction doesn't make sense... If $(B-\lambda I)^nv=0$, then we have to show $(B-\lambda I)^{n+1}v=0$. Since it's true for $n$, we get $(B-\lambda I)(B-\lambda I)^nv=0$, then $(B-\lambda I)=0$, which is obviously not true. – Dia McThrees Nov 30 '14 at 22:33
• No, that's not how the induction goes. The statement is that if $(B-\lambda I)^k v = 0$ then $S v = 0$. For the induction step, you assume that $(B - \lambda I)^k v = 0$ implies $S v = 0$, and you consider $u$ such that $(B - \lambda I)^{k+1} u = 0$. – Robert Israel Nov 30 '14 at 23:35
• @RobertIsrael Ohhh! Thank you! I've been banging my head on this for a long time. – Dia McThrees Nov 30 '14 at 23:37

As I mentioned in a comment to the answer by Robert Israel, one must take the hypothesis on $A,B$ to mean that even over an extension of the (unspecified) field $K$ over which $T$ is a linear operator (i.e., for which it acts on $M_n(K)$) they have no common eigenvalue. With the more strict interpretation of "no common eigenvalues in $K$" there are easy counterexamples; for instance take $K=\Bbb R$, $n=2$, $A$ a rotation matrix (other than $\pm I_2$), and $B=A^{-1}$ (so there are no eigenvalues in$~K$ at all, let alone common eigenvalues), one checks that all orthogonal reflection matrices$~S$ are in the kernel of$~T$ (because $SAS^{-1}=B$).

This suggests that a different formulation of that hypothesis might allow a proof that does not need to extend the field$~K$. Indeed one may replace the hypothesis by the existence of relatively prime annihilating polynomials: $P_A,P_B\in K[X]$ such that $P_A[A]=0=P_B[B]$. This clearly excludes the existence of a common eigenvalue$~\lambda$ in$~\overline K$, since that would make $X-\lambda$ a non-existing common factor in $\overline K[X]$ of both $P_A$ and $P_B$, while conversely in the absence of such common eigenvalues one can take $P_A=\chi_A$ and $P_B=\chi_B$ (or to avoid needing the Cayley-Hamilton theorem, one could also take their minimal polynomials instead of characteristic polynomials).

So here is the proof that $T(S)=0$ implies $S=0$. Iterating the hypothesis $AS=SB$ easily gives $A^kS=SB^k$ for all $k\in\Bbb N$, and by linearity $P[A]S=SP[B]$ for all $P\in K[X]$. In particular $$0=0S=P_A[A]S=SP_A[B].$$ Now the given fact that $P_A$ and $P_B$ are relatively prime means that $P_B$ has an inverse modulo$~P_A$, namely the second Bezout coefficient $V$ in a Bezout relation $\gcd(P_A,P_B)=1=UP_A+VP_B$. Therefore $P_B[A]V[A]=I$, and right-multiplying the equation above by $V[A]$ gives $S=0$, as desired.