The problem goes like this:
Suppose $A, B\in M_n(\mathbb{C})$ and $AB-BA=c(A-B)$ with nonzero $c\in \mathbb C$. Then $A, B$ are simultaneously triangularizable and those upper triangular matrices coincide on the main diagonal.
I've proved that the characteristic polynomials of $A$ and $B$ are identical (by using the identity $(xI-A)((x+c)I-B)=(xI-B)((x+c)I-A)$); however, I am stumped at constructing an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both upper triangular.
Best Answer
It suffices to show that $A$ and $B$ share some eigenpair. The result then follows from recursion.
Since $B$ has only $n$ eigenvalues, it has an eigenpair $(\lambda,x)$ such that $\lambda-c$ is not an eigenvalue of $B$. Let $A_1=A-\lambda I$ and $B_1=B-\lambda I$. Then $B_1x=0$ and $A_1B_1-B_1A_1=c(A_1-B_1)$. Right-multiply both sides by $x$, we obtain $-B_1A_1x=cA_1x$ or equivalently, $(B_1+cI)A_1x=0$. Since $\lambda-c$ is not an eigenvalue of $B$, we see that $-c$ is not an eigenvalue of $B_1$. Hence $B_1+cI$ is nonsingular, $A_1x=0$ and $(\lambda,x)$ is also an eigenpair of $A$.