Question:
Let $A,B$ be $n\times n$ complex matrices, and let $a$ and $b$ be complex numbers such that $$AB-BA=aA+bB.$$
Show that $A$ and $B$ are simultaneously diagonalizable.
My Try:
Case 1: If $a=b=0$, then $AB=BA$. This is true. [Editor's note: In this case, the result is well known.]
Case 2: Without loss of generality, we let $a\neq 0$, so we may assume $a=1$ (let $\dfrac{1}{a}B$ replace $B$).
Let $C=AB-BA=A+bB$, then $$CB-BC=C$$ and I can't continue.
Thank you very much!
Best Answer
Edit: The statement is false. Here is a counterexample: $$ A=\pmatrix{1&0\\ 1&2},\ B=\pmatrix{1&0\\ 0&2},\ AB-BA=\pmatrix{0&0\\ -1&0}=B-A\ne0. $$ Both $A$ and $B$ here are diagonalisable. However, since $A$ and $B$ do not commute, they are not simultaneously diagonalisable.
However, under the assumption that $A,B$ and $AB-BA$ are diagonalisable, the statement is true. In this case, your $C$ is diagonalisable and it suffices to show that $C=0$. Without loss of generality, suppose $B$ is a diagonal matrix of the form $(\lambda_1I_{n_1})\oplus\cdots\oplus(\lambda_k I_{n_k})$, where $\lambda_1,\ldots,\lambda_k$ are distinct and they have increasing real parts. Then $CB−BC=C$ implies that with a conforming partition to $B$, $C$ is a block strictly upper triangular matrix and hence $C$ is nilpotent. Therefore $C$ is a diagonalisable nilpotent matrix and it must be zero.