Proving that $A=A(B−I)$ and $B=B(A−I)$ implies $A$ is diagonalizable.

Question

Let $A, B \in K^{n \times n}$.

I found that $2$ and $0$ are the eigenvalues of $A$. Then I tried to prove that $rank(2I-A)=rank(A)$ if so the $nul(A) + nul(2I-A) = n$ and that means $A$ is diagonalizable. But I cant prove it. Does anyone have any ideas ?

Answer 1

The assumptions can be rewritten as $AB=2A$ and $BA=2B.$ Therefore $Bx=0$ iff $Ax=0.$ Hence $\dim \ker A=\dim \ker B.$ The identity $B(A-2I)=0$ implies that $$\dim \ker B\ge \dim{\rm Im} (A-2I)=n-\dim\ker(A-2I)$$ Hence $$\dim\ker A+\dim\ker (A-2I)\ge n$$ Therefore the whole space can be decomposed into the direct sum of $\ker A$ and $\ker (A-2I).$

Answer 2

If $x\in\ker(A-2I)$ then $Ax=2x$ so $$2x=A(B-I)x=ABx-2x\quad\Rightarrow\quad x=A(\tfrac14Bx)\in{\rm im}(A)\ ,$$ so $\ker(A-2I)\subseteq{\rm im}(A)$. Conversely, your equations imply $A^2=2A$, so if $x\in{\rm im}(A)$ then $$(A-2I)x=(A-2I)Av=(A^2-2A)v=0\ .$$ Thus $\ker(A-2I)={\rm im}(A)$.