Prove that if $AB=BA$, ${\rm Ker}(A)={\rm Ker}(B)$, and $A$ has distinct eigenvalues, then $A$ and $B$ have the same eigenvectors

Question

If $A$ and $B$ are two $n\times n$ matrices such that $AB = BA$ and $\operatorname{Nul}(A) = \operatorname{Nul}(B)$, given that $A$ has $n$ distinct eigenvalues with corresponding eigenvectors, how can I show that $A$ and $B$ have the same eigenvectors? In other words, how do I prove that $B$ is diagonalizable? $A$ is diagonalizable.

Answer 1

Let $v$ be an eigenvector of $A$ with $\lambda$ being the eigenvalue.

If $\lambda = 0$, then $Av=0$. Since $Null(A)=Null(B)$, $Bv=0$ as well. Hence $v$ is an eigenvector of $B$.

If $\lambda \neq 0$,

$$A(Bv)=BAv=\lambda Bv$$

That is, if $Bv\neq 0$, then $Bv$ is an eigenvector of $A$ with $\lambda$ being the eigenvalue. Since the eigenvalues of $A$ are distinct, $Bv=kv$. That is $v$ is an eigenvector of $B$.