We were asked to prove the following:
Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.
It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.
Thanks
Best Answer
From a comment:
If we see that this is true, then we will be able to conclude that $B$ is diagonalized by a basis of eigenvectors for $A$.
It is true because the eigenspace of $A$ for the eigenvalue $\lambda$, $\{x:Ax=\lambda x\}$, is one dimensional by the hypothesis that $A$ has $n$ distinct eigenvalues. The equation $A(Bv)=\lambda Bv$ says that $Bv$ is in this space, which is spanned by $v$. Therefore there exists a scalar $c$ such that $Bv=cv$.