Show that $B$ is diagonalizable if If $AB=BA$ and $A$ has distinct real eigenvalues

Solution 1:

From a comment:

I still don't understand how it is that if $v$ is an eigenvector of $A$ corresponding to some eigenvalue $\lambda$, that $A(Bv)=\lambda Bv$ implies that $v$ is also an eigenvector of $B$.

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 $Bv=cv$.

Solution 2:

So $A$ is diagonalizable with n distinct elements on its main diagonal, $A=diag$ $(a_1,..., a_n)$. Let $B=(b_{ij})$, do the multiplication for both sides of $AB=BA$ and use component-wise correspondence for matrix equality. Now you see the elements off the main diagonal of B must be zero