I am working on the below problem:
Suppose $A$ and $B$ are linear transformations from $\mathbb{R}^n$ to itself, each having $n$ distinct real eigenvalues. Show that $A$ and $B$ are simultaneously diagonalizable.
Here are my thoughts so far:
First, every linear transformation from $\mathbb{R}^n$ to itself corresponds to an $n \times n$ matrix with entries in $\mathbb{R}$. Thus, I will proceed with matrix language. Two $n \times n$ matrices $A,B$ are simultaneously diagonalizable if there is a nonsingular matrix $S$ such that $S^{-1}AS$, $S^{-1}BS$ are diagonal matrices.
Further, it's not too difficult to show that the $A,B$ given in the problem are diagonalizable, since the characteristic polynomial (and therefore, the minimal polynomial) will split as a product of distinct linear factors over $\mathbb{R}$. Thus, I believe that my problem reduces down to showing that $A$ and $B$ share the same eigenvectors. This is where I'm not sure how to proceed.
There is a variation of this problem that deals with commuting matrices $A,B$, but we're not offered that luxury here.
How can I show that $A$ and $B$ have the same eigenvectors ? To this point, could there be a typo -- that is, $A$ and $B$ should both have the same $n$ distinct eigenvalues in order to prove the result? Would that help me ?
This is a problem on an old Algebra preliminary exam from my department that I was working on.
Thanks!