Let $A$ be a real $n \times n$ symmetric matrix with distinct eigenvalues (that is, $A$ has no repeated eigenvalue). Let $B$ be a real $n \times n$ matrix that commutes with $A.$ Then show that the eigenvalues of $B$ are all real.
I couldn't quite able to do this problem. Could anyone please give me some small hint? I started with some like that $:$
Let $\lambda$ be an eigenvalue of $B$ corresponding to an eigenvector $x$ then we have $$\overline {\lambda}\ \|x\|^2 = \langle Bx, x \rangle = \langle x, B^t x \rangle.$$ But since $A$ is symmetric and $B$ commutes with $A$ hence so is $B^t.$ Using commutativity of $A$ and $B^t$ one can easily show that if $x$ is eigenvector of $A$ corresponding to some eigenvalue $\mu$ then $B^t x$ is also an eigenvector of $A$ corresponding to the same eigenvalue. But eigenspaces of $A$ are all one dimensional and hence this proves that $x$ is also an eigenvector of $B^t.$ Now if we can show that $\lambda$ is an eigenvalue of $B^t$ corresponding to the eigenvector $x$ we are through. This is where I got stuck.
Thanks a bunch.
Best Answer
Hint: By the spectral theorem, all eigenvalues of $A$ are real. Let $x \in \Bbb R^n$ be an eigenvector of $A$ with associated eigenvalue $\lambda$. Show that $x$ must also be an eigenvector of $B$. You might find it helpful to separately consider the cases where $\lambda \neq 0$ and $\lambda = 0$.