If $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{n \times n}$ are two symmetrical matrices with same eigenvectors then AB =BA.

Question

I have the following statement "If $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{n \times n}$ are two symmetrical matrices with same eigenvectors then $AB = BA$." (#)

I know that $AB = BA$ => same eigenvectors is a false statement because we can easily find a counterexample. But what's with this (#) statement? A proof is not necessary.

Answer 1

Hint There exists orthogonal matrix $P$ and diagonal $D_1, D_2$ such that $$A=PD_1P^T\\B=PD_2P^T$$

Note The claim still holds if $A,B$ are diagonalisable, $A=PD_1P^{-1}\\B=PD_2P^{-1}$. Symmetric is only used to deduce that the matrices are (orthogonally) diagonalisable.

Answer 2

If they are symmetrical, they can be diagonalized in an orthonormal eigenvector basis but if they share the same eigenvectors then the basis is the same for both.

Take $v$ from the eigenvector basis, suppose its eigenvalue is $\lambda$ for $A$ and $\mu$ for $B$. then $ABv=A\mu v= \lambda \mu v$ and $BAv=B\lambda v=\mu \lambda v$ which shows that they commute for every vector in a basis of the space, therefore they commute in the whole space.