If $A$ and $B$ are square symmetric matrices and have same eigenvectors, do they commute?

Question

If $A$ and $B$ are square symmetric matrices and have same eigenvectors, do they commute?

How can I prove this? I have searched a lot but couldn't find the answer.

Any help regarding way forward?

Answer 1

You need to be a bit careful with the phrase "have the same eigenvectors". What I think you mean is the following:

There is a basis $\{v_1,v_2,\dots,v_n\}$ of $\Bbb R^n$ such that each vector $v_i$ is an eigenvector of both $A$ and $B$

If this is what you mean, then it's easy to show that $A$ and $B$ commute. First, show that $AB(v_i) = BA(v_i)$ for each $i = 1,\dots,n$. With that, we can state that by linearity, $$ AB\left(\sum_{i=1}^n c_i v_i\right) = \sum_{i=1}^n c_i ABv_i = \sum_{i=1}^n c_i BAv_i = BA\left(\sum_{i=1}^n c_i v_i\right) $$ where $c_1,\dots,c_n$ are arbitrary real coefficients. It follows that $AB = BA$.

Answer 2

Another way you might prove this is to say:

There is a basis $\{v_1,v_2,\dots,v_n\}$ of $\Bbb R^n$ such that each vector $v_i$ is an eigenvector of both $A$ and $B.$

Define a transition matrix $$P=\left(\begin{array}{c|c|c} v_1 & \cdots & v_n\end{array}\right).$$

Then $P^{-1}AP =D_A$ and $P^{-1} BP =D_B.$ You can write

$$P^{-1}APP^{-1} BP=\underbrace{D_AD_B=D_BD_A}_{(1)}=P^{-1}BPP^{-1}AP.$$

Where $(1)$ is true since diagonal matrices commute with other diagonal matrices. So we have shown that $P^{-1}APP^{-1} BP=P^{-1}BPP^{-1}AP$ and after canceling $P^{-1}P$ we get $$P^{-1}A BP=P^{-1}BAP$$ and so $$A B=BA,$$ as desired.