I'm asked to prove if $A, B$ are real symmetric matrices that commute then they share a common eigenvector.
My approach: Since $A$ is real symmetric, it has an orthonormal eigenbasis. Now if all of these vectors are in the kernel of $B$ then $B$ has to be $0$ matrix. So WLOG lets take an eigenvalue $\lambda$ of $A$ and $W$ be the corresponding eigenspace for it, which is non trivial under $B$. As $B$ commutes with $A$, $A(Bv)= BA(v)=\lambda Bv$ amd hence $Bv \in W$ for any $v \in W$ so $B(W) \subset W$.
Now let us take the restriction of $B$ to $W$. We can now find a non zero eigenvector (perhaps with some other eigenvalue) $w'$ of $B$ in $W$ since $W$ does not lie in the kernel of $B$. Then this $w'$ is a common eigenvector for $A$ and $B$.
Is this proof correct? Did I use the symmetric information correctly? I don't think we needed both $A$ and $B$ to be symmetric here, since we are using that fact for only $A$.
Best Answer
Your proof is incomplete as you didn't show why $B$ has an eigenvector in $W$. If you let $A$ be the identity matrix then you've shown that all real matrices have real eigenvectors. This is not true. For instance, the nontrivial rotation matrices have no real eigenvectors.
You need to use the fact that $B$ is symmetric to prove that $B|_W$ has an eigenvector. Indeed, you can just show that $B|_W$ is symmetric. Write $B$ in the orthonormal eigenbasis you specified at the start. Since it's orthonormal, this will still be symmetric. Without loss of generality put the eigenvectors in $W$ at the start of this basis. Then as $B[W] \subseteq W$, $B$ will contain a top left block corresponding to $B|_W$. Hence, $B|_W$ is symmetric so it has a real eigenvector.