Assume $S$ and $T$ are diagonalizable maps on $\mathbb{R}^n$ such that $S\circ T$=$T \circ S$. Then $S$ and $T$ have a common eigenvector.
I already have proof, but I just need validation in one part.
My proof:
Let $F$ be an eigenvector of $T$. This means $\exists \; \lambda \in R$ such that $T(v)=\lambda v$. Then, using the fact that $S\circ T$=$T \circ S$, we have
$$ S(T(v)) = (S\circ T)(v)=(T \circ S)(v)=T(S(v)) \Longrightarrow T(S(v))=\lambda S(v)$$
Thus, $S(v)$ is also an eigenvector of $T$. So, $S$ maps eigenvectors of $T$ to eigenvevtors of $T$. Thus, $S$ must have an eigenvector of $T$.
How would one rigorously prove that if $S$ maps eigenvectors of $T$ to eigenvectors of $T$, then $S$ also has an eigenvector of $T$?
Thanks.
Best Answer
You've shown that the eigenspaces of $T$ are invariant under $S$. If $E_\lambda$ is the $\lambda$-eigenspace of $T$ inside $\mathbb{R}^n$, then it makes sense to speak of $S'=S|_{E_\lambda}: E_\lambda \rightarrow E_\lambda$. Then the key fact is that the characteristic polynomial $p'(T)$ of $S'$ is a factor of the characteristic polynomial $p(T)$ of $S$. Since $p(T)$ splits completely, so does $p'(T)$. In particular, $p'(T)$ has a root, corresponding to an eigenvalue of $S'$. That means $S'$ has an eigenvector in $E_\lambda$, but an eigenvector of $S'$ is simultaneously an eigenvector of $S$ and $T$.
To see why the characteristic polynomial of $S'$ divides the characteristic polynomial of $S$, note that $\mathbb{R}^n = E_\lambda \oplus \hat{E_\lambda}$, where $\hat{E_\lambda}$ is the direct sum of all other eigenspaces $E_\mu$ of $T$. Taking a basis of eigenvectors for $T$, and writing the matrix of $S$ with respect to this basis, we see that the characteristic polynomial of $S$ is equal to the characteristic polynomial of $S|_{E_\lambda}$ times the characteristic polynomial of $S|_{\hat{E_\lambda}}$.