Let $V$ be finite-dimensional and suppose $S$ and $T$ are diagonalizable linear operators on $V$. Moreover, suppose $v$ is an eigenvector of $T$ if and only if $v$ is an eigenvector of $S$. Show that $ST=TS$.
I think my approach below is right, but I just wonder why both $S$ and $T$ are diagonalizable in this problem.
My approach:
Since $S$ is diagonalizable, there exists a list of eigenvectors of
$S$, $v_1, …, v_n$ that forms a basis of $V$.By the given condition, $v_1, …, v_n$ are eigenvectors of $T$ as
well.Thus, the following results hold. $$Sv_1= \lambda _1 v_1, Tv_1=
{\lambda}'_1 v_1, …, Sv_n= \lambda _n v_n, Tv_n= {\lambda}'_n v_n.$$Also, since $v_1, …, v_n$ is a basis of $V$, $\forall v \in V$,
$v=a_1v_1+…+a_nv_n$ for some $a_1, …, a_n \in \mathbb{F}$.Thus, $\forall v$, $$STv=a_1 \lambda _1 {\lambda}'_1 v_1 + … + a_n
\lambda _n {\lambda}'_n v_n=TSv$$ This implies, $ST=TS$.
$\blacksquare$
However, as I mentioned in the beginning, I wonder why both linear operators are diagonalizable in this problem, while I only used that $S$ is diagonalizable.
Best Answer
Formally, you're right. The weaker hypothesis $S $ diagonalizable together with $v$ is an eigenvector of $S$ if and only if it is an eigenvector of $T$ would be sufficient.
However, those two hypothesis immediately imply that $T $ is also diagonalizable as you proved it.