Show that if $TS=ST$ then $\operatorname{range}{S}$ and $\ker{S}$ are invariant under $T$. Is the converse true?

Question

Let $V$ be a finite-dimensional vector space and $S$, $T$ $\in \mathcal{L}(V)$ be operators such that $ST=TS$. It is rather straightforward to show that both $\newcommand{\im}{\operatorname{range }} \im{S}$ (respectively $\newcommand{\im}{\operatorname{range }} \im{T}$) and $\ker{S}$ (respectively $\ker{T}$) are invariant under $T$ (respectively $S$).

1. Let $s \in \ker{S}$. Then $0 = T(0) = TS(s) = ST(s) \implies T(S) \in \ker{S}$
2. Let $v \in V$. Then $TS(v)=ST(v) \in \newcommand{\im}{\operatorname{range }} \im{S}$

Is the converse true? Namely if $S \in \mathcal{L}(V)$ and $T \in \mathcal{L}(V)$ are operators such that $\newcommand{\im}{\operatorname{range }} \im{S}$ is invariant under $T$, $\newcommand{\im}{\operatorname{range }} \im{T}$ is invariant under $S$, $\ker{S}$ is invariant under $T$ ad $\ker{T}$ is invariant under $S$, does it follow $ST=TS$? If not, what about a counterexample?

Answer 1

Let $T\in\mathcal L(\mathbb R^2)$ be represented by the matrix $\pmatrix{1&2\\0&1}$ and $S$ by $\pmatrix{1&0\\2&1}$. Then the kernel of both $T$ and $S$ is $\{0\}$ and the range is $\mathbb R^2$. But $T$ and $S$ do not commute.

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Hope this helps.

Answer 2

No, for example two anti-commutative matrices $A,B$ with $AB=-BA$ satisfy your criterion. E.g. consider.

$$ A=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \qquad B=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$