In Exercises 3B Q2:
Suppose $V$ is a vector space and $S,T \in \mathcal{L}(V,V)$ are such that
$$\text{range }S \subset \text{null }T$$
Prove $(ST)^2 = 0$
I found the solution here. I understand the solution, but I think of something different when I try the question by myself and want to know any mistake I have made.
When I saw $\text{range }S \subset \text{null }T$, my thought was the dimension of $\text{range }S \leq $ the dimension of $\text{null }T$. I am not sure whether this is correct.
If this is correct, then
$$(ST)^2 = S(T(S(T(u))))$$
Let's look at $T(S(T(u))$, not sure whether this is correct again
$$\text{dim }S(T(u)) =\text{dim range }T + \text{dim null }T$$
$$\text{dim range }S =\text{dim range }T + \text{dim null }T$$
Because $\text{dim range }S \leq \text{dim null }T$, dim range $T$ must be equal to zero, and $\text{dim range }S = \text{dim null }T$
because dim range S cannot be smaller than dim null T
Not sure whether this is correct again
Best Answer
Using matrix notation if the range of $S$ is in the nullspace of $T$ then $TS$ is zero and so $(ST)^2=(ST)(ST)=S(TS)T=0$ which is the result.