Suppose $U$ is a finite-dimensional vector space, that $S ∈ L(V, W )$, and that $T ∈ L(U, V )$. I'm trying to prove that $\operatorname{dim} \operatorname{null}(ST ) = \operatorname{dim} \operatorname{null}(T ) + \operatorname{dim} (\operatorname{range}(T ) ∩ \operatorname{null}(S))$.
First I realized that for a vector $x$ to be in $\operatorname{null}(ST)$, it can
(1) be in $\operatorname{null} T$, in which case $Sx=0$ (since $S$ is a linear map) and thus $x \in \operatorname{null}(ST)$
(2) get mapped to an element of $\operatorname{range}T$ that happens to be in the null space of $S$.
Hence $\operatorname{null}T$ and $\operatorname{range}(T ) ∩ \operatorname{null}(S)$ combine to make $\operatorname{null}(ST)$, i.e. $$\operatorname{null}T + \operatorname{range}(T ) ∩ \operatorname{null}(S) = \operatorname{null}(ST).$$
Now, taking $\text{dim}$s, I get
$$\operatorname{dim}\operatorname{null}T + \operatorname{dim}(\operatorname{range}(T ) ∩ \operatorname{null}(S)) - \operatorname{dim}(\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S) = \operatorname{dim}\operatorname{null}(ST).$$
It seems like for the theorem to hold, we must have that $\text{dim}(\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S) = 0$, i.e. that $\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S = \{ 0 \}$ for arbitrary linear maps $S, T$. Could someone please point me in the right direction as to how go about conceptualizing and proving this?