Proving rank$(S\circ T)\le$ rank $S$ and null$(S\circ T)\ge$ null $T$

Question

Let $T$:$V\rightarrow W$ and $S$:$W\rightarrow U$ be linear transformations. I need to prove the two statements in the title.

I don't know how to approach this problem. I am quite sure it involves the use of rank-nullity theorem, an example for the first part:

$\text{rank } S=\dim W-\text{null }S$ and $\text{rank }(S\circ T)=\dim V-\text{null}(S\circ T)$

But even when assigning values, it doesn't prove to be enough to give a direct proof. Should I take different cases?

Answer 1

I suggest that you consider the spaces rather than their dimensions. If you prove that $$ range(S \circ T) \subset range(S) $$ then the statement about rank (which is just the dimensions of the two sides) follows.

Now something is in the range of $S\circ T$ if it's $S(T(v))$ for some $v$. Calling $T(v)$ by the name $w$, that means it's $S(w)$, which makes it in the range of $S$. So I've proved that anything in the range of $S \circ T$ is in $range (S)$. So I'm done.

Can you make the parallel argument for the nullspace/nullity?