Rank Nullity and Dimension relation

Question

How would one prove the relations:

$rank S◦T = rankT-dim(kerS ∩ ImT)$

and

$nullity S◦T = nullityT+dim(kerS ∩ ImT)$

I understand that the use of rank nullity theorem is required but am confused by the composition.

Answer 1

It's useful to recognize here that the composition $S \circ T$ is basically (although not really) the restriction of $S$ to $\text{Im}(T)$. Under this restriction, apply the rank-nullity theorem, and the result should come out.