I am trying to solve the below problem.
Let $\phi$ and $\psi: V \to V$ be linear operators on a vector space $V$ of dimension $n$. Show that $$\mathrm{ran}(\phi \circ \psi) \geq \mathrm{rank}(\phi) + \mathrm{rank}(\psi) - n.$$
The only thing I know for sure I can do is use the rank-nullity theorem. I'm interested in the rank of $\phi \circ \psi$, so I can say: \begin{align*} n = \dim V = \mathrm{rank}(\phi \circ \psi) + \dim \mathrm{ker}(\phi \circ \psi). \end{align*} So $\mathrm{rank}(\phi \circ \psi) = n - \dim\mathrm{ker}(\phi \circ \psi)$, and the problem is reduced to showing that \begin{align*} n - \dim\mathrm{ker}(\phi \circ \psi) \geq \mathrm{rank}(\phi) + \mathrm{rank}(\psi) - n. \end{align*} My next instinct is to apply rank-nullity to $\phi$ and $\psi$ individually, but there's no way around having to compare the dimension of the kernel of the composition map to the rank of the individual maps. It seems to me that, because the kernel of the composition map is a subspace of the domain $\psi$, it must have dimension $\leq n$.
Am I on the right track, or is some better way to compare these?