I'm trying to solve the following problem.
Let $\phi$ and $\psi: V \to V$ be linear operators on a vector space $V$ of dimension $n$. Show that $$ \text{rank}(\phi \circ \psi) \geq \text{rank}(\phi) + \text{rank}(\psi) - n. $$
The only machinery I think I have to use for this are the rank nullity theorem and the fact that a subspace of a finite-dimensional vector space is finite-dimensional with dimension less than or equal to the larger space. The rank nullity theorem implies the following: \begin{align*} n = \dim V = \text{rank}(\phi) + \text{ker}(\phi) \\ n = \dim V = \text{rank}(\psi) + \text{ker}(\psi) \\ n = \dim V = \text{rank}(\phi \circ \psi) + \text{ker}(\phi \circ \psi) \end{align*} Also, if $y \in \text{Image}(\phi \circ \psi)$, then $(\phi \circ \psi)(x) = \phi(\psi(x)) = y$ for some $x$, so $y \in \text{Image}(\phi)$, so $\text{rank}(\phi) \geq \text{rank}(\phi \circ \psi)$. That doesn't seem to hlep becaause the inequality sign is reversed. I can say that $\text{ker}(\psi) \leq \text{ker}(\phi \circ \psi)$. The maps have the same domain, so I think I can then say that $\text{rank}(\psi) \geq \text{rank}(\phi \circ \psi)$. So: \begin{align*} \text{rank}(\phi \circ \psi) & \leq \text{rank}(\psi) \\ &= n - \text{ker}(\psi). \end{align*} if I write out the above statement for $\text{rank}(\phi \circ \psi)$, then $$ n - \text{ker}(\phi \circ \psi) = \text{rank}(\phi \circ \psi) \leq n - \text{ker}(\psi) $$ This unfortunately doesn't seem to work.
I'd appreciate any help on how to proceed. I'll keep updating this post as I work on this problem more.
