Bound on the rank of the composition map

Question

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?

Answer 1

One way to handle this is to introduce the operator $T: \psi(V) \to V$ given by $T(v) = \phi(v)$ for all $v \in \psi(V)$. In an informal sense, $T$ is "just $\phi$" but regarded as a map only on $\psi(V)$ instead of all of $V$, but note that $T$ has different rank-nullity theorem consequences attached to it.

Then complete the following program: $$ \begin{align*} \operatorname{rank}(\psi) & = \operatorname{rank}(\phi \circ \psi) + \operatorname{nullity}(T) \\ & \leq \operatorname{rank}(\phi \circ \psi) + \operatorname{nullity}(\phi) \\ & = \operatorname{rank}(\phi \circ \psi) + n - \operatorname{rank}(\phi), \end{align*} $$ The first equality is a consequence of applying rank-nullity to $T: \psi(V) \to V$, which acts on a space of dimension $\operatorname{rank}(\psi)$ and whose range coincides with that of $\phi \circ \psi: V \to V$.

The inequality in the middle is a consequence of the fact that the nullspace of $T$ is contained in the nullspace of $\phi$.

The final equality is a consequence of applying rank-nullity to $\phi: V \to V$.

Answer 2

I think the issue is what is the intersection $$\ker \phi\cap \mathrm{im} \psi$$ The greater this intersection, the smaller is $\mathrm{rank}(\phi\circ \psi)$. So assume worst case, that is $$\ker \phi\subseteq \mathrm{im} \psi$$. Now $$\dim \ker\phi=n-\mathrm{rank} \phi$$ so what remains of $\mathrm{im} \psi$ to be sent to a non zero image will have dimension at least $$\mathrm{rank} \psi-(n-\mathrm{rank} \phi)$$