Problem:
Let $\phi$ and $\psi : V \to V$ be linear operators on a vector space $V$ of dimension $n$. Show that
$$rank(\phi \circ \psi) ≥ rank(\phi) + rank(\psi) − n.$$
My work:
I have already seen this and this questions where the same is answered however I have a different question. I started writing the following proof:
Consider the linear operator $\phi \circ \psi \in \text{Hom}(V,V)$ and let $\dim \text{Ker } \psi = a$ and $\dim \text{Im } \phi = b$, . We want to show $$\dim \text{Im } (\phi \circ \psi) \ge \dim \text{Im } \phi + \dim \text{Im } \psi – n = \dim \text{Im } \phi – \dim \text{Ker } \psi = b – a$$ so we only need to consider the case $b > a$, otherwise we would be finished. Let $(u_1, \ldots, u_a)$ be the basis of $\text{Ker } \psi \subseteq V$ so we can extend it to $(u_1, \ldots, u_a, v_1, \ldots, v_{n-a})$ a basis of $V$. For any $v \in V$ and for some scalars $x_i$, $$(\phi \circ \psi)(v) = (\phi \circ \psi)(u_1x_1 + \ldots + u_ax_a + v_1x_{a+1} + \ldots + v_{n-a}x_{a+n}) = (\phi \circ \psi)(v_1)x_{a+1} + \ldots + (\phi \circ \psi)(v_{n-a})x_{a+n}$$ So $span((\phi \circ \psi)(v_1), \ldots, (\phi \circ \psi)(v_{n-a})) = \text{Im } (\phi \circ \psi)$ and…
As you see it is unfinished since my intention was to prove $\text{Im }(\phi \circ \psi) \ge n – a \ge b – a$ since $\text{Im } \phi \subseteq V$ but for that I needed $(\phi \circ \psi)(v_1), \ldots, (\phi \circ \psi)(v_{n-a})$ to be linearly independent, but I didn't realise that there is no need for them to be linearly independent until I got to that point.
So my doubt is: Is there a way to fix the proof without changing it entirely?
Huge thanks in advance.
Best Answer
Consider $\ker \phi\vert_{\mathrm{Im}\ \psi}.$ We have $$\ker \phi\vert_{\mathrm{Im}\ \psi}=\ker \phi\cap\mathrm{Im}\ \psi.$$ Then$$\dim \mathrm{Im}\ \phi\circ\psi=\dim \mathrm{Im}\ \phi\vert_{\mathrm{Im}\ \psi}=\dim \mathrm{Im}\ \psi-\dim\ker \phi\cap\mathrm{Im}\ \psi\geq \dim \mathrm{Im} \ \psi-\dim \ker \phi=\dim \mathrm{Im} \ \psi-n+\dim \mathrm{Im}\ \phi.$$ that's done.