Inequality for the kernel of the composition of linear maps

How do you see that if $F\colon X\to Y$ and $G\colon Y\to Z$ are two linear maps between the vector spaces $X$ and $Y$, then $$\dim(\ker(G\circ F)) \leq \dim(\ker(F)) + \dim(\ker(G))$$?

I see that $\ker(G\circ F) \equiv F^{-1} (G^{-1}( \{0\} ))$ but I'm not exactly sure how this helps. I can also see that $\ker(G\circ F) \supseteq \ker(F)$ in $X$, that is, there is a larger chunk of $X$ (larger than $\ker(F)$) which can get mapped into $\ker(G)\subseteq Y$.

Solution 1:

As you wrote, $\ker(G \circ F) = F^{-1}(\ker G)$. You can prove more generally that if $U \leq Y$ is a subspace then

$$ \dim F^{-1}(U) \leq \dim U + \dim \ker F. $$

To see this, consider $F|_{F^{-1}(U)} \colon F^{-1}(U) \rightarrow U$ and apply the rank-nullity theorem.