# 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 $\ker(G)\subseteq Y$.

• Jul 9, 2020 at 4:10

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.
Here is a method that applies when we are considering bounded linear maps between (possibly infinite-dimensional) Banach spaces. In this context, since the space $$X$$ may be infinite-dimensional, we cannot apply the rank-nullity theorem.
If $$\dim\ker(F) = \infty$$, then there is nothing to show, so suppose that $$\dim\ker(F) < \infty$$. Since $$\ker(F) \subset \ker (G\circ F)$$, and $$\ker(G\circ F)$$ is a closed subspace of the Banach space $$X$$ (hence a Banach space in its own right), there is a closed complement $$C$$ for $$\ker(F)$$ in $$\ker(G\circ F)$$, i.e., a closed subspace $$C$$ of $$\ker(G\circ F)$$ such that $$\ker(G\circ F) = \ker (F)\oplus C.$$ Now, $$F|_C\colon C\to \ker(G)$$ is a vector space isomorphism onto its image $$F(C)$$, so $$\dim (C) = \dim F(C) \le \dim \ker (G)$$ (where the inequality holds as an inequality of extended real numbers). Hence we have $$\dim\ker (G\circ F) = \dim \ker(F) +\dim(C)\le \dim \ker (F) + \dim \ker (G),$$ as desired.