How do I find the kernel of a composition of functions?

Question

Functions $g$ and $f$ are linear and injective. How do I go about finding the kernel of $g \circ f$?

I'm asking because I want to prove that $\ker(f) = \ker(g \circ f)$.

Answer 1

Hint

If you mean by functions: "endomorphisms" then use (prove) theses results:

• $f$ is injective iff $\ker f=\{0\}$
• if $f$ and $g$ are injective then $g\circ f$ is also injective

Answer 2

If $g$ is injective then $\ker(f) = \ker(gf)$ is certainly true. To prove two sets are equal you show that each is contained in the other. For example if $x \in \ker(f)$ then $f(x) = 0$. But then $gf(x) = g(0) = 0$ so $x \in \ker(gf)$. This proves $\ker(f) \subseteq \ker(gf)$. I'll leave the other direction to you (that direction uses the fact that $g$ is injective).

So if $g$ is injective $\ker(gf) = \ker(f)$ and if $f$ is injective then $\ker(f) = 0$ so if both are injective then $\ker(gf) = 0$. This of course makes sense because the composition of two injective maps is again injective and hence has trivial kernel.

Answer 3

Consider what it means to be injective and what it means to have a nontrivial kernel. Its not possible to have both. So, if the kernels are trivial then $\ker (f) =\{0\}$ and $\ker (g\circ f) =\{0\}$ However, you still need to check that that the $0$ in both of those is the same $0$ vector, ie that they are in the same vector space.

Answer 4

This maybe overkill but there is the kernel-cokernel exact sequence which is very useful in situations like this.

If we have a composition $ A \xrightarrow{f} B \xrightarrow{g} C $ then the following sequence is exact :

$ 0 \to ker(f) \to ker(gf) \to ker(g) \to coker(f) \to coker(gf) \to coker(g) \to 0 $

And so if $ker(g)=0$ then $ker(f)=ker(gf)$.

Answer 5

One has $\def\Im{\operatorname{Im}}\ker(g\circ f)=\{\, v\mid g(f(x))=0\,\}=\{\, v\mid f(x)\in\Im(f)\cap\ker(g)\,\,\}=f^{-1}[\Im(f)\cap\ker g]$ whereas $\ker(f)=f^{-1}[\{0\}]$. Thus $\Im(f)\cap\ker(g)=\{0\}$ is a sufficient condition for being able to conclude that $\ker(g\circ f)=\ker(f)$ (which is of course not valid in general, as you question seems to suggest; just take $f=I$ to convince yourself of that). One can see that the mentioned condition is also necessary to get $\ker(g\circ f)=\ker(f)$: if one would have $0\neq w\in\Im(f)\cap\ker(g)$ and this means (because $w\in\Im(f)$) that there exists $v$ with $f(v)=w$; for this vector $v\in\ker(g\circ f)$ but $v\notin\ker(f)$.