# Kernel of Composition

$U,V,W$ are finite dimensional vector spaces over a field $\mathbb{F}$. Let $S:U\rightarrow V,\; T:V\rightarrow W$ be linear maps.

Prove: $\text{Ker}(S)\subset\text{Ker}({TS})$

Attempt:

$\text{Ker}({TS})=\left\{ u\in U\;:\;({T\circ S})(u)=0\right\} =\left\{ u\in U\;:\; S( u)=0\;\bigvee\; T({S(u)})=0\right\} =\left\{ u\in U\;:\; u\in\text{Ker}(S)\;\bigvee\; S(u)\in\text{Im}(S)\cap\text{Ker}(T)\right\}$

$\iff\text{Ker}({TS})=\text{Ker}(S)\cup({\text{Im}(S)\cap\text{Ker}(T))}\supset\text{Ker}(S)$

I was told however that there is a mistake here. What is my mistake? Thank you.

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To answer your question (about where your mistake lies): In the last line, you conclude that $\text{Im}(S) \cap \text{Ker}(T) \subset \text{Ker}(TS)$, but note that $\text{Im}(S) \cap \text{Ker}(T) \subset V$ whereas $\text{Ker}(TS) \subset U$.

ncmathsadist has provided a quick proof of the result.

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Ah, ignorant mistake. Is the process up to that point valid? Also, is it possible to "amend" what I tried to do? –  Derpson Oct 29 '13 at 15:16
I'd say so, but it depends on what you're after. For the question at hand, once you noticed that $u \in \text{Ker}(TS)$ if $S(u) = 0$, then you're done ($\text{Ker}(S) \subset \text{Ker}(TS)$). Did you want to write $\text{Ker}(TS)$ explicitly in terms of sets determined by $S$ and $T$? One way to do this that's in line with your reasoning is to conclude $\text{Ker}(TS) = \text{Ker}(S) \cup S^{-1}(\text{Ker}(T)) = S^{-1}(\text{Ker}(T))$, which I think is more or less what you were trying to say in your last line. –  Dan Oct 29 '13 at 15:26
Thanks for your help! –  Derpson Oct 29 '13 at 16:05

Let $x\in \ker(S)$. Then $S(v) = 0$, so $T(S(v)) = 0$; hence $v\in \ker(TS)$.

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could you point me to my mistake? –  Derpson Oct 29 '13 at 15:05