# kernel of linear combination

Let $T,S: V\to V$ be 2 linear transformations, and $\ker(T)=\{0\}$. I need to prove why $\ker(T\circ S)=\ker(S)$.
I have no idea how to prove it.

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Hint In general, the way to show two sets $X,Y$ are equal is to show $X \subseteq Y$ and $Y \subseteq X$ – andybenji Jan 3 '13 at 23:02

## 1 Answer

Since $\ker T = \{ 0 \}$, $T$ is injective. Therefore, $(T \circ S)v=0$ if and only if $Sv=0$.

[Both of these results require proofs, but I presume that you've either seen them before or can prove them yourself.]

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Nice job pointing the way without doing it for him. – andybenji Jan 3 '13 at 23:01
Im sorry but I didnt understand the last part – user1816377 Jan 3 '13 at 23:04
@user1816377: What about it don't you understand? – Clive Newstead Jan 3 '13 at 23:05
"Therefore, (T∘S)v=0 if and only if Sv=0." I understand why S(v) has to $0$ so we get result $0$ but I dont understtand why the result is $ker(s)$ in every case – user1816377 Jan 3 '13 at 23:08
@user1816377: The point is that $Sv=0$ if and only if $v \in \ker S$, by definition of the kernel. Likewise, $TSv=0$ if and only if $v \in \ker (T \circ S)$. So proving $\ker (T \circ S) = \ker S$ amounts to showing that: $TSv=0$ if and only if $Sv=0$. And this follows by injectivity of $T$, as I argued above. – Clive Newstead Jan 3 '13 at 23:23