# Proof involving linear maps

Suppose $V$ is a vector space and $S,\ T \in L(V)$ such that range $S \subset$ null $T$.

Prove that $$(ST)^2= 0$$

I have no idea how to go about this could someone maybe explain it in English or give a hint?

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For all $x\in V$, $S(x)\in$ range$S\subset$ null $T$.

Therefore, $TS(x)=0$. Rest will follow as shown by Lee.

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Note that $$TS=0$$ sothat $$(ST)^2=STST = S0T=0$$

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for $S0T = 0$, how does the mapping work in this step? Multiplication of linear maps is like a composition right? So how would it be broken down in this case? Thank you! – Soaps Jul 23 '14 at 2:16
Note that $TS=0$ is a zero matrix. That is all entries are zero. But it is a matrix. So when $A$ is zero matrix then $$\ast\ AB=BA=0$$ is clear for any $B$ ? So note that by $\ast$, $$(ST)^2=(ST)(ST)=S ((TS)T )=S(0T)=S0=0$$ since matrix multiplication is associative. – HK Lee Jul 23 '14 at 2:23
Ah I was thrown off by the fact that they are linear maps, thank you! – Soaps Jul 23 '14 at 2:30

As a quick sketch, simply notice that if $x \in V$, then $$(ST)^2x = STST x = STS (Tx) = ST(S(T(x)) = S(T(S(T(x))) = S(0) = 0$$ where the second to last equality follows because $S(T(x)) \in \mathrm{range}\, S$ and thus is in $\mathrm{null} \, T$.

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By definition, $(ST)^2$ is the linear map defined by $x \mapsto (S(T(S(T(x))))$. Remember that a linear map $Y$ is zero if and only if $Y(x)=0$ for all $x \in V$. So it suffices to check that the above definition of $(ST)^2$ gives zero when applied to an arbitrary element of $V$.

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