Proof : If $S,T:V \rightarrow V$ (V is finite) and $KerS=\{0\}$ then $Im(TS)=Im(T)$

Question

I have this problem :

Proof : If $S,T:V \rightarrow V$ (V is finite) and $KerS=\{0\}$ then $Im(TS)=Im(T)$

My solution :

• Let $v \in ImT$ exist $w \in V$ such that $T(w)=v$.

Since $KerS=\{0\}$ Its injective, then exist such $x \in V$ such that $S(x)=w$, Therefore $T(S(x))=T(w)=v \implies ImT \subseteq ImTS$.

• Let $v \in ImTS$ exist $ w \in V$ such that $TS(v)=T[S(v)]=w$

Let $u \in V$ such that $S(v)=u \implies T(u)=w$.

Therefore $ImT \subseteq ImTS$ and also $ImTS \subseteq ImT \implies$ ImTS=ImT.

I'm not sure with my solution does $u \in V$, implies that $T(u)=w$, Could anyone explain that part of the proof,I don't really understand that since $T$ and $S$ are different transformations

Thanks!

Answer 1

$\text{Im}(TS) \subseteq \text{Im}(T)$ part is fine. There is no problem. But the other inclusion, there is a subtle point. "Since $\text{Ker}(S)=\{0\}$, it's injective. Then exist such $x \in V$ such that $S(x) =w.$" The last condition is for surjective maps, not for injective maps.

Here is a hint: Prove that $S$ is a surjective map also.