Suppose v is in the kernel of T. That is, T(v)= 0. Can you prove that S(v) is also in the kernel of T? That is, can prove that T(S(v))= 0? (Of course, ST= TS means that T(S(v))= S(T(v)).)
So if I can prove S(v) is also in the kernel of T, then that proves S(Ker T) is contained in Ker T?
Progress:
Suppose v is in S(KerT), thus S(T(v))=0
S(T(v))=0
T(S(v))=0 as ST=TS
Since S(v) is in Ker T as T(S(v))=0
T(KerT)=0
T(v)=0 thus v is in Ker T, so S(Ker T) is in Ker T.
Any good?
"S(ker T)" means the set of all vectors S(v) where v is in the kernel of T so, yes.
No! Saying that v is in S(KerT) means that v= S(u) for some u in kerT. "S(T(v))=0" means that v is in the kernel of ST.Progress:
Suppose v is in S(KerT), thus S(T(v))=0
No, you are going "the wrong way". As I said before, if "v is in S(kerT)" then v= S(u) of some u in kerT. Now, Tv= TS(u)= ST(u)= S(0)= 0.S(T(v))=0
T(S(v))=0 as ST=TS
Since S(v) is in Ker T as T(S(v))=0
T(KerT)=0
T(v)=0 thus v is in Ker T, so S(Ker T) is in Ker T.
Any good?