Thread: Linear Maps and Kernels

I need some help with a question. Can anyone give me a hint to start? I'm lost

Let V be a vector space and S,T are in the set of linear maps from V to V. Show that if ST=TS, then S(Ker T) is contained in Ker T.

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?

Originally Posted by manygrams

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?

"S(ker T)" means the set of all vectors S(v) where v is in the kernel of T so, yes.

Progress:

Suppose v is in S(KerT), thus S(T(v))=0

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.

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?

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.

This seems like too easy of an answer... is it right?

T(v)=0
S) S(T(v))=S(0)
T(S(v))=S(0)
=0
S(v) contained in Ker T
S(Ker T) contained in Ker T


Can I be sure that S(0) = 0? Sorry, I'm having a lot of trouble with this class.

S(0)= 0 for every linear map: for any vector v, S(v)= S(v+ 0)= S(v)+ S(0).

You need to add to the beginning of your proof "suppose u is in S(ker T). Then u= Sv for some v in kernel of T ..."