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Math Help - Linear Maps and Kernels

  1. #1
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    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.
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  2. #2
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    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)).)
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  3. #3
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    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?
    Last edited by manygrams; October 28th 2010 at 07:50 PM.
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  4. #4
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    Quote Originally Posted by manygrams View Post
    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.
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  5. #5
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    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.
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  6. #6
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    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 ..."
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