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Another question in linear algebra.

  1. Jun 24, 2007 #1
    I need to prove that if T1 and T2 are linear operators, then
    dim(kerT1)+dim(KerT2)>=dim(Ker(T1oT2)).

    now, i thought to show that Ker(T1oT2) is a subset of ker(T1), and then it obviously follows, but here im stuck: if u in ker(T1oT2) then T1(T2(u))=0 so T2(u) is in Ker(T1) but i need to show that u is in kerT1, if it's correct.

    any hints.
     
  2. jcsd
  3. Jun 24, 2007 #2

    matt grime

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    Can I change it to S and T, rather than T1 and T2, which, let's face it, is a very bad notation.

    ker(ST) is certainly not a subset (or even subspace) of ker(S). There is no reason to suppose that T maps ker(S) into ker(S), which is what you are trying (and therefore failing) to show.

    u is in ker(ST) if either T(u)=0, or T(u) lies in ker(S). Of course T(u)=0 implies that T(u) is in ker(S), thus all you need to do is to work out the dimension of the preimage of ker(S) under T.

    The pre-image of any subspace V under a map T is a subspace of the form U+ker(T), with the sum not necessarily direct, and T mapping U isomorphically to V. (This is just the isomorphism theorems in action.) This completely solves your problem, and was just writing out the definitions.
     
    Last edited: Jun 24, 2007
  4. Jun 24, 2007 #3

    jambaugh

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    You have things reversed.
    [tex]\mathop{ker}(T_2) \subseteq \mathop{ker}(T_1\circ T_2)[/tex]

    But that won't prove your inequality either but it is one step. Also consider the space
    [tex] \mathbf{N}= \mathop{ker}(T_1\circ T_2) - \mathop{ker}(T_2)[/tex]
    and its image:
    [tex]\mathbf{N}'= T_2(\mathbf{N})[/tex]
     
    Last edited: Jun 24, 2007
  5. Jun 25, 2007 #4
    so you mean i need to find the dimension of T^-1(kerS), right?
     
  6. Jun 25, 2007 #5
    ah, ok i see it's only an application of dim(ImT)+dim(KerT)=dimV.
     
  7. Jun 25, 2007 #6

    jambaugh

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    There ya go.
     
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