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Proving an image and annihilator of a kernel are equal

  1. May 19, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose T:V→U is linear and V has finite dimension. Prove that Im Tt = (Ker T)0

    2. Relevant equations
    dim(W)+dim(W0)=dim(V) where W is a subspace of V and V has finite dimension.

    3. The attempt at a solution
    I first proved Im Tt ⊆ (Ker T)0. Let u be an arbitrary element of Ker T and Tt(Φ )∈ Im Tt, then Tt(Φ)(u)=Φ(T(u))=Φ(0)=0. Thus, since all elements of Im Tt map all elements of Ker T into 0, Im Tt ⊆ (Ker T)0.

    I now only need to proof dim((Ker T)0) = dim Im Tt. Since V has finite dimension, dim((Ker T)0) = dim(V)-dim(Ker T), but dim(Im T) = dim(V)-dim(Ker T). Thus, dim((Ker T)0)=dim(Im T). This is where I got stuck, how do I prove dim (Im T) = dim ((Im T)0)?
     
  2. jcsd
  3. May 20, 2017 #2
    this assumption is redundant. Redundant assumptions just obscure a core of the theorem and make proof more complicated than it really is.
     
    Last edited: May 20, 2017
  4. May 20, 2017 #3
    Hint.
    First prove a
    Theorem. Let ##A:V\to U,\quad w:V\to\mathbb{R}## be lenear mappings such that ##\mathrm{ker}\,A\subseteq \mathrm{ker}\,w##. Then there exist a linear mapping ##\Lambda:U\to\mathbb{R}## such that ##w=\Lambda A##.

    It is actually a special case of some another general theorem
     
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