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Direct sum of nullspace and range

  1. Sep 10, 2013 #1
    Is this true? I am studying direct sums and was wondering if the following statement holds? It seems to be true if one considers the proof of the dimension theorem, but I need to be sure, so I can steer my proof toward a particular direction.

    ## N(T) \bigoplus R(T) = V ## where ##V## is the source of the linear map ##T:V → V## ?

    BiP
     
  2. jcsd
  3. Sep 10, 2013 #2

    Office_Shredder

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    It's not. The null space and range have dimensions which add up to the dimension of V, but they can overlap. For example consider the map

    T(x,y) = (y,0).

    This is a linear map from R2 to R2. The null space is (x,0) for x in R and the range is (x,0) for x in R, so N(T)+R(T) is just the x-axis
     
  4. Sep 10, 2013 #3
    I see. Suppose that you knew the statement to be true. Then could you conclude that T is invertible?

    BiP
     
  5. Sep 10, 2013 #4

    Office_Shredder

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    It's not hard to construct a projection in two dimensions which is a counterexample to that claim
     
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