Direct sum of nullspace and range

  • Thread starter Bipolarity
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  • #1
Bipolarity
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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
 

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  • #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
 
  • #3
Bipolarity
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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

I see. Suppose that you knew the statement to be true. Then could you conclude that T is invertible?

BiP
 
  • #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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