Direct sum of nullspace and range

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Discussion Overview

The discussion revolves around the validity of the statement regarding the direct sum of the null space and range of a linear map, specifically questioning whether the null space and range of a linear transformation can be expressed as a direct sum that equals the entire vector space.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the truth of the statement that the null space and range of a linear map form a direct sum equal to the vector space, suggesting it seems true based on the dimension theorem.
  • Another participant asserts that the null space and range can overlap, providing a counterexample with a specific linear map T(x,y) = (y,0), indicating that the null space and range do not form a direct sum.
  • A later reply reiterates the counterexample and poses a hypothetical question about the implications of the statement being true, specifically regarding the invertibility of the transformation.
  • Another participant mentions the ease of constructing a projection in two dimensions as a counterexample to the claim of direct sum.

Areas of Agreement / Disagreement

Participants generally disagree on the validity of the statement regarding the direct sum of the null space and range, with multiple competing views presented without resolution.

Contextual Notes

Participants note that while the dimensions of the null space and range add up to the dimension of the vector space, this does not imply that they are disjoint or form a direct sum.

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

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