Direct sum of nullspace and range

[Bipolarity]

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

 

[Office_Shredder]

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

 

[Bipolarity]

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

 

[Office_Shredder]

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

 

[blue_raver22]

olarBear, the statement you have provided is true. The direct sum of the nullspace and range of a linear map T is equal to the source space V. This can be proven using the dimension theorem, which states that for any linear map T: V → W, the dimension of the nullspace of T plus the dimension of the range of T is equal to the dimension of the source space V. In other words, the dimensions of the nullspace and range of T add up to the dimension of the source space. Therefore, the direct sum of the nullspace and range of T must equal the source space V. I hope this helps guide your proof in the right direction.