# Direct sum of nullspace and range

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
Staff Emeritus
Gold Member
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

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
Staff Emeritus