Is this transformation defined on a basis is a linear map?

Question

Let V be a vector space, and let {a, b, c} be a basis for V.

Define T : V → V as a ↦ 0, b ↦ a, c ↦ a. Also define T(x+y) = T(x) + T(y) and T(sx) = sT(x), for any vectors x, y and scalar s.

Is T a linear map? If so, then shouldn't rank(V) + nullity(V) = dim(V)? But doesn't rank(V) = 1, nullity(V) = 1, and dim(V) = 3?

Please help, I am extremely confused.

Edit (what I asked in the comments): I thought that the nullity is the dimension of the null space, and the null space is the vectors that map to 0. So the basis of the null space is {a}, and the nullity is 1. However, b + (-c) maps to a + (-a) = 0, so would b and c be in the basis of the null space? Would that make the nullity 3? I feel like I'm missing something...

Answer 1

It should be Rank(T) + Nullity(T) = Dim(V)

Rank and nullity are the dimensions of the image space and kernel respectively with respect to a given linear operator.

If {a,b,c} is a basis for V, then so is {a, b - c, c}. Under this basis, you’ll see that the nullity is 2.