What is the basis of T?

Question

Let $T: \Bbb{R^3} \to \Bbb{R^3}$ be a linear transformation:

$T(a,b,c) = (a+b+c,0,0)$

Find a basis for $\text{Nullspace}(T)$.

I found that the basis spans $(-1,1,0)$ and $(-1,0,-1)$, is this true ?

I am concerned that the dimension of $V$ is $3$, but I only have $2$ basis.

Answer 1

hint: $(a+b+c,0,0) = (a+b+c)(1,0,0)$. Can you take it from here?, and for the nullspace of $T$, you need $a+b+c= 0 \to (a,b,c) = (a,b,-a-b) = a(1,0,-1)+b(0,1,-1)$. Can you again take it from here as well?