andrey21 said:
It is the set of vectors in the domain for which T(v) = 0 correct?
Yes, good! So you have
T(x_1, x_2, x_3, x_4)= (x_1+x_2+ x_3, x_3, x_1+ x2)= (0, 0, 0)
which gives the three equations
x_1+ x_2+ x_3= 0
x_3= 0
x_1+ x_2= 0
Since, by the second equation, x_3= 0, the first equation becomes x_1+ x_2= 0, the same as the third equation. We cannot solve a single equation in two unknowns-the best we do is solve for one in terms of the other: x_2= -x_1. Since x_4 does not appear in any of these equations, there is no restriction on it- it can be any number.
With the conditions that x_2= -x_1 and x_3= 0 we can write any such vector as
(x_1, -x_1, 0, x_4)= (x_1, -x_1, 0, 0)+ (0, 0, 0, x_4)= x_1(1, -1, 0, 0)+ x_4(0, 0, 0, 1).
Now, do you see a basis for the kernel?
You could have done this from your reduced form of the matrix by arguing that
\begin{bmatrix}1 & 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x_1 & x_2 & x_3 & x_4\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
results in the same equations: x_1+ x_2= 0 and x_3= 0 but I think it is more instructive to go back to the original definitions than to rely on partially-remembered and not very well understood formulas.
