Dimensions of a linear transformation 
Let $T:\mathbb R^3$ maps to $\mathbb {R}^2$ be a linear transformation defined by 
  $$T(x,y,z)=(x+y,x-z)$$
  Then the dimension of the null space of $T$ is ?

As I solved this:
The possible row reduction gives 2 pivot hence 2 pivot columns I.e the no. Of linearly independent vectors is 2. Those two independent vectors I.e.$(1,1,0)$ and $(0,1,1)$ make the basis hence the dimensions (no. Of basis) is 2.
But answer given is 1. What's wrong with that?
 A: The dimension of the null space (1) added to the dimension of the image (2) gives the number of columns (3). The vectors you found are a basis for the image.
A: The Matrix A associated to T is:
$$A=\begin{bmatrix}
1 &1  & 0\\
1 & 0 & -1
\end{bmatrix}$$
$$rank(A) = 2 \implies null(A)=3-2=1$$
A: Or, a little more directly, the "null space" of a linear transformation, A, is defined as the set of all v such that Av= 0.  It is easy to show that the null space is a subspace of the domain space so has some dimension. 
In the example given, with $A= \begin{bmatrix}1 & 1 & 0 \\ 1 & 0 & -1\end{bmatrix}$ we have $Av= \begin{bmatrix}1 & 1 & 0 \\ 1 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ y \\ x- z\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.  That gives the equations x+ y= 0, x- z= 0.  Two equations cannot be solved for specific values of three unknowns.  What we can do is write y= -x, z= x so any such vector must be of the form $\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ -x \\ x\end{bmatrix}= x\begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}$.  That shows that the null space has basis $\{\begin{bmatrix}1 \\ -1 \\ 1 \end{bmatrix}\}$ and so has dimension 1.
The "rank" of a linear transformation, by the way, is the dimension of the image of the domain space in the range space.  That is, it is the dimension of the space of all $\begin{bmatrix}u \\ v \end{bmatrix}$ such that there exist $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ with $\begin{bmatrix}1 & 1 & 0 \\ 1 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ y \\ x- z\end{bmatrix}= \begin{bmatrix}u \\ v \end{bmatrix}$.  That is, there are x, y, and z, not necessarily unique, such that x+ y= u and x- z= v. Given any u and v, we could take x= u, y= 0, z= u- v.  Since we can find such x, y, z for any u, v, the rank is 2. 
