Let $T: \Bbb R^3 → \Bbb R^2 $be a linear transformation defined by $T( x, y, z) = ( x + y, x - z)$ Problem: Let $T: \Bbb R^3 → \Bbb R^2 $be a linear transformation defined by $T( x, y, z) = ( x + y, x - z)$.  Then dimension of null space of $T$ is 
(A) $1 $
(B) $2 $
(C) $0$
(D) none of these 
Solution : We know that 
$ \dim (\Bbb R^2) =\dim $(null space  ) $+\dim ($range  $T$) 
$2 =$ dim (null space  ) +$2$
$0 =$ dim (null space  ) 
Am I doing right ??
 A: You have
$$ T(x,y,z) = (0,0) \iff x+y = 0 \text{ and } x-z = 0 \\ \iff x=z \text{ and } -x=y \\ \iff (x,y,z) = (x,-x,x)$$
It is now clear that a basis of the null space of $T$ is given by $\{(1,-1,1)\}$, i.e. its dimension is $1$.
A: Rank nullity theorem : For a linear map $T:V\rightarrow W$
$\text{dim}(\text{im} (T)) + \text{dim} (\text{ker} (T)) = \text{dim} (V)$
Just to look in another way :
Given $m,n\in \mathbb{R}$ Can you find $x,y,z\in \mathbb{R}$ such that $$x+y=m;x-z=n$$
Do you think three unknowns with two equations always have solution?
Do you think this helps?
A: Since $T(0,1,0)=(1,0)$ and $T(0,0,-1)=(0,1)$, you know that the image of $T$ is $\mathbb{R}^2$. Then the rank-nullity theorem makes you finish.
Note that the rank-nullity theorem tells you that, for a linear map $T\colon V\to W$,
$$
\dim V=\dim N(T)+\dim\operatorname{Im}(T)
$$
where $N(T)$ is the null space and $\operatorname{Im}(T)$ is the image (or range) of $T$. All you can say involving the dimension of the codomain $W$ is that $\dim\operatorname{Im}(T)\le\dim W$. The relation expressed by the theorem has to do with the dimension of the domain.
