Uniqueness for a linear transformation given nullspace and range Find the representation in a canonical base for: $$\mathcal{T}: \mathbf{R^3} \rightarrow \mathbf{R^3}$$
Such that :$$ \quad\mathcal{N}(\mathcal{T}) = \quad \langle\left(\begin{matrix} 1\\0\\-1 \end{matrix}\right)\rangle$$ and the image of $\mathcal{T}$ is defined by the equation:$$x +2y -z=0$$
Is $\mathcal{T}$ unique?
The only thing I've figured out is:$$Im(\mathcal{T}) =  \quad \langle\left(\begin{matrix} -2\\1\\0 \end{matrix}\right), \left(\begin{matrix} 1\\0\\1 \end{matrix}\right)\rangle $$
and for any vector $\mathbf{x}$ in $\mathbf{R^3}$:$$\mathcal{T}(\mathbf{x})= \alpha\left(\begin{matrix} -2\\1\\0 \end{matrix}\right) + \beta\left(\begin{matrix} 1\\0\\1 \end{matrix}\right)$$
I have absolutely no idea on how to proceed, any ideas...

Suppose we have a linear map from $T:U\rightarrow V$ and we have a basis for $U$. If I tell you where $T$ takes each basis vector of $U$ then you know $T$ exactly, and therefore these conditions define $T$ uniquely. Now suppose instead of telling you where specific basis vectors go, I just tell you a vector in $U$ gets mapped to $(-2,1,0)$ and some other vector gets mapped to $(1,0,1)$ and another vector gets sent to $(0,0,0)$, without specifying more information you do not know $T$ exactly. 
For example if we use the standard basis, you could say $e_1$ goes to $(-2,1,0)$ and $e_2$ goes to $(1,0,1)$, or you could say $e_2$ goes to $(-2,1,0)$ and $e_1$ goes to $(1,0,1)$. Both of these maps would satisfy the conditions I gave. Is it clear how the conditions I gave are the same as the ones you were given above? The key is to remember these are linear transformations so $T(c_1e_1+c_2e_2)=c_1T(e_1)+c_2T(e_2)$. 
Hope that helps! 
