What is a linear transformation $T:\mathbb{R}^3\longrightarrow\mathbb{R}^4$, whose null space is generated by $(0,1,-3)$ and $(0,-3,4)$? What is a linear transformation to $T:\mathbb{R}^3\longrightarrow\mathbb{R}^4$, whose null space is generated by $(0,1,-3)$ and $(0,-3,4)$?
So, I know that the given pair of vectors generate the subspace $\{(0,y,z):\ y,z\in\mathbb{R}\}$, as they are linearly independent. By the Rank-nullity theorem, the image space in $\mathbb{R}^4$ is of dimension $3-2=1$.
But I can't seem to go further than this. Any help will be appreciated.
 A: Note that in general, given a linear mapping $T\colon V\to W$, $T$ is completely determined by its action on a basis $\beta$ of $V$. (Proof included below).
In your example, given any basis $\beta=\left\{v_1,v_2,v_3\right\}$ of $\mathbb{R}^3$, $T$ is completely determined by $T(v_1),T(v_2)$ and $T(v_3)$.
Note that $v_1=(0,1,-3)$ and $v_2=(0,-3,4)$ are linearly independent and you definitely want $T(v_1)=0=T(v_2)$. Now choose $v_3$ such that $\beta$={v_1,v_2,v_3} is a basis. You don't want $T(v_3)=0$ as this would make the null space larger than you want.
You can now simply plug any values that fit this data!
Proof of claim: Let $v\in V$ be any vector and let $\beta$ be a basis of $V$. As $\beta$ is basis, we can write $$v=\sum_{i=1}^n\lambda_iv_i$$ where $\lambda_i\in \mathbb{R}$ (or whatever field you are working over) and each $v_i$ belongs to $V$. Then $$T(v)=T(\sum_{i=1}^n \lambda_iv_i)=\sum_{i=1}^n\lambda_iT(v_i)$$
is determined by the values $T(w)$ for all $w\in \beta$.
A: Complete $\{(0,1,-3),(0,-3,4)\}$ to a basis.  One easy way is to add the first standard basis vector $e_1=(1,0,0)$.  Then define $T((0,1,-3))=T((0,-3,4))=0$ and set $T((1,0,0))=(1,0,0,0)$, say.  Then just demand that $T$ be linear.
