Finding a linear transformation with a given null space The problem statement is,
Find a linear transformation $T: \mathbb R^3 \to \mathbb R^3$ such that the set of all vectors satisfying $4x_1-3x_2+x_3=0$ is the (i) null space of $T$ (ii) range of $T$.
For (i),
I found out the basis of the null space for the system,
$$\begin{pmatrix}4&-3&1\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =0$$
viz, $\mathbf v_1 = \begin{pmatrix}3\\4\\0\end{pmatrix} , \mathbf v_2 = \begin{pmatrix}-1\\0\\4\end{pmatrix}$
So, basically, I have to find linear transformation such that $T\begin{pmatrix}3\\4\\0\end{pmatrix} =0$ and $T\begin{pmatrix}-1\\0\\4\end{pmatrix}=0$ such that vector $\mathbf v \in span\{\mathbf v_1, \mathbf v_2\}$ satisfies $T\left(\mathbf v\right) =0$
Now, I'm stuck at this point. I'm not able to describe the such linear transformations.
Further, for (ii), I don't know how to approach the problem.
 A: For (i), consider the matrix
$$ A = \begin{pmatrix} 4 & -3 & 1 \\ 4 & -3 & 1 \\ 4 & -3 & 1 \end{pmatrix}. $$
Denoting by $T_A \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ the corresponding linear map $T_A(v) = Av$, we have
$$ T_A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4x_1 - 3x_2 + x_3 \\ 4x_1 - 3x_2 + x_3 \\ 4x_1 - 3x_2 + x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
so $\ker(T_A)$ is precisely the solution subspace of the equation $4x_1 - 3x_2 + x_3$.
For (ii), consider the matrix
$$ B = \begin{pmatrix} 3 & -1 & 0 \\ 4 & 0 & 0 \\ 0 & 4 & 0 \end{pmatrix} $$
whose columns are $\bf{v_1},v_2$ and the zero vector. The image of $T_B$ is spanned by the columns of $B$ and thus again is equal to the solution subspace of your equation.
A: Any linear transformation is completely determined by its action on a basis.
For the first question extend the set $\{v_1, v_2\}$ to a basis for $\mathbb{R}^3$ by adjoining a vector $v_3$ and specify a linear transformation with the required kernel by setting $T(v_3)$ to be a vector all of whose components are nonzero.
