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A matrix $A\in\mathbb{R}^{m,n}$ defines a linear mapping $f:\mathbb{R^n}\rightarrow\mathbb{R}^m$ by regarding elements of $\mathbb{R}^n$ as column vectors and setting $f(\textbf{v})=A\textbf{v}, \textbf{v}\in\mathbb{R}^{n,1}$. Since the linear function I am looking for is an endomorphism of $\mathbb{R}^3$, then $A\in\mathbb{R}^{3,3}$.

From the definition of null space one has: $\ker(f)=\{\textbf{v}\in V|f(\textbf{v})=\textbf{0}\}$ where $\textbf{0}$ is the zero-vector with the respect to the codomain basis. Hence, I though of the matrix: \begin{equation} A=\begin{pmatrix} 1&-4&0\\0&0&1 \end{pmatrix} \end{equation} It is not a $3\times3$ square matrix though so it is not correct as I need one more row vector.

Is my line of reasoning correct? How can the solution be found?

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    Try $(0,0,2)$ or $(2,-8,0)$ as the other row, which are linearly dependent of the others so the kernel will be the same – 2017-01-27
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    $x-4y-z=0$ is the equation of a plane not a line, so what do you mean? – 2017-01-27
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    There are many such matrices corresponding to different choices for the image of this map. – 2017-01-27
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    @mannav I thought about it but shouldn't the matrix be non singular? If there's a linear dependence the determinant would be zero. – 2017-01-28

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Hint: Consider a plane passing through the origin and orthogonal to your given line.

Let $W$ be such a plane.

Consider the map, $f:\Bbb R^3 \to \Bbb R^3 $ to be a projection onto $W$

The matrix corresponding to $f$ will serve your purpose.