Find a linear mapping $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ such that its kernel is $\mathrm{ker}(f) = \{ x-y+z=0 \}$ I know that, given a linear function $f:V\rightarrow W$, its kernel or null space is defined as: $ker(\varphi)=\{v\in V|\varphi(v)=0\}\subset V$ and the rank-nullity theorem. 
The answer to this question is $ f(x, y, z) = (x − y + z, 0, 0)$ but I cannot see how you come to this conclusion. 
If anyone can clarify this to me, I'd be very thankful.
 A: First of all you can look for a basis of the nullspace of $f$, i.e. $\ker(f)\subset\mathbb{R}^3$. 
Since the associated homogeneous system yields $\infty^2$ solutions, it follows that $\mathcal{B}_{\ker(f)}:(1,1,0),(-1,0,1)$ and $\dim[\ker(f)]=2$. 
To have a basis of $f$, then complete $\mathcal{B}_{\ker(f)}$ by choosing a third linear independent vector, e.g. $(1,0,0)$.
Since, by definition of $\ker(f)$ it is known that $f(1,1,0)=f(-1,0,1)=(0,0,0)$, it is left to associate a non-zero vector to the map of the lasty introduced vector. In general, 
\begin{equation} f(1,0,0)=(a,b,c)\neq\mathbf{0} \end{equation}
Then, having chosen for simplicity the canonical basis of $\mathbb{R}^3$ $\mathcal{K}_3=(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$, its vector are written as linear combination of those of $\mathcal{B}_f$. 
After denoting the three vectors of $\mathcal{B}_f$ as $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$, i.e. 
\begin{equation}\mathbf{v}_1=(1,1,0)\end{equation}
\begin{equation}\mathbf{v}_2=(-1,0,1)\end{equation}
\begin{equation}\mathbf{v}_3=(1,0,0),\end{equation} 
one has:
\begin{equation}f(\mathbf{e}_1)=f(1,0,0)=f(\mathbf{v}_3)=(a,b,c)\end{equation}
\begin{equation}f(\mathbf{e}_2)=f(0,1,0)=f(\mathbf{v}_1)-f(\mathbf{v}_3)=(0,0,0)-(a,b,c)=(-a,-b,-c)\end{equation}
\begin{equation}f(\mathbf{e}_3)=f(0,0,1)=f(\mathbf{v}_2)+f(\mathbf{v}_3)=(0,0,0)+(a,b,c)=(a,b,c)\end{equation}
The values inferred this way are the columns of tha matrix associated to $f$ through the canonical base:
\begin{equation}M_{\mathcal{K}}^f=\begin{pmatrix} a&-a&a\\b&-b&b\\c&-c&c\\
\end{pmatrix}\end{equation}
Finally, arbitrarly pick $(a,b,c)=(1,0,0)$.
\begin{equation}M_{\mathcal{K}}^f=\begin{pmatrix} 1&-1&1\\0&0&0\\0&0&0\\
\end{pmatrix}\end{equation}
corrisponding to the function you looked for: $f(x,y,z)=(x-y+z,0,0)$.
A: There is not one solution there are an infinity. Take any $0\neq\mathbf{v}\in\Bbb{R}^3$. One has
$$\begin{align}\varphi :\Bbb{R}^3&\to\Bbb{R}^3\\(x,y,z)&\to f(x,y,z)\cdot\mathbf{v}\end{align}$$
$\varphi$ is obviously linear and $\ker{\varphi}$ is $\{(x,y,z)\in\Bbb{R}^3,\, f(x,y,z)=0\}$
The answer given by the O.P corresponds to $\mathbf{v}=(1,0,0)$
