Find a formula for a linear transformation Find an example of a linear transformation formula $\varphi$ so that:
$$\ker\varphi = \{(x,y,z,t) \in \mathbb{R}^4 : x-y+6z+2t=0\},$$
$$\operatorname{im}\varphi = \operatorname{span}((2,3,1))$$
How to approach such problem? The most standard way to is appreciated.
 A: $\varphi$ is a linear transformation $\mathbb{R}^4\rightarrow\mathbb{R}^3$, so the matrix $A$ representing $\varphi$ (with respect to standard basis) is $3$ by $4$. Now, if
$$\ker\varphi=\{(x,y,z,t)\in\mathbb{R}^4:x-y+6z+2t=0\}$$
then everything in the kernel of $A$ is orthogonal to $(1,-1,6,2)$, so let's set
$$A=\begin{bmatrix}1&-1&6&2\\ ?&?&?&?\\?&?&?&?\end{bmatrix}.$$
We are not done yet, because we haven't specified the remaining entries. But this is not difficult, because we know
$$\text{im}\varphi=\text{span}((2,3,1))$$
which implies that all the column vectors are scalar multiples of $(2,3,1)$. So for example, the first column is just $1/2$ times $(2,3,1)$, which gives
$$A=\begin{bmatrix}1&-1&6&2\\ 3/2&?&?&?\\1/2&?&?&?\end{bmatrix}.$$
Continuing this logic, we can fill out the last three columns similarly, giving us
$$A=\begin{bmatrix}1&-1&6&2\\ 3/2&-3/2&9&3\\1/2&-1/2&3&1\end{bmatrix}.$$
Now we are done.
A: Observe that $\{(x,y,z,t) \in \mathbb{R}^4 : x-y+6z+2t=0\}$ is the set of all vectors of the form $$(y-6z-2t,y,z,t) = y(1,1,0,0) + z(-6,0,1,0) + t(-2,0,0,1)$$
where $y,z$ and $t$ runs over all the real numbers. So, choose a linear map $\varphi : \mathbb R^4 \to \mathbb R^3$ such that
$$\varphi(1,1,0,0) = \varphi(-6,0,1,0) = \varphi(-2,0,0,1) = 0$$
and $\varphi(v) = (2,3,1)$ for some $v \in \mathbb R^4$ which is not in the span of $$\{(1,1,0,0),(-6,0,1,0),(-2,0,0,1)\}.$$
A: The following matrix describes such a one:  $\begin{pmatrix} 2&-2&12&4\\3&-3&18&6\\1&-1&6&2\end{pmatrix}$.
A: In standard basis, we  need to find a $4\times 3$ matrix  which has rank $1$, such that all column vectors are collinear to the vector $^{\mathrm t\mkern-1.5mu}(2,3,1)$ and also such that all row vectors (which are collinear) are collinear to the vector $(1,-1,6,2)$.
To make it simple, we're  looking for a matrix of the form
$$\begin{bmatrix}
1&-1&6&2 \\
. & . & . & 3 \\
. & . & . & 1
\end{bmatrix}$$
Therefore, we can start from the first row and multiply it by the relevant coefficients to obtain the last column:
To make it simple, we're  looking for a matrix of the form
$$\begin{bmatrix}
1&-1&6&2 \\
\frac 32 & -\frac32  & 9 & 3 \\
\frac12 & -\frac12  & 3 & 1
\end{bmatrix},\quad\text{or}\quad\begin{bmatrix}
2 & -2 & 12 & 4 \\
 3 & -3  & 18 & 6 \\
 1 & -1  & 6 &  2
\end{bmatrix}\quad\text{if we prefer integer coefficients.}$$
