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.
$\varphi$ 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
$$A=\begin{bmatrix}1&amp;-1&amp;6&amp;2\\ ?&amp;?&amp;?&amp;?\\?&amp;?&amp;?&amp;?\end{bmatrix}.$$
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
$$A=\begin{bmatrix}1&amp;-1&amp;6&amp;2\\ 3/2&amp;?&amp;?&amp;?\\1/2&amp;?&amp;?&amp;?\end{bmatrix}.$$
Continuing this logic, we can fill out the last three columns similarly, giving us
$$A=\begin{bmatrix}1&amp;-1&amp;6&amp;2\\ 3/2&amp;-3/2&amp;9&amp;3\\1/2&amp;-1/2&amp;3&amp;1\end{bmatrix}.$$
Now we are done.
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
$$\begin{bmatrix}
1&amp;-1&amp;6&amp;2 \\
. &amp; . &amp; . &amp; 3 \\
. &amp; . &amp; . &amp; 1
\end{bmatrix}$$
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
$$\begin{bmatrix}
1&amp;-1&amp;6&amp;2 \\
\frac 32 &amp; -\frac32  &amp; 9 &amp; 3 \\
\frac12 &amp; -\frac12  &amp; 3 &amp; 1
\end{bmatrix},\quad\text{or}\quad\begin{bmatrix}
2 &amp; -2 &amp; 12 &amp; 4 \\
 3 &amp; -3  &amp; 18 &amp; 6 \\
 1 &amp; -1  &amp; 6 &amp;  2
\end{bmatrix}\quad\text{if we prefer integer coefficients.}$$