How to find all transformations whose kernels are $\operatorname{span}(\begin{bmatrix} -1 & 1 & 2 \end{bmatrix}^{\textrm{T}})$? I saw this post with a similar problem, however, the answers don't touch on how to actually find the matrix beyond just eye-balling it (which is trivial for this example). So my question is: how do I algorithmically find all such transformations?
Moreover, the responses only had matrices that either had 2 or 3 rows. Do the standard matrices have to be $2 \times 3$ or $3 \times 3$, or can it be $3 \times n$, where $n \ge 1$?
For matrix $A \in \mathbb{R}^{1 \times 3}$ and $\vec{v} = \operatorname{span}(\; \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} \;)$, we can have:
\begin{equation*}
A = k\begin{bmatrix} 1 & 1 & 0 \end{bmatrix} \ \ \ \mathrm{or} \ \ \
A = k\begin{bmatrix} 1 & 0 & 1/2 \end{bmatrix} \ \ \ \mathrm{or} \ \ \
A = k\begin{bmatrix} 0 & 1 & -1/2 \end{bmatrix} \\  \text{for all} \ k \in \mathbb{R}\ \text{such that} \ A\vec{v} = \vec{0}.
\end{equation*}
This shows that it seems to work for $A \in \mathbb{R}^{1 \times 3}$? To continue, for $B \in \mathbb{R}^{2 \times 3}$, I'm assuming we basically do the same thing by simply placing all combinations of $A$ onto their own rows in $B$?
 A: I could be wrong, but I think if your matrix isn't square, it isn't a linear transformation, since it changes the dimension of the vector it's acting on.
To answer your other question, you need a square matrix to be a linear transformation, so construct a square matrix, say $$A := \begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{bmatrix}.$$
Next, as you said, if $\vec{v}$ is the vector which spans the kernel, then the kernel itself is the set of all $k\vec{v}, k\in\mathbb{R}$. So set $A(k\vec{v}) = [0&0&0]^T$ and you'll get signs on all the $a_{ij}$s that should hopefully give you a hint of what your matrix $A$ should look like.
A: The ideas the following: when we want to find a linear transformation, all we need to know is which starting vector space to which target vector space we want the mapping to be made, a basis for the starting vector space and to know how the linear transformation affects these basis vectors. If we know these components, the idea is to use the definition directly to make the linear transformation.
Since your question seems to be an attempt to understand this idea, there are still things you need to specify (as the target vector space). In the specific case that you want to find a linear transformation $A:{\bf R}^3\to {\bf R}^2$ such that $\ker(A)={\rm span}\{(-1,1,2)\}$ so first we need a basis (not the basis) for ${\bf R}^3$ but since we have the kernel and we know that the kernel is always subspace of the starting vector space we can use this vector to add our basis of ${\bf R}^{3}$ call it $$\beta=\{(-1,1,2),v_2,v_3\}$$ and noticed that we have vectors $v_2,v_3$ in $\beta$ its raison to be is that in order to have a basis for ${\bf R}^3$ we need three vectors spanning and linearly independents. That is all that its exercise imposes as condition, so we have some freedom to select these vectors $v_2$ and $v_3$ for convenience we take the canonical ones and we have $$\beta=\{(-1,1,2),(0,1,0),(0,0,1)\}$$
Well, we already have a base of ${\bf R}^3$ that also satisfies the condition that it imposes. Why? Well, by definition we know that $A(-1,1,2)=(0,0)$ (why?). However, for now we do not know $A(0,1,0)$ and $A(0,0,1)$ and since your problem doesn't add more restrictions, we again have some freedom but now a bit conditioned  since we can't even send these vectors to the kernel (why?), so we setting
$A(0,1,0)=(1,0)$ and $A(0,0,1)=(0,1)$.
Summarizing everything, then we have

*

*Linear transformation: $A:{\bf R}^3\to {\bf R}^2$.

*A basis: $\beta=\{(-1,1,2),(0,1,0),(0,0,1)\}$.

*Conditions: $A(-1,1,2)=(0,0), A(0,1,0)=(1,0)$ and $A(0,0,1)=(0,1)$.

Using all it, now we can use the definition of linear transformation to make the rule for $A$, finding $\alpha_1,\alpha_2$ and $\alpha_3$ in
$$\begin{bmatrix}a\\b\\c\end{bmatrix}=\alpha_{1}\begin{bmatrix}-1\\1\\2\end{bmatrix}+\alpha_2\begin{bmatrix}0\\1\\0\end{bmatrix}+\alpha_{3}\begin{bmatrix}0\\0\\1\end{bmatrix}$$
I will let you solve the linear system and find $\alpha_1,\alpha_2$ and $\alpha_3$.
Finally, the definition the linear transformation give
$$A\begin{bmatrix}a\\b\\c\end{bmatrix}=\alpha_{1}A\begin{bmatrix}-1\\1\\2\end{bmatrix}+\alpha_2A\begin{bmatrix}0\\1\\0\end{bmatrix}+\alpha_3\begin{bmatrix}0\\0\\1\end{bmatrix}$$
I think you can conclude.
Now I think you can see that there is nothing special in ${\bf R}^2$ and that the same idea can be used with another target vector space as ${\bf R}^{n}, M_{m\times n}({\bf R}), P_{n}({\bf R})$ etc.
