Solving Av=w for A

Question

Let $\mathbf{v}$ and $\mathbf{w}$ be nonzero vectors in $\mathbb{R}^n$, with $n>1$.
Clearly, a matrix $\mathbf{A}$ satisfying $\mathbf{Av}=\mathbf{w}$ need not be unique.

For example:
$$\mathbf{A}[0.5,0.5]^{\top}=[-0.5,0.5]^{\top}\text{ holds for both a reflection matrix }\mathbf{A}\text{ and a rotation matrix }\mathbf{A}.$$

Can anything of interest be said about the set of matrices satisfying $\mathbf{Av}=\mathbf{w}$ for fixed, nonzero vectors $\mathbf{v}$ and $\mathbf{w}$? What if we restrict to just invertible matrices? This seems like a natural question to ask, but I'm not sure how to approach it.

Answer 1

Maybe, an interesting point of view can be the following. The product $Av$ is a linear combination of the columns of $A$. Furthermore, $v$ and $w$ have the same dimension. Then if $A$ has to be invertible, you are looking for the basis of $\mathbb{R}^n$ with respect to the vector $w$ is represented by the coefficients in $v$.

Answer 2

Suppose $A$ is an $m\times n$ matrix, and $v\in\mathbb{R}^n$, $w\in\mathbb{R}^m$.

The map $A\mapsto Av$ is linear and onto from $\mathbb{R}^{m\times n}\to\mathbb{R}^m$. Its kernel $K$ is the space of all matrices whose rows are orthogonal to $v^T$. Every fiber is a coset of this subspace. It suffices to find one element of the fiber to use as a coset representative. We may use $A=(w{v}^T)/|v|^2$ as one such representative. Thus,

$$ A=(u_1~\cdots~u_m)^T+(wv^T)/|v|^2 $$

for any vectors $u_1,\cdots,u_m\perp v$ characterizes all such $A$.