Is the set of all underdetermined matrices A that solve Ax = b (for a given x and b) dense? Let's say I have two fixed vectors $x \in \mathbb{R}^{n}$ and $b \in \mathbb{R}^{m}$, where $m < n$. There are an infinite number of matrices $A \in \mathbb{R}^{m \times n}$ that can solve this underdetermined system. My question is: are these solutions dense over the set of all matrices in $\mathbb{R}^{m \times n}$?
Please be gentle, I am not very mathematical.
 A: Take $\vec{x}=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}, \vec{b}=\begin{pmatrix}1\\ 0\end{pmatrix},$ and $A=\begin{pmatrix}1&1&1\\ 1&1&1\end{pmatrix}$. Let $\{A_n\}_{n=1}^{\infty}$ be any sequence of $2\times 3 $ matrices such that $A_n\vec{x}=\vec{b}$ for all $n \geq 1$. Assume $A_n \longrightarrow A$ with respect to $L^2$ norm $\|\cdot \|$. Find $N$ so that $n \geq N$ implies $\|A_n -A\|<1$. Then for any $n\geq N$ $$\begin{eqnarray*}1&=&\Bigg\|\begin{pmatrix}1\\ 0\end{pmatrix}-\begin{pmatrix}1\\ 1\end{pmatrix}\Bigg\| \\ &=& \|A_n\vec{x}-A\vec{x}\| \\ &
= & \|(A_n-A)(\vec{x})\| \\ &\leq & \|A_n-A\|\|x\|\\ &=& \|A_n-A\| \\&<& 1\end{eqnarray*}$$ which is a contradiction, so we don't necessarily have density.
A: No it is not dense.
The simplest example is to choose $\vec{b}$ to have very large magnitude, and to take $\vec{x}$ to have very small magnitude. Then the matrix $A$ will have at least one large entry, so there is no matrix close to the $0$ matrix.
To be more precise, it is helpful to reshape this system in the form
$$
\begin{bmatrix}
\vec{x}^T & \vec{0}^T & \vec{0}^T & \dots \\
\vec{0}^T & \vec{x}^T & \vec{0}^T & \dots \\
\vec{0}^T & \vec{0}^T & \vec{x}^T & \dots \\
\vdots & \vdots & \vdots & \ddots
\end{bmatrix}
\begin{bmatrix} a_1 \\ a_2 \\ \vdots \end{bmatrix}
= \vec{b}
$$
Where $a_j^T$ is the $j$th row of the matrix $A$. This is helpful because it poses the unknown variables as the unknown column vector in an underdetermined linear system, so the solution space is a hyperplane in $\mathbb{R}^{(nm)}$ and can't be dense.
A: The set of all $A$ that satisfy $Ax=b$ for fixed $x,b$ is closed so is only dense if it equals all matrices (e.g. when $x=0$ and $b=0$).
A: The set of $m\times n$ matrices is a vector space, and the map $A\mapsto Ax$ is linear. Hence the set of solutions (in $A$) of $Ax=b$ is an affine subspace of the set of $m\times n$ matrices, and it is not dense, unless it is the whole space (which is the case iff $x=0$ and $b=0$).
