Does there exist any method that can be used to calculate or evaluate $A$ given $Ax=y$? I have a simple equation like as $Ax=y$, where $A\in\mathbb{R}^{m\times n},x\in\mathbb{R}^n,y\in\mathbb{R}^m$. Given that $x,y$ are known a prior, does there exist any method that can be used to calculate or evaluate $A$? ($A$ is not a sparse matrix)
 A: There are infinitelly many matrices $A$ which satisfy your condition. This is because $Ax=y$ is a system of $m$ linear equations, but $A$ has $m\times n$ variables. So no, you cannot calculate $A$ from $Ax=y$.
A: We can generate an arbitrary number of matrices for which this equation is true.  To show this, let $G_n \in GL(n)$ be any invertible $n \times n$ matrix whose first column is the vector $x$;  this implies that
$$
G_n \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = x. 
$$
Similarly, let $G_m \in GL(m)$ be any invertible $m \times m$ matrix whose first column is $y$.  Putting these two equations together, we get
$$
A G_n \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = G_m \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}
$$
or
$$
\underbrace{G_m^{-1} A G_n}_{\equiv B} \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.
$$
This implies that the first column of the matrix $B = G_m^{-1} A G_n$ must be $[ 1 \: 0 \: 0 \dots 0 ]^T$, but its other $n - 1$ columns can be anything you like.  
Thus, we can conclude that for any three matrices $B \in \mathbb{R}^{m \times n}, G_m \in GL(m), G_n \in GL(n)$ such that:


*

*the first column of $G_n$ is $x$; 

*the first column of $G_m$ is $y$; and 

*the first column of $B$ is $[ 1 \: 0 \: 0 \dots 0 ]^T$,


the matrix $A = G_m B G_n^{-1}$ will satisfy the equations $Ax = y$.
