I have the matrix equation of real coefficients
$Ax = y$
where $y$ and $x$ are of different size. Is it possible to do a simple operation which will turn $A$ into a matrix function of both $y$ and $x$? I guess I can multiply on the right side with $x^{\top}$ and divide by the norm to yield the identity-matrix, but is there a smoother way to proceed?
On
Of course. Plug \begin{equation} x=\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0\end{bmatrix}, \end{equation}
where $1$ is in the $j$-th position, into the equation. Now $y$ gives you the $j$-th column of $A$. You will need to do this for as many times as the columns of $A$, of course.
On
For an $n\times m$ matrix, the system
$$y=Ax$$
has $nm$ unknowns (the $a_{ij}$) for $n$ equations. This is blatantly insufficient.
(By the way, $xx^T$ cannot be invertible as all columns are proportional to each other and the rank is at most $1$.)
On
No, you can't isolate $A$. As a concrete example, take $$A_1 = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} \quad A_2 = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \quad x = y = \begin{pmatrix}1 \\ 0 \end{pmatrix}.$$ Note that $y = A_1x$ and $y = A_2x$. So, knowing that $y = A x$ cannot tell you a specific matrix, as it could equally be $A_1$ or $A_2$ (or indeed, any $2 \times 2$ matrix whose first column is $x$).
Multiplying both sides by $x^\top$ will not help, in general. The matrix $xx^\top$ is a $2 \times 2$ rank $1$ matrix (in this case, coincidentally equal to $A_2$). It has no inverse. So, while it's true that $A(xx^\top) = yx^\top$, there is no unique way to solve for $A$, as the example shows.
However, what you can do is reconstruct $A$ if you know the value of $Ax$ on a basis. So, if you know the value of $A\begin{pmatrix}1 \\ 0 \end{pmatrix}$ and $A\begin{pmatrix}0 \\ 1 \end{pmatrix}$, then you know the whole matrix (indeed, in the case of the standard basis, the former is just the left column of $A$, while the latter is the right column of $A$).
In general, if $(v_1, \ldots, v_n)$ is a basis of $\Bbb{R}^n$ (or $\Bbb{C}^n$, or whatever other field), then the matrix $$B = \left(\begin{array}{c|c}&&&\\v_1 & v_2 & \dots & v_n \\ &&&\end{array}\right)$$ is an $n \times n$ invertible matrix. Moreover, $$AB = \left(\begin{array}{c|c}&&&\\Av_1 & Av_2 & \dots & Av_n \\ &&&\end{array}\right),$$ so if you know the value of $Av_i$ for all $i$, then you know the value of $AB$ and the value of the invertible matrix $B$. Compute the inverse of $B$, and you can easily isolate $A$.
On
As others have noted, the solution $A$ will not be unique. That being said, here's a nice formula for an $A$ that works in terms of $x$ and $y$ using a scaled Householder transformation. If $x$ and $y$ are non-zero vectors with $\hat x = \frac{x}{\|x\|}$ and $\hat y = \frac{y}{\|y\|}$, then we can take $$ A = \begin{cases} \frac{\|y\|}{\|x\|}(I - 2\frac{vv^T}{v^Tv}) &\hat x \neq \hat y\\ \frac{\|y\|}{\|x\|}I & \hat x = \hat y \end{cases} $$ where $v = \hat x - \hat y$.
Does this help?!
Let $ A\begin{bmatrix} 2 \\ 3 \end{bmatrix} =\begin{bmatrix} 10 \\ -2 \\ 5 \end{bmatrix}$, then we can find $A$ by assuming $$A=\begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix}$$ we get 3 equations $$2a+3b=10,~ 2c+3d=-2,~ 2e+3f=5$$ as there are 6 unknowns one will get many solutions and hence many matrices.