Given an unknown square matrix $H$ and known vectors $x$ and $y$, is it possible to find $H$ if $H\vec{x}=\vec{y}$?

Question

If $H\vec{x}=\vec{y}$ can we find $H$ by inverting $\vec{x}$? No idea how to invert $\vec{x}$ though or if this equation is even solvable.

Answer 1

Let $H_{ij}$ denote the elements of the $m\times n$-matrix $H$. Then you are asking to solutions of the following linear system of equations:

$$\sum_{k=1}^n H_{1k}x_k = y_1$$ $$\sum_{k=1}^n H_{2k}x_k = y_2$$

and so on, with the final equation being

$$\sum_{k=1}^n H_{mk}x_k = y_m$$

Obviously, we have $n\times m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.

Answer 2

For every vector $\vec{x}$ exist a non zero vector $\vec{z}$ such that $\vec x \cdot\vec z=0$. Then we can build a new matrix $K$ with as many rows as number of $\vec y$ coordinates and each row of $K$ coincides with $\vec z$. Then $$ K\vec x=0\\ (H+K)\vec x=\vec y\\ H+K\neq H $$ then your statement is false because for every $\vec x, \vec y$ there aren't a single $H$ (there're infinitely many of them).

For example $$ H=\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right)\\ \vec x=\left(\begin{array}{c} 1\\ 2 \end{array}\right)\\ \vec y=\left(\begin{array}{c} 3\\ 2 \end{array}\right)\\ \vec z=\left(\begin{array}{c} 2\\ -1 \end{array}\right)\\ K=\left(\begin{array}{cc} 2 & -1\\ 2 & -1 \end{array}\right)\\ H+K=\left(\begin{array}{cc} 3 & 0\\ 2 & 0 \end{array}\right)\\ (H+K)\vec x=\left(\begin{array}{c} 3\\ 2 \end{array}\right)=\vec y $$