Krdytkyu

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

Let $H_{ij}$ denote the elements of the $mtimes n$-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 $ntimes m$ unknown elements (the entries of the matrix $H$) and only $m$ equations. This is, in general, not solveable uniquely.

For every vector $vec{x}$ exist a non zero vector $vec{z}$ such that $vec x cdotvec z=0$. 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
$$
Kvec x=0\
(H+K)vec x=vec y\
H+Kneq H
$$
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
$$
H=left(begin{array}{cc}
1 & 1\
0 & 1
end{array}right)\
vec x=left(begin{array}{c}
1\
2
end{array}right)\
vec y=left(begin{array}{c}
3\
2
end{array}right)\
vec z=left(begin{array}{c}
2\
-1
end{array}right)\
K=left(begin{array}{cc}
2 & -1\
2 & -1
end{array}right)\
H+K=left(begin{array}{cc}
3 & 0\
2 & 0
end{array}right)\
(H+K)vec x=left(begin{array}{c}
3\
2
end{array}right)=vec y
$$