How to undo linear combinations of a vector If $v$ is a row vector and $A$ a matrix, the product $w = v A$ can be seen as a vector containing a number of linear combinations of the columns of vector $v$. For instance, if
$$ 
v = \begin{bmatrix}1, 2\end{bmatrix}, \quad 
A = \begin{bmatrix}0 & 0 & 0 \\ 1 & 1 & 1\end{bmatrix}, \quad
w = vA = \begin{bmatrix}2, 2, 2\end{bmatrix} 
$$
read by columns, the matrix $A$ is saying: make 3 combinations of the columns of vector $v$, each of which consists of taking 0 times the first column and 1 time the second column.
Now, the goal is to reconstruct, to the extent of possible, vector $v$ from $A$ and $w$, in other words to find a vector $v'$ such that $$v'A = w .$$ 
Two things to consider:


*

*The matrix $A$ can have any number of columns and may or may not be square or invertible.

*There are times when elements of the original vector can't be known, because $w$ contains no information about them. In the previous example, this would be the case of $v_1$. In this case, we would accept any value of $v'_1$ as correct.
How would you approach this problem? Can $v'$ be found doing simple operations with $w$ and $A$ or do I have to invent an algorithm specifically for the purpose?
 A: Given $A$ and $b$, you are trying to solve for $x$ in $xA=b$. This is just solving a system of linear equations, and the usual methods apply. Well, usually the problem is presented as $Ax=b$, but that just means instead of doing elementary row operations you'll be doing elementary column operations. Or, you could take the transpose and solve $A^tx^t=b^t$, and then you'd be back in line with the rest of us. 
A: Clearly
$$  \begin{bmatrix}0 & 2\end{bmatrix}
    \begin{bmatrix}0 & 0 & 0 \\ 1 & 1 & 1\end{bmatrix}
  = \begin{bmatrix}2 & 2 & 2\end{bmatrix}$$
So $v'=[0 \; 2]$ is a solution.
So we can suppose than any other solution can look like
$v'' = v' + [x \; y]$.
\begin{align}
   (v' + [x \; y])A &= [2 \; 2 \; 2] \\
   v'A + [x \; y]A &= [2 \; 2 \; 2] \\
   [2 \; 2 \; 2] + [x \; y]A &= [2 \; 2 \; 2] \\
   [x \; y]A &= [0 \; 0 \; 0] \\
   [y \; y \; y] &= [0 \; 0 \; 0] \\
   y &= 0
\end{align}
So the most general solution is
$v'' = v'+  \begin{bmatrix}x & 0\end{bmatrix} = \begin{bmatrix}x & 2\end{bmatrix}$
