I have matrix $A$ type $m\times n$, which has endless solutions (the original words are: the system $A(x)=b$ have solution to any $b$ vector). Now I need to prove that there is $Q$ matrix so: $AQ = I_m$ ($I$ type $m\times m$). Of course that $Q$ needed to be $n\times m$, but beyond that I couldn't find any pattern here.
Does any of you can help?
Here is one way to think about how to find a solution:
If you take an $n\times k$ matrix $B$ with columns $b_1, b_2, \ldots, b_k$, then the columns of the product $AB$ are $Ab_1, Ab_2, \ldots, Ab_k$.
You want $AQ$ to be the $m\times m$ identity matrix. That means that if $q_1, q_2, \ldots, q_m$ are the columns of $Q$, you want, for instance, that $$ Aq_1 = \begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix} $$ You are told that $A$ is such that a $q_1$ exists. Any other column of $Q$ can be found (or just shown to exist) similarily.