if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that $\displaystyle BA=I_n$? why or why not?
the second question is identical just switch m<n.
I dont understand this question at all. any help?
If $\displaystyle A$ is such a matrix so that $\displaystyle A^TA$ is invertible then it can be shown $\displaystyle B= (A^TA)^{-1}A^T$ satisfies $\displaystyle BA = I_n$.
For example, let $\displaystyle A=\begin{bmatrix}1&0\\0&1\\1&1 \end{bmatrix}$ and follow the computations above to convince yourself.
There is a similar formula for the second part as well.