Let $A \in \mathcal M_{m×n}(F)$. Prove that if $\text{rank}(A) = m$, then there exists $B \in \mathcal M_{n×m}(F)$ such
that $AB = I_m$.
I think I need to prove that $A^{-1}$ exists, such that $B=A^{-1}$, but I thought that only square matrices are invertible. Is there another way of approaching this problem? Any guidance is appreciated. Thank you!
Best Answer
Here's one way to think about the problem: note that we necessarily have $m \leq n$ (why?). When $m = n$, this is a simple question of matrix inversion. When $m<n$, we may state that $A$ has some set of $m$ columns that are linearly independent.
Let $i_1,i_2,\dots,i_m$ be the (indices of the) columns of $A$ corresponding to a linearly independent set of columns (without repetition). Let $e_i$ be the $i$th standard basis vector. Noting that $A e_i$ is the $i$th column of $A$, we note that $$ A\pmatrix{e_{i_1}&e_{i_2}&\cdots & e_{i_m}} $$ Gives us an $m \times m$ matrix with full rank. Since this is a square matrix with full rank, it is invertible.
Let $B$ be given by $$ B = \pmatrix{e_{i_1}&e_{i_2}&\cdots & e_{i_m}} \left[ A\pmatrix{e_{i_1}&e_{i_2}&\cdots & e_{i_m}}\right]^{-1} $$ Confirm that $AB = I_m$