Thread: Product of two non-square matrices is identity

1. Product of two non-square matrices is identity

Question: Suppose $A$ is $m \times n$ with rank $m$. Prove there exists $n \times m$ matrix $B$ such that $AB = I_m$.

I am kind of stuck on this. Any hints on which properties or theorems on rank of matrices that I should appeal to? Thanks!

2. Originally Posted by Last_Singularity
Question: Suppose $A$ is $m \times n$ with rank $m$. Prove there exists $n \times m$ matrix $B$ such that $AB = I_m$.

I am kind of stuck on this. Any hints on which properties or theorems on rank of matrices that I should appeal to? Thanks!
let $\mathcal{B}=\{\bold{e}_i, \ 1 \leq i \leq m \}$ be the standard basis for $\mathbb{R}^m$ and let $T: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ be the linear map with $[T]_{\mathcal{B}}=A.$ this map is onto because $\text{rank}(A)=m.$

so for any $1 \leq i \leq m,$ there exists $\bold{x}_i \in \mathbb{R}^n$ such that $T(\bold{x}_i)=\bold{e}_i.$ now define $S: \mathbb{R}^m \longrightarrow \mathbb{R}^n$ by $S(r_1 \bold{e}_1 + \cdots + r_m \bold{e}_m)=r_1 \bold{x}_1 + \cdots + r_m \bold{x}_m.$ see that

$S$ is a linear map and $TS = \text{id}.$ let $B=[S]_{\mathcal{B}}.$ we have $AB=I_m$ because $TS = \text{id}.$