Suppose $P$ is a matrix with a full column rank. Why there is a matrix $C$ such that $CP=I$?
Let's say the rank of $P$ is $r$. I understand that since the column space $C(P)$ equals the column space $C(I_r)$. Then each column of $P$ can be transferred to each column of $I_r$, i.e. $$[p_1, p_2, \ldots, p_r](\alpha_{11},\alpha_{21},\cdots,\alpha_{r1})^{T}=(1,0,\cdots,0)^T,$$where $p_i$ means the $i^{\rm th}$ column of $P$. As a result, there should be a matrix $A$ such that $PA=I_r$, which is actually a right inverse. I don't understand why $P$ has a left inverse.
Best Answer
I’m assuming, for the sake of simplicity, the scalars to be real numbers.
Let $P$ be a $m \times n$ matrix with full column rank, so $n$ is the rank of $P$.
In the canonical bases, $P$ represents an morphism $u: \mathbb{R}^n \rightarrow \mathbb{R}^m$. Since $P$ has full column rank, $u$ is injective, that is, $(u(e_1),u(e_2),\ldots,u(e_n))$ is free, so there are vectors $s_1,\ldots,s_{m-n} \in \mathbb{R}^m$ completing that real family into a basis.
We can define a morphism $v:\mathbb{R}^m \rightarrow \mathbb{R}^n$ by specifyig its values on the aforementioned basis: let us require $v(s_k)=0$ and $v(u(e_l))=e_l$.
Then the matrix $C$ representing $v$ in the canonical bases is a left inverse for $P$.