# Why matrix with full column rank has a left inverse?

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.

• Is $P$ a square matrix? Commented Nov 14, 2019 at 21:09
• Not necessarily. Let me write down the whole sentence. $A$ be a $m\times n$ matrix with rank $r$. Then $(P, Q)$ is said to be a rank factorization of $A$ if $A = PQ$, where $P$ has full column rank and $Q$ has a full row rank. Since $P$ has a full column rank and $Q$ has a full row rank, there are matrices $C$ and $D$ such that $CP=I$ and $QD=I$. Commented Nov 14, 2019 at 21:10

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 $$P$$.
• Thank you for the clear explanation. Also, I noted that $(P^TP)^{-1}P^T$ is a left inverse of $P$. Commented Nov 14, 2019 at 21:33
• That only works for real numbers, where $A$ and $A^TA$ always have the same kernel dimension (thus the same rank). In particular, if $A$ is a real matrix with full column rank, then $A^TA$ is a square matrix with full rank so is invertible. Then $((A^TA)^{-1}A^T)A=(A^TA)^{-1}(A^TA)=I$. Commented Nov 14, 2019 at 21:37