# Partitioning rectangular matrix to get invertible square matrix?

It seems like a very stupid question but I'm somehow not able to show this.

Let $A$ be an $m\times n$ real matrix, $m < n$, with all its $m$ rows linearly independent. In this case can I always find a set of $m$ columns, say, $C_1,...,C_m$ such that $[C_1 ... C_m]$ is an invertible square matrix? If yes how do I show this?

Context:

I'm trying to describe the null space of $A$. If the above is true I can find it as follows:

$Ax=0 \Rightarrow (A_m \; A_{-m}) \begin{bmatrix}x\\y\end{bmatrix} = 0$,

where $A_m$ is the aforementioned square matrix, if it exists, $A_{-m}$ is the rest of matrix $A$ (i.e. $A$ is partitioned into $A_m$ and $A_{-m}$), and $x \in \mathbb{R}^m, y \in \mathbb{R}^{(n-m)}$.

If $A_m$ is invertible the null space can be described as:

$x= \begin{bmatrix}-A^{-1}_mA_{-m}\\I_{n-m}\end{bmatrix}c, \; c \in \mathbb{R}^{(n-m)}$

Note that since $A$ has $m$ rows linearly independent so $\text{Rank }A=m$ and it is always possible to find $m$ linearly independent columns since
$\text{Rank A=Row Rank (A)=Column Rank(A)=m}$
• No that can't be.Take $\Bbb R^2$ for example; $\dim \Bbb R^2=2$ does it imply that any two vectors in $\Bbb R^2$ are linearly independent. Commented Oct 24, 2017 at 15:46