# Thread: proof using invertible and spanning

1. ## proof using invertible and spanning

Let A be an n x n matrix. Denote it's columns b v1, v2, ... vn. Keep in mind these are column vectors in Rn.

Prove that A is invertible if and only if {v1, v2, ... vn} spans Rn.

2. ## Re: proof using invertible and spanning

I have no idea what you know and can use. But here is a quick proof using things I feel ought to be done before proving this theorem.

$A$ spans $\mathbb{R}^n$

iff

$Ax=b$ is consistent for all $b \in \mathbb{R}^n$

iff

$A$ has $n$ pivot rows (and also $n$ pivot columns since $A$ is square)

iff

$A$ is row equivalent to the $n\times n$ identity matrix

iff

$A$ is invertible.

3. ## Re: proof using invertible and spanning

First show that if you take A times $\begin{bmatrix}1 \\ 0 \\ 0 \\ \cdot\cdot\cdot \\ 0\end{bmatrix}$ you get the first column of A, that if you take A times $\begin{bmatrix} 0 \\ 1 \\ 0 \\ \cdot\cdot\cdot \\ 0 \end{bmatrix}$ you get the second column, etc.

In other words, A maps the standard basis for Rn into the columns of A. Those vectors span the range of A.