Let A be an n x n matrix. Denote it's columns b v_{1}, v_{2}, ... v_{n}. Keep in mind these are column vectors in R^{n}.

Prove that A is invertible if and only if {v_{1}, v_{2}, ... v_{n}} spans R^{n}.

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- Apr 25th 2013, 04:12 PMwidenerl194proof using invertible and spanning
Let A be an n x n matrix. Denote it's columns b v

_{1}, v_{2}, ... v_{n}. Keep in mind these are column vectors in R^{n}.

Prove that A is invertible if and only if {v_{1}, v_{2}, ... v_{n}} spans R^{n}. - Apr 25th 2013, 05:24 PMGusbobRe: 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.

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

iff

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

iff

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

iff

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

iff

$\displaystyle A$ is invertible. - Apr 25th 2013, 05:26 PMHallsofIvyRe: proof using invertible and spanning
First show that if you take A times $\displaystyle \begin{bmatrix}1 \\ 0 \\ 0 \\ \cdot\cdot\cdot \\ 0\end{bmatrix}$ you get the first column of A, that if you take A times $\displaystyle \begin{bmatrix} 0 \\ 1 \\ 0 \\ \cdot\cdot\cdot \\ 0 \end{bmatrix}$ you get the second column, etc.

In other words, A maps the standard basis for R^{n}into the columns of A. Those vectors span the range of A.