i need to prove the following:

let $\displaystyle A\in \mathbb{R}^{n\times n}$.

$\displaystyle rank(A)=n \Leftrightarrow det(A)\neq 0$

this is what i've tried so far:

($\displaystyle \Rightarrow $)

$\displaystyle rank(A)=n$ iff A is invertible.

since A is invertible, it can be presented as a product of elementary transformations:

$\displaystyle A=E_1\cdot E_2\cdot ...\cdot E_k$

and then: $\displaystyle det(A)=det(E_1\cdot E_2\cdot ...\cdot E_k)=det(E_1)\cdot det(E_2)\cdot ...\cdot (E_k)$

now, since E is either $\displaystyle E_{i,j}(t)$, $\displaystyle E_{i,j}$, or $\displaystyle E_i(\lambda )$ its determinant is 1, -1 or $\displaystyle \lambda$ respectivly, then:

$\displaystyle det(E_1)\cdot det(E_2)\cdot ...\cdot det(E_k)\neq 0$

my problem is with my last sentence: i'm kinda using what i need to prove in my proof...

that's cheating.

i can really use a guidance here.

plus, how can i prove the other direction? ($\displaystyle \Leftarrow $)

thanks in advanced!