Proving Rank A=n Implies Basis of Rn for Matrix A

[stunner5000pt]

Let A be an m x n matrix with columns C1, C2, ... Cn. If rank A = n show taht $$\{ A^{T}C_{1}, A^{T}C_{2}, ... , A^{T}C_{n} /}$$ is a basis of Rn.

ok $$\mbox{rank} A^{T} = n$$
the columns of A are rows of A transpose
im not sure how to proceed though...
a column times itself with $$C_{1}^2 + C_{2} C_{1} + ... + C_{n}C_{1}$$ for the first term of $$A^{T} C_{1}$$ is the rank maintained through this multiplication? What justifies that?

help is greatly appreciated!

thank you!

 

[StatusX]

What (product of) matrix (-x + ces) has the A^TC_i as its columns? And what property does an n x n matrix have to have for its rows to form a basis of R^n?

 

[stunner5000pt]

what do u mean (-x + ces) ?
arent the rows of a square matrix A linearly independent if they form a basis for Rn?

 

[StatusX]

Sorry, it was supposed to be like "giraffe(s)," but that's not as easy when the word ends in an x. Anyway, right, they form a basis if the matrix containing them as columns has rank n.

 

[stunner5000pt]

Sorry, it was supposed to be like "giraffe(s)," but that's not as easy when the word ends in an x. Anyway, right, they form a basis if the matrix containing them as columns has rank n.

ok...
im still not sure to do with the -x + Ci part.