Originally Posted by
rubixcircle Let β={u1, u2, ... , un} be a subset of F^n containing n distinct vectors and let B be an nxn matrix in F having uj as column j.
Prove that β is a basis for Fn if and only if det(B)≠0.
For one direction of the proof I discussed this with a peer:
Since β consists of n vectors, β is a basis if and only if these vectors are linearly independent, which is equivalent to the map L_B being one-to-one. Since the matrix B is square, this is in turn equivalent to B being invertible, hence having a nonzero determinant.
However I do not understand the transition from the vectors being linearly independent to being one to one. Why is this true? Also, how do I prove the reverse direction?