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Basis and Determinant Proof

  1. Jul 12, 2011 #1
    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?
     
  2. jcsd
  3. Jul 13, 2011 #2

    Mark44

    Staff: Mentor

    To answer your first question, β is assumed to be a basis for Fn, and the matrix B = [(u1) (u2) (u3) ... (un)] has as its columns the vectors in β.

    Let c be a row vector of scalars, {c1, c2, ..., cn}, and consider the equation Bc = 0.

    Since β is a basis (by assumption) of an n-dimensional space, its n vectors must be linearly independent, so the only solution to the equation Bc = 0 is c [itex]\equiv[/itex] 0. (I.e., c1 = 0, c2 = 0, ..., cn = 0.)

    From this, we see that ker(B) = {0}, and since B represents a map from Fn to Fn, B must be invertible, so |B| [itex]\neq[/itex] 0.
     
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