Basis and Determinant Proof

  • #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?
 

Answers and Replies

  • #2
35,393
7,270
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?

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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