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Answer to Question #3459 in Linear Algebra for Mike

Question #3459
Let β={u1, u2, ... , un} be a subset of Fn 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 this is something that I found :

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 LB 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? Is there a better way to prove this? Also, how do I prove the reverse direction?
Expert's answer
We have that β=(u1,u2,…,un) is a basis for Fn. Let’s prove that det(B) ≠ 0.
Suppose we have an equation:
c1 u1+c2 u2+...+cn un=0 where c1,c2,...,cn - unknown variables.
It represents a system of n equations for each component. The matrix of this system is matrix B. We know that ci=0 i = (1,n) is a solution of the system. But as β must be a basis than the system must have no other solutions except c_i=0 i=(1,n).
So det(B) ≠ 0 (as a result of Cramer’s rule).


We have that det(B)≠0. Let’s prove that β=(u1,u2,…,un) is a basis for Fn.
Suppose that u1,u2,…,un are linearly dependent. Than the equation c1 u1+c2 u2+...+cn un = 0 has non-zero solution.
Let c1 ≠ 0. So
u1= - c2/c1 u2 -…- cn/c1 un. When we calculate det(B), we calculate it as a sum of determinants:
because each of the determinants = 0 (each has 2 equal columns).
But det(B) must be non-zero. So our assumption was wrong and u1,u2,…,un are linearly independent.
As we have n linearly independent vectors in Fn, β is a basis of Fn.

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