We have that β=(u1
) is a basis for Fn. Let’s prove that det(B) ≠ 0.
Suppose we have an equation:
=0 where c1
- 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
) is a basis for Fn.
Suppose that u1
are linearly dependent. Than the equation c1
= 0 has non-zero solution.
≠ 0. So
= - c2
. 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
are linearly independent.
As we have n linearly independent vectors in Fn, β is a basis of Fn.