# 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?

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

**Necessity:**

We have that β=(u

_{1},u

_{2},…,u

_{n}) is a basis for Fn. Let’s prove that det(B) ≠ 0.

Suppose we have an equation:

c

_{1}u

_{1}+c

_{2}u

_{2}+...+c

_{n}u

_{n}=0 where c

_{1},c

_{2},...,c

_{n}- unknown variables.

It represents a system of n equations for each component. The matrix of this system is matrix B. We know that c

_{i}=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).

Sufficiency:

Sufficiency:

We have that det(B)≠0. Let’s prove that β=(u

_{1},u

_{2},…,u

_{n}) is a basis for Fn.

Suppose that u

_{1},u

_{2},…,u

_{n}are linearly dependent. Than the equation c

_{1}u

_{1}+c

_{2}u

_{2}+...+c

_{n}u

_{n}= 0 has non-zero solution.

Let c

_{1}≠ 0. So

u

_{1}= - c

_{2}/c

_{1}u

_{2}-…- c

_{n}/c

_{1}u

_{n}. 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 u

_{1},u

_{2},…,u

_{n}are linearly independent.

As we have n linearly independent vectors in Fn, β is a basis of Fn.

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