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Proving linear independence

the.cookie.monster

I'm getting a bit confused on the theory.
So to prove a vector set is linearly independent
1. Prove the determinant is non-zero so that means the only solution to Ax=0 is the trivial solution
2. Reduce A and show that there is a leading one in every column- how does this show that the vectors are linearly independent though??

Reply 1

jassi1

This is how I see it:

1. Given vectors

v_1, v_2, ... v_n \in K^n

. Let A be the matrix made up of rows

v_i

.

Then

det(A) \neq 0 \iff A

is non-singular

\iff rank(A) = n

. Hence the row rank of A is n and so it follows

v_i

are linearly independent*.

2. Row reducing A doesn't change the row space of A (and hence the row rank) so if the row reduced form of A has a leading 1 in each column the resulting matrix is forced to be the identity and since the rank is the number of non-zero rows in the row reduced form of A. rank(A) = n, and thus similarly the rows are linearly independent.

* Recall the row rank is equal to the size of the largest linearly independent subset of rows thus if rank(A) = n the rows are lin indep.

Hope this helps.

Reply 2

BananaPie

If a vector set were linearly dependent, then what you will get after row of reduction is a column or row of zeros. Evaluate the determinant of this and you will realize that the determinant is 0. Hence as you mentioned det=/=0 if it's linearly independent.
This follows on by saying that if you RREF your vector matrix and end up with 1's in every column, this will only be possible if the determinant is not 0 and this is only possible if each vector within your matrix cannot be expressed as a sum of other vectors.