Proof on vector spaces

leftfootwonder7 · #1 $Old$ November 5th, 2009, 03:01 PM

Im stuck on the following question:

Let V be a vector space. Prove that if it is possible to find m vectors in V which are linearly independent, and n vectors which span V, then m must be less than, or equal to, n.

Any help/hints would be great

davismj · #2 $Old$ November 5th, 2009, 03:48 PM

I posted a response but I think I read the question wrong.

Jameson · #3 $Old$ November 5th, 2009, 03:50 PM

Quote:

Originally Posted by leftfootwonder7 $View Post$

Im stuck on the following question:

Let V be a vector space. Prove that if it is possible to find m vectors in V which are linearly independent, and n vectors which span V, then m must be less than, or equal to, n.

Any help/hints would be great

Call the linearly independent vectors $L={v_0, v_1...v_m}$ and the set of n vectors which spans V $S={u_0, u_2...u_n}$ .

Assume m>n. Then there is at least one more vector in L than in S. L is a subset of V so must be in the span(S)=V. Since any vector in v is linearly independent, it cannot be expressed by a combination of other vectors in V, or of span(S). Even if all vectors in S are independent, the number of independent vectors in L is always greater than those in S, thus at least one vector in L is not in the span(S), which implies not in V, making a contradiction.

Kind of messy but that's the basic idea.

leftfootwonder7 · #4 $Old$ November 6th, 2009, 06:59 AM

Yes i see now, proof by contradiction. Thanks