# Homework Help: Vector space

1. Jan 24, 2010

1. The problem statement, all variables and given/known data
Find the dimnesion and a basis of vector space V

2. Relevant equations
V is the set of all vectors (a,b,c) in R^3 with a+2b-4c=0

3. The attempt at a solution
(4c-2b,b,c) = b(-2,1,0) + c(4,0,1)
so {(-2,1,0),(4,0,1)} is the basis of the SUBSPACE of V right?

how do I get the 3rd linearly independent vector so that it forms the basis of V?
Is it (1,0,0)? (0,1,0) seems possible... but the basis of a vector space isn't unique right?

2. Jan 24, 2010

### rasmhop

This is a good start, and the right idea needed to solve the problem.

THE subspace? V has many subspaces and you're not asked to find bases of them. What you want is a basis of V. Remember that a basis of V is a set of vectors that span V and is linearly independent. Now consider:
1. For any vector of the form (4c-2b,b,c) (i.e. a vector in V) can you express it as a linear combination of (-2,1,0) and (4,0,1). If the answer is yes, then (-2,1,0) and (4,0,1) span V and therefore we need not find more basis elements.
2. Is {(-2,1,0),(4,0,1)} linearly independent?
If you can answer both 1 and 2 in the affirmative, then {(-2,1,0),(4,0,1)} is a basis of V.

Why do you believe you need a 3rd vector? You would only need a third vector if the dimension of V is 3, but it could be smaller (2 for instance).

Neither (1,0,0) nor (0,1,0) is actually in V so they can't be used. No a basis isn't unique, but your set from the first part of your answer is a good guess for a basis.

3. Jan 24, 2010

I think I've read somewhere that if there is m linearly independent vectors in R^n where m<n, these vectors form the subspace

since we have R^3 now and found 2 linearly independent vectors, so I thought these 2 vectors form a subspace...

does that mean basis of V is JUST {(-2,1,0) , (4,0,1)} in this case?

4. Jan 24, 2010

### rasmhop

I guess I was thrown of by your use of the. It's true that m linearly independent vectors span a subspace of R^n. In your case V is the subspace of R^n.

They do, but they do form the subspace V or another?

Yes, but you need to confirm it by proving that they span V and that they are linearly independent. Of course this implies that V has dimension 2. So in this case {(-2,1,0), (4,0,1)} span a subspace of R^3, and this subspace is V.