# Linear Algebra - Number of vectors in a basis

1. Jul 18, 2010

### jinksys

As I read one linear algebra book I have, I am told that "If a vector space V has a basis with 'n' vectors, then every basis in vector space V has 'n' vectors.

So every basis in R3 has 3, every basis in R4 has 4, etc.

However, I have a problem that says:

Let S = { "five vectors" } be a set of vectors in R4.
Find a subset of S that is a basis for W = span S.

The solution goes through putting the matrix into row-echelon form, and it turns out v1, and v2 of the set S are a basis for W = Span S.

I'm confused, I thought bases of R4 had four vectors? Could someone clear this up for me?

2. Jul 18, 2010

### lanedance

you're correct, a basis for R4 will have 4 vectors

the diminesion of span{S} is the maximal number of linearly independent vectors in S, in your case this is 2

As such, no combination of vectors in S can be used as a basis for R4.

W = span{S} is a 2 dimensional subspace of R4

3. Jul 18, 2010

### Staff: Mentor

Here's a simpler example in a lower-dimension space, R2.

Let W = {<-10, -5>, <2, 1>, <6, 3>}

No basis for R2 can have more than two vectors. W is a subset of R2, and as it turns out, W is a one-dimensional subspace of R2. The span of W, written as span(W), is the set of all linear combinations of the vectors in W. Geometrically, span(W) is a line through the origin, and passing through the point (2, 1).

Again, span(W) is a one-dimensional subspace even though the vectors in W have two components.

4. Jul 18, 2010

### jinksys

I understand, thanks!

So if I had the problem:

S = { (1, 0, 0, 1), (0, 1, 1, 0), (1, 1, 1, 1), (-1, 1, 1, -1) }

and I am to find a basis for the subspace W = span S of M2,2...

I set up a matrix A where each vector of S is a column of A.
Reduce to row-echelon form.

I get a leading 1 in row 1 and 2, so the basis of W is {(1,0,0,1), (0,1,1,0)}.

Is this the correct way to approach this problem?

5. Jul 18, 2010

### lanedance

yep, that looks good to me,

note any vector in S can be written in terms of your 2 basis vectors

also worth noting your basis vectors are orthogonal as a result of the matrix reduction