Linear Algebra - Number of vectors in a basis

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Homework Help Overview

The discussion revolves around the concept of bases in vector spaces, specifically in the context of linear algebra. Participants explore the implications of having a set of vectors in R4 and the dimensionality of their span.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of a basis and the dimensionality of vector spaces, questioning how a set of five vectors can lead to a basis with fewer than four vectors in R4. There are attempts to clarify the relationship between the number of vectors in a basis and the dimension of the span of a set of vectors.

Discussion Status

The conversation is active, with participants providing insights into the dimensionality of spans and the nature of subspaces. Some have offered examples to illustrate points, while others seek confirmation of their understanding regarding the approach to finding a basis.

Contextual Notes

There is an ongoing exploration of the definitions and properties of vector spaces, particularly regarding linear independence and the implications of having more vectors than the dimension of the space. Participants are also navigating the constraints of homework rules and the specific problem setups presented.

jinksys
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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?
 
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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
 
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.
 
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?
 
jinksys said:
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?

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
 

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