How to test if vectors span a space?

Click For Summary

Homework Help Overview

The discussion revolves around testing whether a set of vectors spans a vector space, specifically in the context of a 5-dimensional space. The original poster seeks clarification on methods to determine spanning and linear independence, having forgotten the relevant concepts during the summer break.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between linear independence and spanning, with one noting that 5 linearly independent vectors span a 5-dimensional space. There are mentions of using determinants and row reduction as methods for testing these properties.

Discussion Status

The discussion has progressed with participants providing insights into the relationship between linear independence and spanning. Some have offered methods for testing these properties, while others have shared personal experiences related to the topic. There is an acknowledgment of the original poster's curiosity about the implications of using specific units in their problem.

Contextual Notes

The original poster is working under the constraint of not having access to their linear algebra textbook and is seeking reminders about fundamental concepts. There is also a mention of potentially dealing with infinite dimensional spaces, which adds complexity to the discussion.

ramparts
Messages
44
Reaction score
0
Hey folks - I left my linear book at home for the summer and am having trouble finding this on the Google, so I'm hoping you'll be good for an easy question :) I have a set of vectors that I think spans a vector space, but I've forgotten how to test it. I can make a set of 5 simultaneous equations (it's a 5-dimensional space) but that seems a bit much. How do I do this again?

I'd also appreciate a reminder about how to test for linear independence, though it seems with this particular set of vectors that part is pretty trivial.

Thanks!
 
Physics news on Phys.org
If you have 5 vectors that are linearly independent, then they span a 5-dimensional vector space. (n-1) vectors can never span an n-dimensional space. One test for linear independence, besides using the definition, is to find the determinant of the matrix of the n vectors in question. If it is 0, the list of vectors is linearly dependent.
 
I didn't know it was that simple, thanks! If the determinant of the n-by-n matrix is non-zero, presumably that means the vectors are linearly independent?
 
Sorry, I left out something obviously important - I'm looking to see if these vectors span R^5.
 
Note that you can still have a set of linearly dependent vectors spanning a given space (you'd have more vectors than the dimension of that space). The sufficient condition is to express each of the space's basis elements as linear combinations of the set of vectors you are considering. If you have "too many" vectors then there will be more than one way to do this.

Note that you may for example be dealing with an infinite dimensional space. For example does the set of polynomials in x span the space of analytic functions (of x) on the real number line? Answer is yes, proof involves showing every analytic function has a power series expansion i.e. is a linear combination of monomials.
 
Yep - I was looking for a basis set, but it was pretty trivial that it was linearly independent so the main thing was to find span. Just had to sharpen my linear algebra, I put the vectors in a matrix and row reduced, and yeah, it spans.

It was actually a mildly interesting problem, I was curious if full-on Planck units (setting G, h-bar, c, e_0 and k to 1) uniquely covered any combination of base units (they do).
 
Oh, that 5-dimensional vector space :wink: On that note, this thread (link) may interest you.
 

Similar threads

Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K