PROOF: Independent vectors and spanning vectors

Click For Summary

Discussion Overview

The discussion revolves around the concepts of vector independence and spanning sets in the context of linear algebra. Participants explore definitions, proofs, and methods related to the dimension of vector spaces and the criteria for vectors to form a basis.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that at least m vectors are needed to span an m-dimensional space and that m vectors spanning such a space form a basis.
  • There is a request for clarification on the definition of finite dimension in vector spaces, with emphasis on understanding the concept before defining it.
  • One participant suggests that a basis consists only of independent vectors, implying that n independent vectors are necessary to span an n-dimensional space.
  • Another participant poses a question about determining the linear independence of two specific vectors, U and V, and seeks guidance on applying the independence condition.
  • Responses include contrasting methods for checking linear independence, with one participant advocating for row reduction of a matrix formed by the vectors, while another emphasizes using the definition involving a linear combination equating to zero.
  • There is a discussion about the validity of using the linear combination method versus matrix row reduction, with participants expressing differing preferences for proof techniques.

Areas of Agreement / Disagreement

Participants express differing views on the best methods to demonstrate linear independence and the definitions involved in the discussion. No consensus is reached on a single approach or interpretation of the concepts.

Contextual Notes

Some definitions and assumptions regarding vector spaces and linear independence are not fully explored, and the discussion includes various interpretations of the foundational concepts.

Ella087
Messages
3
Reaction score
0
Proof:
1. why you need at least m vectors to span a space of dimension m.
2. If m vectors span an m-dimensional space, then they form a basis of the space.
 
Physics news on Phys.org
do you know how to do proofs?
 
What does it mean to say that a vector space has finite dimension? (NOT just that "its dimension is finite. You have to have 'finite' dimension before you can define dimension.) What is the definition of "dimension n".
 
Try reading some definitions in your textbook.

And reading some proofs of some easy theorems. You'll see how to apply definitions to prove statements like you have.
 
these are proved in my free linear algebra notes on my web page. the full details are not given, but i believe the proofs can be filled in without huge difficulty, if you try and understand the definitions.
 
lets start from the definition that a basis consists only from
independent vectors
now in order to span some space in "n" dimension you need to have "n" independent vectors
which are a basis for this dimension.
 
Find the independence

Find if these vectors are lineary independent :..

U=(1 2 9) . v=(2 3 5) .


I know the condition of the lineary independent is

au+bv=0 but how I can use this condition here .,.,.,

Please If you know the answer u can send it at >> ahmedtomyus@yahoo.com
 
Last edited:
noooooooooooooo
dont use that formula
stack them one upon the other as matrix and make a row reduction
if you dond have a line of zeros in the resolt then they are independent
 
tomyus said:
Find if these vectors are lineary independent :..

U=(1 2 9) . v=(2 3 5) .


I know the condition of the lineary independent is

au+bv=0 but how I can use this condition here .,.,.,

Please If you know the answer u can send it at >> ahmedtomyus@yahoo.com
Normally I would not respond to a question that has nothing to do with the original question but since transgalactic responded to it...

transgalactic said:
noooooooooooooo
dont use that formula
stack them one upon the other as matrix and make a row reduction
if you dond have a line of zeros in the resolt then they are independent
Why not? au+ bv= a(1, 2, 9)+ b(2, 3, 5)= (a+2b, 2a+ 3b, 9a+ 5b)= (0, 0, 0) seems easy enough. We must a+ 2b= 0, so a= -2b. Then 2a+ 3b= -6a+ 3b= -3b= 0. b must equal 0, so a= -2b= 0 also. Since a and b must both be 0 the two vectors are, by definition, independent.

Yes, if we have 10 vectors or 10000 then "row reduction" would be simpler but I believe it is better practice to use the basic definitions.
 

Similar threads

Undergrad Characterizing linear independence in terms of span
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Vector subspace and basis vectors in the context of data science
  • · Replies 6 ·
Replies
6
Views
4K
Undergrad Dimension and solution to matrix-vector product
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Why is the dual of Z^n again Z^n ?
  • · Replies 13 ·
Replies
13
Views
2K
Undergrad Confused about small detail in rank-nullity theorem
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Row space, Column space, Null space & Left null space
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Showing ##k[x_1,\ldots,x_n]/\mathfrak{a}## is finite dimensional
  • · Replies 40 ·
2
Replies
40
Views
4K
MHB Statements with linearly independent vectors
  • · Replies 10 ·
Replies
10
Views
2K
Undergrad Standard basis and other bases...
  • · Replies 9 ·
Replies
9
Views
4K