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Affine independence in terms of linear independence

  1. Oct 19, 2015 #1
    This question mostly pertains to how looking at affine independence entirely in terms of linear independence between different families of vectors. I understand there are quite a few questions already online pertaining to the affine/linear independence relationship, but I'm not quite able to find something that helps my particular problem, nor am I able to make the connection on my own.

    I want to try and understand how the linear independence of a family of ##n## difference vectors from any arbitrary 'origin' vector, say ##(\overrightarrow{a_i a_0}, \ldots, \overrightarrow{a_i a_j}, \ldots \overrightarrow{a_i a_n})## where ##a_i\ and\ a_j \in \mathbb{R}^{n}## and ##j \neq i## for any arbitrary 'origin' ##i \in I##, implies the linear independence of the whole family of ##(n+1)## vectors ##(\hat{a_0}, \ldots, \hat{a_n})## where ##\hat{a_j} = (1, a_j)##

    I am able to understand this from the perspective of using families of points, but I am unable to visualize how I would construct this only using families of vectors. I've tried looking at the vectors as position vectors, but I think that way of thinking would not necessarily be correct.
     
    Last edited: Oct 19, 2015
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  3. Oct 19, 2015 #2

    andrewkirk

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    Hello and welcome to physicsforums.

    I'm afraid your notation is quite unusual.

    What does ##\overrightarrow{a_i a_0},## represent? Given that you've said ##a_i\in\mathbb{R}## that would suggest that ##\overrightarrow{a_i a_0}=a_0-a_i\in\mathbb{R}##, which is a scalar. You can think of that as a vector if you like, but ##\mathbb{R}## as vector space has only one dimension, so you can't have more than one linearly independent vector in it..

    What does the right hand side of ##
    \hat{a_j} = (1, a_j)
    ## represent?
     
  4. Oct 19, 2015 #3
    Thank you for your welcome.

    I apologize. I meant to write ##\mathbb{R}^{n}##, not ##\mathbb{R}##. Yes, the notation ##\overrightarrow{a_i a_0},## is just used to represent ##(a_i - a_0) \in \mathbb{R}^{n}##. We can keep it in the latter form if it makes more sense.

    And ## \hat{a_j} ## is just that. A vector ##\in \mathbb{R}^{n+1}## comprising of ##(1, a_j)##. I would like to know the intuition as to why the linear independence of these forms are equivalent.
     
  5. Oct 19, 2015 #4

    andrewkirk

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    Is it a proof, or a visualization, that you are missing? If it's a visualization, why not take a small concrete example.
    The easiest that still has vector structure is n=2. Take for instance a0=(1,1), a1=(2,2), a2=(1,2). Draw a picture of these in ##\mathbb{R}^2## and then another of what you get with the move into ##\mathbb{R}^3##.
     
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