# Affine independence in terms of linear independence

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1. Oct 19, 2015

### Wiseguy

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
2. Oct 19, 2015

### andrewkirk

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?

3. Oct 19, 2015

### Wiseguy

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.

4. Oct 19, 2015

### andrewkirk

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 $\mathbb{R}^3$.