Linear independence of basis vectors

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Homework Help Overview

The discussion revolves around proving the linear independence of standard basis vectors in vector spaces, specifically in the context of \(\mathbb{R}^3\) and potentially \(\mathbb{R}^n\). The original poster expresses confusion regarding the proof process despite understanding the definition of linear independence.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to apply the definition of linear independence to standard basis vectors. Some participants suggest using contradiction to explore the implications of linear dependence. Others prompt the original poster to consider specific cases in \(\mathbb{R}^3\) and generalize to \(\mathbb{R}^n\).

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the proof. The original poster indicates a growing understanding of the topic, while others encourage further exploration of the implications of linear dependence.

Contextual Notes

The original poster mentions personal challenges that may affect their comprehension of the material, which could influence their engagement with the discussion.

cappygal
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How do I prove the linear independence of the standard basis vectors? My book is helpful by giving the definition of linear independence and a couple examples, but never once shows how to prove that they are linearly independent.
I know that the list of standard basis vectors is linearly independent if:
The only choice of a_1, a_2, ... a_m that makes a_1v_1+a_2v_2+...+a_mv_m=0 is a_1=a_2=...=a_m=0.
But i don't know where to go from there .. any help would be appreciated :confused:
 
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Use contradiction. If they were linearly dependent, then by your definition, at least one of the vectors could be written as a linear combination of others. Is this true of the standard basis ?
 
For example, in [itex]\mathbb{R}^3[/itex], the standard basis is:

[tex]\left\{ {\left( {1,0,0} \right),\left( {0,1,0} \right),\left( {0,0,1} \right)} \right\}[/tex]

This basis is linearly independent if, as you say:

[tex]a\left( {1,0,0} \right) + b\left( {0,1,0} \right) + c\left( {0,0,1} \right) = \left( {0,0,0} \right)[/tex]

implies that [itex]a = b = c = 0[/itex].

Well, solve the condition for a, b and c and see if you can find anything else besides the solution we expect.
Can you now see how it will be for [itex]\mathbb{R}^n[/itex] in general?
 
Thank you so much .. it actually makes sense now .. Something about being out of school with a broken pelvis means that it's harder to understand what they do in class without you ... thanks!
 
No problem :smile:
 

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