Are x1, x2, and x3 linearly dependent in R^n?

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Homework Help Overview

The discussion revolves around the linear dependence of vectors x1, x2, and x3 in R^n, with a focus on derived vectors y1, y2, and y3. Participants explore the implications of linear independence and dependence through various mathematical approaches.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up matrices to analyze linear independence, considering row echelon form and determinants. There is also a mention of the relationship between the derived vectors y1, y2, and y3.

Discussion Status

The discussion includes attempts to validate methods for determining linear dependence. Some participants express uncertainty about the correctness of their approaches, while others affirm the validity of the methods discussed.

Contextual Notes

There is an emphasis on the implications of free variables in the context of linear dependence, as well as the potential for confusion regarding the relationships between the original and derived vectors.

DMOC
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y1 = x2 - x1

y2 = x3 - x2

y3 = x3 - x1

x1, x2, x3 are linearly independent in R^n

Is that true for y1 y2 and y3?

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Well, normally with linear indepence problems, I can set up a matrix and check to see if the row echelon form has free variables or not, or I can calculate the determinant of a square matrix. Here, I'm just given vectors x1 x2 and x3. What I did was set up a matrix like this with x1 as the first row, x2 as the second, and x3 as the third:

-1 1 0 = 0
0 -1 1 = 0
-1 0 1 = 0

And I end up with the following matrix

1 0 -1 0
0 1 -1 0
0 0 0 0

So I assume these vectors are linearly dependent (not independent) due to the free "variable" of x3?
 
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Or you might note y3 = y1 + y2.
 
Yes, that should be correct ... my bad. But would the longer method be valid? i.e. mathematically legal?
 
DMOC said:
Yes, that should be correct ... my bad. But would the longer method be valid? i.e. mathematically legal?

Yes.
 
Thank you.
 

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