Linear Algerba - Finding linearly independent vectors

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Discussion Overview

The discussion revolves around solving a system of equations derived from the matrix equation ATx=0, where A is a known matrix. Participants are attempting to find linearly independent vectors from the resulting equations, which are expressed in terms of several variables. The scope includes mathematical reasoning and exploration of linear algebra concepts.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a system of equations and seeks help in finding linearly independent vectors from them.
  • Another participant suggests that the equations represent parametric equations describing surfaces, indicating that the solution involves finding the intersection of these surfaces.
  • A different participant notes that the equations allow for a 3D subspace of solutions, emphasizing that fixing certain variables can yield specific solutions.
  • One participant elaborates on the simplification process of the equations and provides specific solutions in vector form, claiming they form a basis for the solution space.
  • Another participant expresses confusion about the simplification process and the choice of variables to set to specific values, seeking clarification on the reasoning behind these choices.
  • One participant explains their approach to solving the equations, stating that any three variables can be expressed in terms of the others and that setting certain variables to 1 or 0 helps ensure the independence of the resulting vectors.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and confusion regarding the methods used to simplify the equations and find linearly independent vectors. While some acknowledge the validity of the proposed solutions, others question the reasoning behind the simplification and variable selection processes. The discussion remains unresolved with multiple viewpoints presented.

Contextual Notes

Participants have not reached a consensus on the best approach to simplify the equations or the criteria for selecting variables to achieve linearly independent vectors. There are also varying interpretations of the nature of the solutions and their geometric implications.

Moomax
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Hi guys,

I am solving a problem in the form: (ATx=0 where A is a matrix of known numbers and I am solving for x. After performing reduction and multiplying ATx, I am left with the following equations:


-X1 + X4 - X5 = 0

-X2 + X4 = 0

-X3 + X4 -X5 + 28X6 = 0

From these equations, I am trying to find the linearly independent vectors however, I am not sure how to do this. I tried pulling out a linear book but I couldn't make straight forward sense of the procedure. Can anyone help me out? Thanks!
 
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I may be missing the scampi for the bees, but those aren't vectors, they're parametric eqn's, which describe surfaces.

Solving these equations is equivalent to finding the intersection of the three surfaces. I count 6 unknowns and 3 eqns, in which case the solution will be a 3D relation.
 
I'm not sure where linear independence comes into it but if X1, ..., X6 are the components of your vector then like christianjb says you will only be able to fix a 3D subspace, where all vectors in it satisfy your equation.

If you just need one specific solution, try fixing X1, X2 and X3. You'll then be able to calculate X4, X5 and X6 from your equations.
 
Moomax said:
Hi guys,

I am solving a problem in the form: (ATx=0 where A is a matrix of known numbers and I am solving for x. After performing reduction and multiplying ATx, I am left with the following equations:


-X1 + X4 - X5 = 0
So X5= X4- X1

-X2 + X4 = 0
and X2= X4

-X3 + X4 -X5 + 28X6 = 0
X3= X4- X5+ 28X6
= X4- (X4- X1)+ 28X6
= X1+ 28X6

So X5, X2, and x3 can be written in terms of X1, X4, and X6.

Now, take X11= 1, X4= x6= 0. Then X5= -1, X2= 0, and X3= 1. One solution is <1, 0, 1, 0, -1, 0>.

Take X4= 1, X1= X6= 0. Then X5= 1, X2= 1, and X3= 0. Another solution is <0, 1, 0, 1, 1, 0>.

Take X6= 1, X1= X4= 0. Then X5= 0, X2= 0, and X3= 28. A third solution is <0, 0, 28, 0, 0, 1>.

The three vectors, <1, 0, 1, 0, -1, 0>, <0, 1, 0, 1, 1, 0>, and <0, 0, 28, 0, 0, 1> form a basis for the solution space.

From these equations, I am trying to find the linearly independent vectors however, I am not sure how to do this. I tried pulling out a linear book but I couldn't make straight forward sense of the procedure. Can anyone help me out? Thanks!
 
HallsofIvy is on the right track with what I mean. But I am still a little bit confused about the actual proceedure.

Like what made you simplify the equations the way you did? When I was playing around with them trying to simpify it never occurred to me to set them equal the way you did.

Also after that was done, what made you pick the variables you did to equal 0 while others equal to 1?
 
I was just "solving" the equations. It would be nice if we could solve the equations for specific numbers (in which case the solution space would be the trivial subspace, containing only (0,0,0)) but here we can't. So we do the best we can. Since there are three (independent) equations, we can solve for three values- in terms of the other three. In this problem, ANY three can be solved for in terms of the other three. I just chose the easiest.

As for setting one variable equal to 1 and the others 0, that guarantees that the vectors you get will be independent.
 

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