Basis and Dimension of Solution Space

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Homework Help Overview

The discussion revolves around finding a basis and the dimension of the solution space for a homogeneous system of equations involving multiple variables. Participants are analyzing the implications of their row-reduced matrices and the relationships between the variables.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to reduce the system of equations to row echelon form and are discussing how to parametrize the variables. Questions arise regarding the presence of zero components in the basis vectors and how that affects the spanning of the solution space.

Discussion Status

There is an ongoing exploration of the relationships between the variables based on the reduced matrix. Some participants are providing insights into the implications of having certain variables equal to zero and how that relates to the basis vectors. Clarifications are being sought regarding the correctness of the basis representation.

Contextual Notes

Participants are navigating the complexities of linear combinations and the structure of the solution space, with specific attention to the role of zero components in the basis vectors. There is a noted confusion regarding the interpretation of the solution set versus the original space.

B18
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Homework Statement


Find a basis for and the dimension of the solution space of the homogenous system of equations.

2x1+2x2-x3+x5=0
-x1-x2+2x3-3x4+x5=0
x1+x2-2x3-x5=0
x3+x4+x5=0

Homework Equations

The Attempt at a Solution


I reduced the vector reduced row echelon form. However the second row doesn't have a 1 in the x2 position which has thrown me off. Do I parametrize here like I did? Something just doesn't seem correct.
Here is my work thus far:
 
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Based on your final matrix, which I didn't check, you have
x1 = -x2 - x5
x2 = x2
x3 = -x5
x4 = 0
x5 = x5
If you stare at this long enough you should see that the above represents all possible linear combinations of two vectors.

BTW, it's confusing to have that last column of zeroes in your matrix. It would be better to eliminate that column since it can't possibly change.
 
Well I got the following equations from the matrix:
x1+x2+x4=0
x3+x5=0
x4=0
Not really sure how I can parametrize these to solve for the basis of the solution space.
 
If those are correct, since x_4= 0, your first equation is x_1+ x_2= 0 which means that x_2= -x_1. And, of course, your second equation s the same as x_5= -x_3. That means we can write <x_1, x_2, x_3, x_4, x_5>= <x_1, -x_1, x_3, 0, -x_3>= <x_1, -x_1, 0, 0, 0>+ <0, 0, x_3, 0, -x_3>= x_1<1, -1, 0, 0, 0>+ x_3<0, 0, 1, 0, -1>. Now do you see what basis vectors for the solution space are?
 
This is the solution I got. However I am a bit confused on my basis. How can the x4 slot both have zeros in the basis vectors and still span the space? Did I make a mistake along the way?
Thanks for the help everyone.

ImageUploadedByPhysics Forums1426716928.795578.jpg
 
B18 said:
This is the solution I got. However I am a bit confused on my basis. How can the x4 slot both have zeros in the basis vectors and still span the space? Did I make a mistake along the way?
Thanks for the help everyone.

View attachment 80599
There's no reason that the x4 component shouldn't be zero. You can do a quick check by multiplying each of the two vectors by your matrix. If you don't get a zero vector for each product, that's a sign that you did something wrong.
 
B18 said:
This is the solution I got. However I am a bit confused on my basis. How can the x4 slot both have zeros in the basis vectors and still span the space? Did I make a mistake along the way?
Thanks for the help everyone.

View attachment 80599
Have you forgotten what it is spanning? Not the original space- these vectors are to span the solution set for these equations- and as you have calculated x4 is 0!
 
Both of your replies helped me understand what was going on here much better. Thank you.
 

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