Basis and Dimension of Solution Space

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SUMMARY

The discussion focuses on finding a basis and the dimension of the solution space for a homogeneous system of equations represented in matrix form. The equations provided lead to a reduced row echelon form, revealing relationships among the variables, specifically that x1 and x2 are dependent, as well as x3 and x5. The basis vectors for the solution space are determined to be <1, -1, 0, 0, 0> and <0, 0, 1, 0, -1>, confirming that x4 can indeed be zero without affecting the span of the solution space.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly homogeneous systems of equations.
  • Familiarity with reduced row echelon form (RREF) and its application in solving systems of equations.
  • Knowledge of vector spaces and basis vectors in the context of linear transformations.
  • Ability to parametrize solutions based on linear combinations of vectors.
NEXT STEPS
  • Study the properties of homogeneous systems of equations in linear algebra.
  • Learn about the process of obtaining reduced row echelon form (RREF) using Gaussian elimination.
  • Explore the concept of vector spaces and how to determine the basis and dimension of a vector space.
  • Investigate the implications of zero components in basis vectors and their role in spanning a solution space.
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Students and educators in linear algebra, mathematicians working with vector spaces, and anyone seeking to understand the structure of solution spaces in homogeneous systems of equations.

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 &lt;x_1, x_2, x_3, x_4, x_5&gt;= &lt;x_1, -x_1, x_3, 0, -x_3&gt;= &lt;x_1, -x_1, 0, 0, 0&gt;+ &lt;0, 0, x_3, 0, -x_3&gt;= x_1&lt;1, -1, 0, 0, 0&gt;+ x_3&lt;0, 0, 1, 0, -1&gt;. 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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