Basis and Dimension of Solution Space

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The discussion focuses on finding the basis and dimension of the solution space for a homogeneous system of equations. Participants are troubleshooting their approach to reducing the matrix and parametrizing the solutions. Confusion arises around the role of the variable x4, which is consistently zero in the basis vectors, leading to questions about how it can still span the space. Clarifications emphasize that the basis vectors are meant to span the solution set, not the original variable space, validating the presence of zero components. Ultimately, the contributors gain a clearer understanding of their calculations and the nature of the solution space.
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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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