Basis and Dimension of Solution Space

[B18]

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:

 

[B18]

 

[Mark44]

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.

 

[B18]

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.

 

[HallsofIvy]

If those are correct, since [itex]x_4= 0[/itex], your first equation is [itex]x_1+ x_2= 0[/itex] which means that [itex]x_2= -x_1[/itex]. And, of course, your second equation s the same as [itex]x_5= -x_3[/itex]. That means we can write [itex]<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>[/itex]. Now do you see what basis vectors for the solution space are?

 

[B18]

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.

 

[Mark44]

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.

 

[HallsofIvy]

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!

 

[B18]

Both of your replies helped me understand what was going on here much better. Thank you.