Am i setting up equations right? Matrices, finding nullspace

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Homework Help Overview

The discussion revolves around setting up equations related to finding the nullspace of a matrix. The original poster presents a matrix and attempts to derive the nullspace through a series of equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to express variables in terms of a parameter and questions whether their setup is correct. Other participants provide feedback on the correctness of the equations and the interpretation of the nullspace.

Discussion Status

Participants are actively engaging with the original poster's reasoning, offering corrections and clarifications regarding the setup of the equations and the nature of the nullspace. There is a recognition that the nullspace consists of a set of vectors rather than a single solution.

Contextual Notes

There is mention of differing interpretations of the nullspace based on different matrix representations, as well as a concern about the accuracy of the original poster's expressions compared to those of their professor.

mr_coffee
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Hello everyone I'm confused if I'm setting these equations up right:
i have:
1 0 1
0 1 -2
0 0 0

so i said:
x + z = 0;
y - 2z = 0;
z = ? because its a whole row of 0's, so u have no info about what z could be
so i said let z = a;
x + a = 0;
y -2a = 0;
z = a;

x = a;
y = 2a;
z = a;

so let a = 1;
my nullspace is:
1
2
1

am i doing that right?
 
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You lost a minus sign, it should be x=-a.

The nullspace is given by all solutions to this system, not a specific one. That is all vectors (-a,2a,a) make up the nullspace. You can look at a specific non-zero one if you're after a basis of the nullspace (since it has dimension one in this case).
 
i c, thank u! but the synatx i used is correct? -a, 2a, a correct? so if i factorted out an a i would get
a[-1,21] because my professor got a different answer: he got:

from this matrix:
which is a not simplified version of the matrix i gave above:
2 1 0
0 -1 2
0 0 0

2x + y = 0;
-y + 2z = 0;
x = -a/2;
z = a/2
y = a;

null(A) = [-1/2,1/2]
and i thought he did it wrong.
 
Strictly as written, null(A)=[-1/2,1,1/2] is wrong, the nullspace is not one vector, but a set of vectors (even if it's just the zero vector).

null(A)=span{[-1/2,1,1/2]} would be a way to write it.
 
Yep that makes sense to me, thanks shmoe
 

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