Solution space to homogenous equations.

  • Thread starter CalculusSandwich
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In summary: I think you made an error somewhere. Substituting your answer into the first equation doesn't give 0. How did you get (-z, 3z, 3z, z) ?
  • #1
CalculusSandwich
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Ok so my question was:

The directions state:

Find the solution space of the following systems of linear homogeneous equations:

x-y+z-w=0
2x+y-z+2w=0
2y+3z+w=0

Is the same as finding the solutions, because I did that and got (-z, -3z, 3z, z). So i said
the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace.
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And someone responded:

I don't think that solution set is right.

The vector -1,-3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. That means when you write your free variable in terms of the other variables, there should be a 0 in that solution set.

The solution (-1,-3,3,1) does not satisfy ax=0.

----------------------------------------------------------------------------------------

Am i right or am i wrong here?
 
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  • #2
I think you made an error somewhere. Substituting your answer into the first equation doesn't give 0. How did you get (-z, 3z, 3z, z) ?
 
  • #3
CalculusSandwich said:
Ok so my question was:

The directions state:

Find the solution space of the following systems of linear homogeneous equations:

x-y+z-w=0
2x+y-z+2w=0
2y+3z+w=0

Is the same as finding the solutions, because I did that and got (-z, -3z, 3z, z). So i said
the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace.
I suggest you try that again. That is not at all what I get.

---------------------------------------------------------------------------------------
And someone responded:

I don't think that solution set is right.

The vector -1,-3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. That means when you write your free variable in terms of the other variables, there should be a 0 in that solution set.

The solution (-1,-3,3,1) does not satisfy ax=0.

----------------------------------------------------------------------------------------

Am i right or am i wrong here?
 

FAQ: Solution space to homogenous equations.

What is the solution space to homogenous equations?

The solution space to homogenous equations is the set of all possible solutions to a system of linear equations where all the constant terms are equal to zero. In other words, it is the set of all vectors that satisfy the equations when all variables are set to zero.

How is the solution space to homogenous equations different from the solution space to non-homogenous equations?

Unlike homogenous equations, non-homogenous equations have a non-zero constant term and therefore, have a specific solution. The solution space to non-homogenous equations is a single point, while the solution space to homogenous equations is a subspace of the vector space.

How can the dimension of the solution space to homogenous equations be determined?

The dimension of the solution space to homogenous equations is equal to the number of variables in the system of equations. This is because each variable corresponds to a dimension in the vector space, and setting all variables to zero creates a basis for the solution space.

Can the solution space to homogenous equations be empty?

Yes, it is possible for the solution space to homogenous equations to be empty. This occurs when the system of equations has no solutions, meaning that the equations are inconsistent or contradictory.

How can the solution space to homogenous equations be represented geometrically?

The solution space to homogenous equations can be represented as a subspace in the vector space. In two-dimensional space, it would be a line passing through the origin, while in three-dimensional space, it would be a plane passing through the origin. In higher dimensions, it would be a hyperplane passing through the origin.

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