Set of vector which are solution to 2 homogeneous systems

In summary, the solution to this problem is to find the vector that is in both S1 and S2, and that is the vector that is the intersection of the two sets.
  • #1
Philip Wong
95
0
hi guys,
I have two homogeneous systems S1 and S2. The solution for
NS(S1) = {[-2,1,0], [-1,0,1]},
NS(S2) = {[-2,1,0], [-3,0,1]}.

I know that in a system if u and v are vectors, the sum of u+v is also a solution in the homogeneous system. i.e. S1=span{[-2,1,0], [-1,0,1]} then [-2,1,0] + [-1,0,1] is also a solution in that system (close under addition if I am correct).

But what about if I want to find the set of vectors which are solution to both S1 and S2? Do I use the same methods to it?
i.e. Solution = span {[-2,1,0]+[-2,1,0] ; [-1,0,1]+[-3,0,1]}

I think this is wrong, but I'm not sure how
 
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  • #2
You say you want "the set of vectors which are solution to both S1 and S2". That is the intersection of the two sets, vectors that are in both. It is NOT the "direct sum" which is what you have. What you have is the set of vectors that solve either S1 or S2, not both.

Actually, for this problem, the solution is simple: notice that [-2, 1, 0] is in both sets. And the other two vectors, [-1, 0, 1] and [-3, 0, 1] are not multiples of each other.
 
  • #3
HallsofIvy said:
You say you want "the set of vectors which are solution to both S1 and S2". That is the intersection of the two sets, vectors that are in both. It is NOT the "direct sum" which is what you have. What you have is the set of vectors that solve either S1 or S2, not both.

Actually, for this problem, the solution is simple: notice that [-2, 1, 0] is in both sets. And the other two vectors, [-1, 0, 1] and [-3, 0, 1] are not multiples of each other.

Oh! So I would be right if I were to solve for S1 OR S2 only?

Arr! I think I might be hooking onto something here, so it should be span{[-2, 1, 0]}. This is where both systems intersects at this line. Right?
 
Last edited:
  • #4
Yes! That is exactly right.
 
  • #5
thanks!
 

1. What is a set of vectors that are solutions to 2 homogeneous systems?

A set of vectors that are solutions to 2 homogeneous systems is a collection of vectors that satisfy both systems of equations when plugged in. In other words, when the vectors are substituted into the equations, they result in a true statement or a zero vector.

2. How can a set of vectors be a solution to 2 homogeneous systems?

A set of vectors can be a solution to 2 homogeneous systems if they satisfy both systems of equations. This means that when the vectors are substituted into the equations, they result in a true statement or a zero vector. This shows that the vectors are a valid solution to both systems.

3. Can a set of vectors be a solution to 2 non-homogeneous systems?

No, a set of vectors cannot be a solution to 2 non-homogeneous systems. This is because non-homogeneous systems have at least one non-zero constant term, which means that the equations are not satisfied by a zero vector. Therefore, a set of vectors that results in a zero vector cannot be a solution to non-homogeneous systems.

4. How do you know if a set of vectors is a solution to 2 homogeneous systems?

A set of vectors is a solution to 2 homogeneous systems if when they are substituted into the equations, they result in a true statement or a zero vector. This can be checked by plugging in the vectors and solving the equations. If the equations are satisfied, then the set of vectors is a solution to the systems.

5. Why is it important to find a set of vectors that are solutions to 2 homogeneous systems?

It is important to find a set of vectors that are solutions to 2 homogeneous systems because it can help solve a variety of problems in mathematics, physics, and engineering. These solutions can also be used to find other important properties, such as linear independence and the basis of a vector space. Additionally, finding a set of vectors that are solutions to 2 homogeneous systems can lead to a better understanding of linear algebra concepts and their applications.

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