# Set of vector which are solution to 2 homogeneous systems

## Main Question or Discussion Point

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

## Answers and Replies

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HallsofIvy
Homework Helper
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.

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:
HallsofIvy