# Determine whether or not something is a subspace

My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct.

By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x):
1) (x+y) $$\in$$ S and
2) kx $$\in$$ S.

Also, the solution set of a homogeneous system is always a subspace.

When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:

Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.

Is my way of solving this problem correct?

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You are right in that you must check your the two conditions. But you must do it for arbitrary vectors, you didn't seem to do this correctly. Consider

$$(1,0,0), (0,1,0) \in \{(x_{1},x_{2},x_{3})|x_{1}x_{2}=0\}.$$

If you add them you obtain $$(1,1,0)$$ which clearly does not have $$x_{1}x_{2} = 0$$. Also the way you present your vectors doesn't seem standard.

My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct.

By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x):
1) (x+y) $$\in$$ S and
2) kx $$\in$$ S.

Also, the solution set of a homogeneous system is always a subspace.

When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:

Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.

Is my way of solving this problem correct?
That was a nice attempt but your steps were wrong.
{(x1,x2,x3)|x1x2=0} is a subset of $$\mathbb{R}^3\[tex] which satisfies the property that x1x2 = 0. but since x1, x2 are in [tex]\mathbb{R}[tex], then either x1=0 or x2=0 or both equal zero. To proof that it is a subspace, let y and s be vectors in {(x1,x2,x3)|x1x2=0} such that y = (y1,y2,y3) and s = (s1,s2,s3), then y+s = (y1+s1,y2+s2,y3+s3) check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0 clear in general; y1s2+s1y2 is not equal to zero. this implies that it does not satisfy the property x1x2 = 0. Hence the set {(x1,x2,x3)|x1x2=0} is not a subspace of [tex]\mathbb{R}^3\[tex]. we don't need to proof the second property My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct. By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x): 1) (x+y) [tex]\in$$ S and
2) kx $$\in$$ S.

Also, the solution set of a homogeneous system is always a subspace.

When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:

Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.

Is my way of solving this problem correct?
That was a nice attempt but your steps were wrong.
{(x1,x2,x3)|x1x2=0} is a subset of R3 which satisfies the property that x1x2 = 0. but since x1, x2 are in R3 then either x1=0 or x2=0 or both equal zero.
To proof that it is a subspace, let y and s be vectors in {(x1,x2,x3)|x1x2=0} such that y = (y1,y2,y3) and s = (s1,s2,s3), then
y+s = (y1+s1,y2+s2,y3+s3)
check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0
clear in general; y1s2+s1y2 is not equal to zero. this implies that it does not satisfy the property x1x2 = 0. Hence the set {(x1,x2,x3)|x1x2=0} is not a subspace of R3. we don't need to proof the second property