Determine whether or not something is a subspace

[kesun]

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?

 

[ThirstyDog]

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.

 

[de_brook]

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\which satisfies the property that x1x2 = 0. but since x1, x2 are in \mathbb{R}, then either x1=0 or x2=0 or both equal zero.<br /> 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<br /> y+s = (y1+s1,y2+s2,y3+s3)<br /> check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0<br /> 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 \mathbb{R}^3\. we don't need to proof the second property

 

[de_brook]

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 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