Determine whether or not something is a subspace

[kesun]

I guess this kind of topic should belong here. :|

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?

 

[yyat]

I guess this kind of topic should belong here. :|

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.

So far so good...

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

The equations also need to be linear.

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;

You mean the vectors have three components. The set contains an infinite number of vectors.

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?

No, the set $$S=\{(x_1,x_2,x_3)|x_1x_2=0\}$$ is not a subspace. Try finding $$x,y\in S$$ such that $$x+y$$ is not an element of S.

 

[kesun]

OHHH DAMN! Is it that (x₁+y₁)(x₂+y₂) = x₁x₂+y₁x₂+x₁y₂+y₁y₂? No that definitely ain't the subspace of it..! Am I correct about that?

 

[Dick]

Be concrete. v1=(1,0,0) is in the subspace, and so is v2=(0,1,0). Is v1+v2?

 

[kesun]

Oh, I see. I may need to confirm a few more examples to be sure that I am apply this theorem correctly.

Now suppose {x$$\in$$R⁵ | ||x||² $$\geq$$ 0}. First of all, it concerns R⁵. Since ||x|| is the norm of the vector, which is the same thing as the distance of the vector, so it will always be greater or equal to 0. Since this does not require the outcome to be a specific value(like it did in the previous one, which was 0), is it safe to conclude that this is a subspace of R⁵?

 

[Dick]

Unless I'm misunderstanding your notation, for ANY vector x in R^5, ||x||^2>=0. So your subspace is ALL of R^5.

 

[kesun]

Unless I'm misunderstanding your notation, for ANY vector x in R^5, ||x||^2>=0. So your subspace is ALL of R^5.

That's exactly how it stated in this book..So this IS a subspace of R⁵, right?

 

[Dick]

R^5 is a subspace of R^5. Sure.