# 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

Homework Helper
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

Homework Helper
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

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

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving