# Subspace of a Sphere/Vector

• RedXIII
In summary, for problem 1, S is not a subspace of R^n because it does not satisfy the definition of a subspace. In problem 2, the vectors 2v1 + 4v2, v2 + 2v3, and 5v3 form a basis of S because they satisfy the conditions of spanning S and being linearly independent, which are the requirements for a set of vectors to form a basis of a subspace.

## Homework Statement

1) Is S a subspace of R^n?

1.1) Given n=4 and a vector is in S if it is in the span of e1, e2 or in the span of e3, e4 where e1...e4 is the canonical basis of R^4

1.2) Given n=3 and S is a sphere of radius 1.

2) Let S be a subspace of R^10 with basis v1; v2; v3. Show that the vectors 2v1 + 4v2; v2 + 2v3; 5v3 also form a basis of S.

## The Attempt at a Solution

1.1) No, because if the vector only spans e1, e2 or e3, e4, then it would only generate a plane in R^4?

1.2) I was thinking yes because a sphere is a 3-d object, thus would be a subspace of itself in R^3. However, if the radius is one then would that be false because if I were to add the vectors of the sphere, it would be outside the sphere?

2) From what I read, I think I would have to prove that they are linear independent then prove that they can generate R^10 (both of which are already proven if v1,v2,v3 is the basis of R^10) but how would I show it without any actual numbers?

Thanks!

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Anybody?

What's the definition of a subspace? For problem 1, you want to show that those cases either satisfy or don't satisfy the definition, so if you don't know the definition, that's the first thing you need to look up.

For problem 2, you need to be bit more careful. The problem doesn't say v1, v2, and v3 form a basis for R10. It says they form a basis for S, a subspace of R10. Now you're given three vectors — let's call them w1, w2, and w3 — and you want to show they form a basis for S. You need to show two things:

1. w1, w2, and w3 span S. That is, if x is in S, then you can find constants c1, c2, and c3 such that x=c1w1+c2w2+c3w3.
2. w1, w2, and w3 are linearly independent. That is, if c1w1+c2w2+c3w3=0, then c1=c2=c3=0 is the only solution.