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This is a question from Pinter's A Book of Abstract Algebra p 289

He asks for a basis of the subspace of $\mathbb{R}^3$ spanned by the set of vectors $(x,y,z)$ that satisfy $x^2+y^2+z^2=1$

I don't really understand this questions but it seems like since the sphere is a surface it should be a basis with two elements that are linearly independent.

So can we just choose $(1,0,0),(0,1,0)$ as our basis? They are in the space and linearly independent but how would we ever write say $(\sqrt{(1/3)},\sqrt{(1/3)},\sqrt{(1/3)})$.

Edit: This is exercise C.6 of Chapter 28 (Vector Spaces) of Pinter's book. The exact wording of the exercise is:

Find a basis for the subspace of $\mathbb R^3$ spanned by the set of vectors $(x,y,z)$ such that $x^2 + y^2 + z^2 = 1$.

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    $(\sqrt{(1/3)},\sqrt{(1/3)},\sqrt{(1/3)})=\sqrt{(1/3)}(1,0,0)+\sqrt{(1/3)}(0,1,0)+\sqrt{(1/3)}(0,0,1) $.2012-01-07
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    @Nikhil see you had to use all three basis elements though2012-01-07
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    What kind of basis are we talking about? Presumably not a linear basis, given that this isn't a linear subspace!2012-01-07
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    The $0$-vector is in any subspace of $\mathbb{R}^3$. It is not on the sphere. Subspaces are, apart from the two trivial cases the $0$-vector and the whole space, lines and planes through the origin.2012-01-07
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    Ok I have edited the question to more directly reflect the wording in the book I must have misinterpreted the statement of the question.2012-01-07
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    @user9352 we cant express vectors with a non-zero z-component using only $(1,0,0)$ and $(0,1,0)$, so all three vectors must be in the basis. By definition, basis is the set of linearly independent vectors such that all vectors in a vector space can be expressed as a linear combination of them. your set of 2 vectors doesnt satisfy this condition for $\mathbb{R^3}$, so it cant be a basis for $\mathbb{R^3}$.2012-01-07
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    So I think now that the vectors in question must span all of $R^3$ since the standard basis is included among them. So the standard basis will answer the question.2012-01-07
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    @user9352: Yes, "spanned" makes all the difference. Space spanned is $\mathbb{R}^3$, since the standard basis vectors are on the sphere. So any basis of $\mathbb{R}^3$ will do, and there is nothing cheaper.2012-01-07
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    Please typeset in the question as it actually appears, including punctuation. What you typed is grammatically ambiguous at best.2012-01-07
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    Note that this exercise actually asks you to find an element in $GL(3)$. (set of 3x3-matrices with determinant $\pm 1$)2012-01-08

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Since $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ all satisfy $x^2+y^2+z^2=1$ and those three vectors span all of $\mathbb{R}^3$, the subspace of $\mathbb{R}^3$ for which a basis is sought is all of $\mathbb{R}^3$. Thus, you need $3$ basis vectors, such as the three given above.