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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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    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

1 Answers 1

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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.