# Basis of a Subspace

by Caspian
Tags: basis, linear algebra, subspace
 P: 15 My book made the following claim... but I don't understand why it's true: If $$v_1, v_2, v_3, v_4$$ is a basis for the vector space $$\mathbb{R}^4$$, and if $$W$$ is a subspace, then there exists a $$W$$ which has a basis which is not some subset of the $$v$$'s. The book provided a proof by counterexample: Let $$v_1 = (1, 0, 0, 0) ... v_2 = (0, 0, 0, 1)$$. If $$W$$ is the line through $$(1, 2, 3, 4)$$, then none of the $$v$$'s are in $$W$$. Is it just me, or does this not make any sense? First of all, (1,2,3,4) is a linear combination of (1,0,0,0)...(0,0,0,1), isn't it? I'm very confused... Any help would be greatly appreciated :).
 Emeritus Sci Advisor PF Gold P: 9,224 Yes, but that's not what they're saying. They're saying that none of the four $v_i$ vectors are in W. They're also saying that a basis vector of W (which must be a multiple of (1,2,3,4)) can't be equal to one of the $v_i$. When they talk about the set of "v's" they really mean a set that only has four members, not the subspace they span.
 P: 15 Ok, ok... I get it. (1,2,3,4) is in the vector space spanned by (1,0,0,0)...(0,0,0,1), but these vectors aren't a basis for the subspace which is the line through (1,2,3,4) because these vectors span too much space. Ok, I guess this was a dumb question. Thanks for your help :).