# Basis of a Subspace

#### Caspian

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 :).

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#### Fredrik

Staff Emeritus
Gold Member
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.

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#### Caspian

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 :).

#### Landau

I don't think you fully get it yet.
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.
It should be obvious that B={(1,0,0,0),...,(0,0,0,1)} isn't a basis for W, since B spans the whole space R^4! So they can't be a basis for any proper subspace of V (indeed, they span "too much space").

But that's not what your book is asserting. They are only talking about a subset of B={(1,0,0,0),...,(0,0,0,1)}. So they're saying that even some subset of B cannot be a basis of W. Remember that a basis of W first of all consists of elements of W. None of the vectors in B are in W.