# Thread: Vector space and its subspace

1. ## Vector space and its subspace

Let $\displaystyle V$ be a vector space, and let $\displaystyle W \subset V$ be its subspace. Is it true, that there exists a space $\displaystyle W'$ such that $\displaystyle V = W \oplus W'$?

It is clear for finite-dimensional spaces (you can work with bases, and matrices)! However i have no clue how to do this in the general case? Is it really true, or is there any counterexapmle?

2. ## Re: Vector space and its subspace

do you believe in the axiom of choice?

if so, then every vector space has a basis. this includes W. so we may assert we have a basis B for W, and can extend this to a basis C for V.

define W' = span(C\B).

(there are some vector spaces for which such bases have never been adequately described. the usual example is the real numbers as a vector space over Q. R is uncountable, so any basis has to be uncountable, so explicitly listing one is out of the question).

3. ## Re: Vector space and its subspace

Originally Posted by Deveno
do you believe in the axiom of choice?

if so, then every vector space has a basis. this includes W. so we may assert we have a basis B for W, and can extend this to a basis C for V.

define W' = span(C\B).

(there are some vector spaces for which such bases have never been adequately described. the usual example is the real numbers as a vector space over Q. R is uncountable, so any basis has to be uncountable, so explicitly listing one is out of the question).
Yes, I know, that when assuming axiom of choice every v. s. has a base. But I haven't seen in any book, the theorem allowing me to extend the base of the subspace to some base of space. Naturally this would allow us to prove the above theorem.

4. ## Re: Vector space and its subspace

well, it's clear we can pick a basis B for W. consider the set L of all linearly independent sets of V that contain B. B is in L, so L is non-empty. L is also partially ordered by inclusion.

for each chain C in L, define C' = UC. clearly, C' is an upper bound for C. let {x1,x2,...,xn} be any finite set of vectors in C'.

since each xj is in C, there is some element Cj in C with xj in Cj, thus some finite union in C, C1 U C2 U....U Cn containing {x1,x2,...xn}
such that xj is in Cj for 1 ≤ j ≤ n.

since C is totally ordered, there is some Ck where 1 ≤ k ≤ n, such that Ck = C1 U C2 U...U Cn, so {x1,x2,...,xn} ⊆ Ck, and since Ck is in L, {x1,x2,...,xn} is linearly independent.

but this means C' is in L (since any finite subset of it is linearly independent).

so by the axiom of choice (ok, zorn's lemma, whatever) L has a maximal element, M, which is a basis for some subspace of V.

suppose y is in V - span(M). let {v1,v2,....,vn} be any finite set of vectors in span(M), and suppose that:

a1v1 + a2v2 +....+anvn - by = 0. if b = 0, for every such set, (and choice of coefficients) then a1 = a2 =...= an = 0, by the linear independence of M,
but then M U {y} is linearly independent, contradicting its maximality.

but if for some such set b ≠ 0, then y = (a1/b)v1 + (a2/b)v2 + ...+ (an/b)vn,

contradicting our choice of y. so there must not be any such y, so V - span(M) = Ø, so M is a basis for V, that contains B as a subset.

5. ## Re: Vector space and its subspace

Thank you, that is great