# Smallest subspace of a vector space

1. Jun 30, 2015

### strobeda

I may have a bad day, or not enough coffee yet.
So,

"If A is a nonempty subset of a vector space V, then the set
L(A) of all linear combinations of the vectors in A is a subspace, and it is
the smallest subspace of V which includes the set A.

If A is infinite, we obviously can't use a single finite listing. However, the
sum of two linear combinations of elements of A is
clearly a finite sum of scalars times elements of A."

Question:
If A is infinite, how can the sum of its linear combinations be finite (even choosing a finite sum of scalars)?

Thank you.

2. Jun 30, 2015

### Orodruin

Staff Emeritus
The elements in A need not be linearly independent. On the contrary, if V is of finite dimension n, A contains at most n linearly independent elements.

3. Jun 30, 2015

### HallsofIvy

What, exactly, is the definition of "linear combination" in your text? I suspect that a "linear combination" of vectors is required to be a sum of a finite sum of scalars times vectors. Also, this does not say "the sum of its linear combinations", it says "the sum of two linear combinations". If a linear combinations is required to be a sum of a finite number of vectors, selected from A, then a sum of two (or any finite number) of linear combinations still involves only a finite number of vetors from A.

4. Jul 1, 2015

### Fredrik

Staff Emeritus
That sentence is pretty hard to understand. You're asking about the sum of "its" linear combinations, where "it" refers to A. So that would be the sum of all linear combinations of elements of A. The statement you quoted doesn't mention that sum.

I suspect that the issue is what Halls is talking about. A linear combination is by definition a finite sum. Even if A is an infinite subset of an infinite-dimensional vector space X, the set of linear combinations of elements of A (i.e. the span of A) is equal to the smallest vector space that has A as a subset (i.e. the vector space generated by A). This subspace may actually be infinite-dimensional, if X is. (Suppose e.g. that X is the set of all polynomials, and that A is the set of all odd-powered monomials $x,x^3,x^5,\dots$).

This isn't hard to prove. The key observation is that every subspace of X that has A as a subspace must contain all linear combinations of elements of A. So no subspace of X that has A as a subspace can be a proper subspace of the set of all linear combinations of A.

5. Jul 1, 2015

### strobeda

Thank you all!
It was the coffee; I didn't see the summation limits!

"If A is infinite, we obviously can't use a single finite listing. However, the
sum Σ1Σn of two linear combinations of elements of A is
clearly a finite sum of scalars times elements of A."

Last edited: Jul 1, 2015
6. Jul 1, 2015

### strobeda

Thank you all!

It was the coffee!

I didn't see the summation indices (1...n) and (1...m) for the two sums of linear combinations!
From the infinite indexed set A, only one finite listing at a time is taken because there is no finite spanning set.

7. Jul 2, 2015

### WWGD

It may not be part of your course, but if you are interested, you can read on Hamel bases and Schauder bases; these last can deal with infinite sums, but then you need to deal with convergence issues, for which you need to deal with a topology.

8. Jul 2, 2015

### strobeda

Thank you, WWGD.

I will search.