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I am trying to prove the statement in the title, that

Every vector space is isomorphic to the set of all finitely nonzero functions on some set.

A finitely nonzero function from $X \rightarrow \mathbb{F}$ (the field of scalars) is one for which there are finitely many $x \in X$ s.t. $f(x) \neq 0$. Also, this set of functions is made a vector space by defining linear combinations the natural way.

I think the solution hinges on the fact that each $v \in V$ is a linear combination of the basis $\mathcal{B}$, and linear combinations are by definition finite (even if $V$ is inf. dimensional).

My attempt: Let $\mathcal{B}$ be a basis. Each $v \in V$ is a (finite) linear combination of elements of $\mathcal{B}$. As such the isomorphism can take each $v \in V$ to $f_v : \mathcal{B} \rightarrow \mathbb{F}$ where $f_v$ takes each basis element to its coefficient in the (unique) linear combination representing $v$. $f_v$ is finitely nonzero because only finitely many basis elements have a nonzero coefficient. We could then prove bijectivity....

Is my solution correct? Also, why is "finite" in the definition of "linear combination"? Wouldn't it make sense to have a vector in $V$ equal to, say, the sum of all basis elements, even if $V$ is inf. dimensional? I think I know one reason-- a polynomial vector space could have $e^x$ as an element otherwise-- but I'd like to hear others.

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  • $\begingroup$ It is important to address the vector space structure of said set $X$, as "isomorphic" only make sense in vector space context (well an algebraic structure). $\endgroup$ Commented Jul 28, 2015 at 5:14

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Your attempt is correct. However, I guess it might be slightly easier to consider the inverse homomorphism $(f\colon \mathcal B\to \mathbb F)\mapsto \sum_{b\in\mathcal B}f(b)b$.

We use "finite" in the definition of linear combination because infinite sums (except when all but finitely many summands are zero - the expressoin $\sum_{b\in\mathcal B}f(b)b$ is of this kind) are not even defined in general. Even if we restrict to the case of a countable index set, or in fact to good-old well-ordered $\mathbb N$, not all "sums" (which are called series instead of sums for a purpose) work: Within $\mathbb Q$ you cannot "sum" $\sum_{n\in\mathbb N}\frac1{n^2}$, within $\mathbb R$ you cannot "sum" $\sum_{n\in\mathbb N}n$ or $\sum_{n\in\mathbb N}(-1)^n$. Without well-order on the index set, the situation becomes worse, not to mention the case of uncountable index set.

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Your solution is correct.

Infinite linear combinations are not such an easy concept as you might think. Your polynomial example is a good one -- though the set of all polynomials with coefficients in $\mathbb{R}$ is a vector space over $\mathbb{R}$, and though $1, x, x^2, x^3, \ldots$ are all independent, an infinite linear combination of these will not be a polynomial. Another example would be the set of continuous functions $f: \mathbb{R} \to \mathbb{R}$. Then many infinite sums of continuous functions are not continuous or even well-defined.

You may be interested to read about Schauder bases, which are (countably) infinite bases for certain kinds of vector spaces. For this notion to work your vector space needs to be Banach -- i.e. it has a norm, and the space is complete under that norm. But unlike with the bases you are used to (called "Hamel bases"), in which everything is a finite linear combination of the basis elements, NOT every space has a Schauder basis. Another caveat is that regardless of how you define infinite sums, not all infinite linear combinations will be defined. For example, take any nonzero vector $v$, and the infinite linear combination $1 v + 1v + 1 v + 1 v + \cdots$ does not have meaning.

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