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One of the definition of basis for a finite dimensional vector spaces is:

Let $V$ be a vector space over a field $F$ and $B = \{v_1, \dots, v_n\} \subset V$. Then, $B$ is a basis for $V$ if for all $v \in V$ there exists unique $(\alpha_1, \dots, \alpha_n) \in F^n$ such that $$ v = \sum_{i=1}^{n} \alpha_i v_i. $$

How does this definition change in infinite dimensional vector spaces? I have checked other posts here but that did not help me.

Here's what I'm thinking: Let $V$ be a vector space over $F$. $B$ is a basis for $V$ if $\forall v \in V$ ($v$ is a linear combination of vectors in $B$) and $\forall u \in B$ the coefficient of $u$ appearing in any linear combination (finite linear combination) of members of $B$ (that gives $v$) is the same. Is this the correct reformulation of the above definition?

user1382316
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3 Answers3

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A finite linear combination makes sense by extending the sum of two vectors by induction (and associativity of addition): $$ v_1 + \cdots + v_{n-1} + v_n := (v_1 + \cdots + v_{n-1}) + v_n $$ It bears repeating: this only makes sense for finite sums. So, a linear combination is always a finite linear combination. The basis set will still be an infinite collection, but only a finite number of them will have nonzero coefficients in the expression of any vector in the space.

A good example to think about here is the vector space over any field $\mathbb{F}$ of polynomials in one indeterminate: $\mathbb{F}[X]$. It has the (monomial) basis $$ \{ 1,\; X,\; X^2,\; X^3,\; \dots \} $$ which is clearly an infinite set, but any particular polynomial (element of the vector space) has a well-defined degree, hence finitely many nonzero terms.

The definition of a basis doesn't change. It's a linearly independent spanning set for the vector space.

Note: There is a notion of a topological vector space in which certain infinite sums are permitted, but this complicates matters as it introduces issues of convergence. Think of power series rather than polynomials.

Sammy Black
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Let $F^{(B)}$ be the set of all $B$-tuples $(\alpha_b)_{b \in B}$ of elements of $F$ such that $\{b \in B : \alpha_b \neq 0\}$ is a finite set.

Then $B$ is a basis for $V$ if and only if for all $v \in V$ there exists a unique $(\alpha_b)_{b \in B} \in F^{(B)}$ such that $$ v = \sum_{b \in B} \alpha_bb. $$ In the case of $B = \{v_1,\dots,v_n\}$ we have that $F^{(B)} = \{(\alpha_{v_1},\dots,\alpha_{v_n}) : \alpha_{v_1},\dots,\alpha_{v_n} \in F\} = F^n$.

azif00
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It won't change, here is the definition from Wikipedia.

For exmaple, consider field of all polynomial with coefficients in $\mathbb{N}$, denote it as $P$. Consider $B=\{1,x,x^2,\ldots\}$ which is infinite. $\forall p(x)\in P$, $p(x)$ has form of $a_nx^n+\cdots+a_{1}x+a_0$ for some unique $a_n,\ldots,a_0\in\mathbb{N}$. In addition, $B$ is linearly independent, so $B$ is a basis.

From the example, we can see that the definition can be also applied to infinite dimension.

lIlIlIlIlIlIlIlIlI
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