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I am struggling to understand the definition of a basis in an infinite dimensional vector space. Specifically, the definition I know says: A subset $B$ of a vector space $V$ is a basis for $V$ if every element of $V$ can be written in a unique way as a finite linear combination of elements from $B$.

However, for any non-empty subset $X$ of a vector space $V$, the zero element of the space can be written in more than one way as a finite linear combination of elements from $X$. For example, $0 = 0v = 0w$, where $v \neq w$ are from $X$. So therefore, no subset $X$ of a vector space $V$ could be a basis for $V$.

What am I missing? What exactly does the definition mean?

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    $\begingroup$ You $0v$ and $0w$ are the same linear combination because both of them are $0v+0w$. The idea is you want to write $0=c_1v+c_2w$, once you choose $c_1=0$ you still need to specify what is $c_2$. $\endgroup$ – Anurag A Mar 29 '16 at 13:32
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The language used is a bit sloppy, but it's common not to be too fussy in these definitions.

Let $S$ be a subset (finite or infinite, it doesn't matter) of $V$. A choice of coefficients for $S$ is a function $f\colon S\to F$ (where $F$ is the base field) such that $\sigma(f)=\{x\in S:f(x)\ne0\}$ is finite.

We observe that, if $T$ is a finite subsets of $S$ such that $\sigma(f)\subseteq T$, then $$ \sum_{x\in\sigma(f)}f(x)x=\sum_{x\in T}f(x)x $$ with the convention that $$ \sum_{x\in\emptyset}f(x)x=0 $$ Since the summation doesn't depend on the finite subset we choose, we set, for a choice of coefficients $f$, $$ \sum_{x\in S}f(x)x=\sum_{x\in\sigma(f)}f(x)x $$ and call this vector a linear combination of $S$. Note that the summation is actually finite and we can use whatever subset $T$ we want, instead of $\sigma(f)$, provided $\sigma(f)\subseteq T$ and $T$ is finite. This is mostly useful for doing computations with choices of coefficients, when other properties are being investigated.

Then we call $S$ a basis for $V$ if

  1. For every $v\in V$, there exists a choice of coefficients $f$ for $S$ such that $$ \sum_{x\in S}f(x)x=v $$ (we can abbreviate this condition by saying that $S$ is a spanning set for $V$).

  2. For every $v$, if $f$ and $g$ are choices of coefficients for $S$ and $$ \sum_{x\in S}f(x)x=\sum_{x\in S}g(x)x $$ then $f=g$ (we can abbreviate this condition by saying that $S$ is linearly independent).

Note that condition 2 can be rewritten as “If $f$ is a choice of coefficients for $S$ and $$ \sum_{x\in S}f(x)x=0 $$ then $f(x)=0$, for every $x\in S$”.

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$\newcommand{\Reals}{\mathbf{R}}$Let $(V, +, \cdot)$ be a real vector space. One commonly says an ordered set $S = (v_{i})_{i \in I}$ indexed by a set $I$ is a basis of $V$ if the following hold:

  • $S$ spans $V$, i.e., for every $v$ in $V$, there exists a function $c:I \to \Reals$, non-zero for only finitely many $i$, such that $v = \sum_{i} c(i) v_{i}$.

  • $S$ is linearly independent, i.e., if $c:I \to \Reals$ is a function, non-zero for only finitely many $i$, and if $0 = \sum_{i} c(i) v_{i}$, then $c(i) = 0$ for all $i$.

In this framework, "uniqueness of representation" is a (simple) theorem: If $v \in V$, there exists a unique function $c:I \to \Reals$, non-zero for at most finitely many $i$, such that $v = \sum_{i} c(i) v_{i}$.

The point is, you can (and for uniqueness, should) view a linear combination from $S$ as a sum over all elements of $S$, but in such a way that at most finitely summands have a non-zero coefficient.

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When you say "in a unique way", it implies that you disregard the zeroes. Otherwise of course no space would have a basis, infinite-dimensional or not (except the zero space...).

This being said, if you want a more subtle definition : a set $S\subset X$ is a basis of $X$ if for any vector space $Y$ and any function $f:S\to Y$ there is exactly one linear map $X\to Y$ that extends $f$.

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  • $\begingroup$ It's not really that you disregard the zeroes, it's that only a finite number of coefficients may be nonzero, and vectors with a zero coefficient don't affect the sum, so you might as well not write them. But they're still there in the linear combination. $\endgroup$ – Najib Idrissi Mar 29 '16 at 13:59
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The expression for $0$ you written are not two; they are same: $$0=0v=0v+0w \mbox{ and } 0=0w=0v+0w .$$


Let $B$ be a subset of $V$. Let $v\in V$ be written as a finite linear combination of elements of $B$: $$v=a_1v_1+a_2v_2+\cdots + a_nv_n =b_1w_1+b_2w_2+\cdots + b_mw_m, \,\,\,\, v_i,w_j\in B.$$

Question: When do we say that these expressions are same?

To answer this, note that $v$ is expressed as linear combination of elements of the set $\{v_1,\cdots,v_n\} \cup \{w_1,\cdots,w_m\}$, and some $v_i$'s and $w_j$'s could be same. We can rewrite the two expressions as below: $$ v = a_1v_1+\cdots + a_nv_n + 0w_1 + \cdots + 0w_m.$$ and $$ v = 0v_1+\cdots + 0v_n + b_1w_1 + \cdots + b_mw_m.$$ Since some $v_i$'s and $w_j$'s could be same, so let $$\{v_1,\cdots, v_n\} \cup \{ w_1,\cdots , w_m\}=\{u_1,u_2,\cdots, u_l\}$$ where all the $u_i$'s are distinct. Then the above two expressions for $v$ can be expressed in terms of $u_i$'s only, using rules in scalar multiplication: $$v=c_1u_1 +\cdots + c_lu_l = c_1'u_1 + \cdots + c_l'u_l.$$ Now these two expressions for $v$ are said to be same if $c_i=c_i'$ for all $i$.

Illustration: Let $v=a_1v_1+a_1v_2$ and $v=b_1w_1 + b_2w_2$, and suppose, $v_2=w_2$ Then the set $\{v_1,v_2\} \cup \{w_1,w_2\}$ contains only three elements: $v_1,v_2,w_1$; relabel them as $$v_1=u_1, \,\,\,\, v_2=u_2, \,\,\,\, w_1=u_3.$$ Note that $w_2=v_2$ and $v_2$ is labeled as $u_2$, so $w_2=u_2$.

The two expressions of $v$ now become: $$v=a_1v_1 + a_2v_2 =a_1u_1 + a_2u_2= a_1u_1 + a_2u_2 + 0u_3$$ and $$ v =b_1w_1 + b_2w_2= b_1u_3 + b_2u_2 = 0u_1 + b_2u_2 + b_1u_3.$$ Now can you say when the last two expressions for $v$ are same?

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