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Let $V$ be a vector space over a field $F$ and $S$ be a linearly independent subset of $V$.

How do i prove that every vector in $\text{span} S$ has a unique representation?

It is trivial when $S$ is finite, but i don't know how to prove this when $S$ is infinite.

In other words, let $t\in \text{span} S$ and assume that $t=\sum_{i=1}^n a_i u_i = \sum_{j=1}^m b_j v_j$. Then how do i prove that $n=m, a_i=b_i$ and $u_i=v_i$?

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up vote 1 down vote accepted

Let the set $S$ be indexed by elements in an indexing set $\Lambda$. So $S=\{u_{\lambda}|\lambda\in\Lambda\}$. Suppose $t$ can be written as a sum in two distinct ways $t=\sum_{\lambda\in\Lambda} a_{\lambda}u_{\lambda}$ and $t=\sum_{\lambda\in\Lambda} b_{\lambda}u_{\lambda}$. That is, for some $\lambda_0\in\Lambda$ we have $a_{\lambda_0}\neq b_{\lambda_0}$.

But then $0=t-t=\sum_{\lambda\in\Lambda} a_{\lambda}u_{\lambda}-\sum_{\lambda\in\Lambda} b_{\lambda}u_{\lambda}=\sum_{\lambda\in\Lambda} (a_{\lambda}-b_{\lambda})u_{\lambda}$ which implies that $(a_{\lambda}-b_{\lambda})=0$ for all $\lambda\in\Lambda$ because $S$ is a set of linearly independant vectors. In particular, $(a_{\lambda_0}-b_{\lambda_0})=0\Rightarrow a_{\lambda_0}=b_{\lambda_0}$. A contradiction.

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I cannot handle this kind of problem fluently, since I feel very uncomfortable to write infinite sum of $a_i$ where only finite of them are nonzero, because i cannot give topology on it to define limit. What is the definition for $\sum_{\lambda\in \Lambda} a_{\lambda} u_{\lambda}$?Is it defined to be $A$ which satisfies, "there exists a finite subset $F$ of $\Lambda$ such that [$G$ is finite subset of $\Lambda$ and $F\subset G$] implies $\sum_{\lambda \in G}= A$"? – Jj- Feb 14 '13 at 14:18
If $t$ can be written as a finite linear sum over vectors $u_i$, then it can certainly be written as an infinite sum where the coefficient $a_{\lambda}$ of $u_{\lambda}$ is non-zero for finitely many $\lambda$. I used the notation $\lambda\in\Lambda$ simply to take care of the possibility that the vector space $V$ had uncountable dimension. If $V$ had countable dimension, you might like to use the more standard $n\in\mathbb{N}$ notation if that makes it any clearer. – Dan Rust Feb 14 '13 at 14:30

Hint: if you subtract those two sums you will get the zero vector as a linear combination of linearly independent vectors. So...

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Of course, I tried that. That argument works when $S$ is finite, but i don't know how that works when $S$ is infinite. Would you give me some details? – Jj- Feb 14 '13 at 13:51
@Katlus, even if $S$ is infinite, any vector is a finite linear combination. Any vector space has a basis and every vector is a finite linear combination of elements of the basis. – Sigur Feb 14 '13 at 13:57
What do you think is the definition of "linearly independent" when a set is infinite? – GEdgar Feb 14 '13 at 13:57
@GEdgar Definition of linearly independent is clear to me, which is opposite of linearly dependent. Would you see my comment above? – Jj- Feb 14 '13 at 14:20

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