Q:

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

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 $u_i=v_i$?

A:

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.

