I am to prove that for some linear function $L:$ $R^n$ --> $R$, there is a unique vector $\chi$ in $R^n$ such that $L(x)=\chi*x$ where $*$ denotes the dot product.
First, to prove the existence of such a vector $\chi$, let $\chi=(L(e^1),...,L(e^n))$ where $(e^1,...e^n)$ is the standard basis. Because $x=\sum_{i=1}^nx_ie^i$, it follows that $L(x)=L(\sum_{i=1}^nx_ie^i)=\sum_{i=1}^nx_iL(e^i)=x*\chi=\chi*x $.
Where I'm possibly a little confused is in showing that $\chi$ is unique. Here's what I have at the moment:
Assume there exists another vector in $R^n$ called $\chi'$ such that $L(x)=\chi'*x$. Then $\chi*x=\chi'*x$, i.e. $x_1(\chi_1'-\chi_1)+...+x_n(\chi_n'-\chi_n)=0$ for all $x$ in $R^n$. Therefore, $\chi_i'=\chi_i$ for all $i=1,...,n$, i.e. $\chi'=\chi$. Is this a logical proof of uniqueness?
In view of uniqueness, take the unit vectors $x=e_i$, $1\leq i\leq n$, and then consider $\langle \chi,x\rangle = \langle \chi',x\rangle$.