Let $X$ be a finite set and $C(X)$ the set of functions $f:X\rightarrow \mathbb{R}$. The bilinear map $Q:C(X)\times C(X) \rightarrow \mathbb{R}$ is called a form. The claim I want to prove is, that there exists a unique $l:X\times X \rightarrow \mathbb{R}$ such that $$Q(f,g)=\sum\limits_{x,y \in X}l(x,y)f(x)g(y)$$
Uniqueness follows since $1_{x},1_{y} \in C(X)$ where $1_{x}$ denotes the function which takes the value $1$ for x and $0$ elsewhere. Thus for arbitrary $x,y \in X$ $$Q(1_{x},1_{y})=\sum\limits_{u,v \in X}l(u,v)1_{x}(u)1_{y}(v)=l(x,y)=l^{'}(x,y)=\sum\limits_{u,v \in X}l'(u,v)1_{x}(u)1_{y}(v)$$ Since $l(x,y)=l^{'}(x,y)$ for all $x,y \in X$ it follows that $l=l^{'}$.
My first thought for the proof of the existence of $l$ was to define $l(x,y):=Q(1_{x},1_{y})$ and use $f(x)=\sum\limits_{u \in X}f(u)1_{x}(u)$ together with the bilinearity of $Q$. But this only yields $$\sum\limits_{x,y \in X}l(x,y)f(x)g(y)=\sum\limits_{x,y \in X}Q(1_{x},1_{y})f(x)g(y)= Q\left(\sum\limits_{x \in X}f(x)1_{x},\sum\limits_{y \in X}g(y)1_{y}\right)$$
Any suggestions how I could prove the existence? Is my idea correct and I'm only making a mistake in the calculation or do I need a different approach?