Correspondence between bilinear forms and linear operators. Let $V$ be a finite set. We define $l(V)=\{\text{functions } f \mid f:V\rightarrow \mathbb{R}\}$, this is a vector space with the usual sum and scalar product. This vector space has an inner product: $u,v\in l(V)$
$$\langle u,v \rangle =\sum_{p\in V}u(p)v(p) $$

We consider the set $B(V)$ of all the symmetric bilinear forms $\mathcal{E}$ on $l(V)$ such that
(1) $\mathcal{E}(u,u)\geq 0$ for every $u\in l(V)$.
(2) $\mathcal{E}(u,u)=0$ if and only if $u$ is a constant function.

In a similar fashion,

We consider the set $L(V)$ of all the symmetric linear operators $H:l(V)\rightarrow l(V)$ with the following properties
(1) $H$ is non-positive definite (i.e $\langle u,Hu \rangle\leq 0$ for every $u\in l(V)$).
(2) $Hu=0$ if and only if $u$ is constant.

I want to know why the assignment $H\mapsto \mathcal{E}_H$ defined as $\mathcal{E}_H(u,v)=\langle u,Hv \rangle$ is a bijective map between $L(V)$ and $B(V)$.
I could not prove that if $\mathcal{E}_H(u,u)=0$ then $u$ is constant. This is part of the property (2) of a bilinear form in $B(V)$.
Edit: Symmetric linear operator mean that $\langle u,Hv \rangle=\langle u,Hv \rangle$.
 A: I'm going to assume that you meant $H$ to be non-negative definite, otherwise the map $\newcommand{\myE}{\mathcal{E}} H\mapsto \myE_H$ isn't even a map $L(V)→ B(V)$. Lets call this map $\newcommand{\myA}{\mathcal A}\myA:L(V)→ B(V)$.
$\myA$ is clearly injective. To prove it is surjective, pick $B∈ B(V)$. Choose an orthonormal basis $e_1… e_n$ of $l(V)$ and define the linear map $\newcommand{\myB}{\mathcal B}\myB:l(V)→ l(V)$ by $\myB(v) = \sum_i B(e_i,v)e_i$. Note that $⟨ e_j, \myB(e_k)⟩ = B(e_j,e_k)$, proving all required properties by using the linearity in each argument. So every $B$ is some $\myE_\myB$, and $\myA$ is surjective.
The non-negativity of $\myE_H$ is proven similarly (see Khanickus's answer for a full write-up)
A: To show that $\epsilon_H$ is in $B(V)$ it follows from that every symmetric operator can be diagonalized with an orthonormal basis.
The $Hu=0\iff u$ is constant is just saying that $H$ has a a one dimensional kernel $(H(1)=0$ where 1 is the constant function). So 1 is an eigenvector and the only one with zero as an eigenvalue. Complete it to an orthonormal basis $e_2,...,e_n$ with $He_j=\lambda_je_j$ for some $\lambda_j>0$ (since $\langle u,Hu\rangle \ge 0$ for all $u$).
Now for(sum over upper and lower indeces) $u=u_11+u^je_j$:
$\epsilon_H(u,u)=\langle u^ie_i, u^j\lambda_j e_j\rangle=u^iu^j\lambda_j\delta_{ij}=\lambda_j(u^j)^2=0\iff u^j=0$ for all $j\iff u=u_11$ is constant.
To show the map is onto given $\epsilon\in B(V)$ take any orthonormal basis $e_1,...,e_n$ and set $H(e_j)=\sum_i\epsilon(e_i,e_j)e_i$.
This $H$ is symmetric so diagonalize with some other orthonormal basis $f_1,...,f_n$ with eigenvalues $\lambda_i\ge 0$.
Now $\epsilon(f_i, f_j)=\lambda_j\delta_{ij}$. Since $\epsilon(u,u)=0\iff u$ is constant wlog $\lambda_1=0$, $f_1=1$ and $\lambda_j>0$ for $j>1$.
To show one-to-one take the basis for $l(V)$ where $e_i(v_j)=\delta_{ij}$ (write $V=\{v_1,...,v_n\}$).
Then if $\epsilon_{H}=\epsilon_{K}$ you have on this basis that:
$He_j (v_i)=\epsilon_{H}(e_i,e_j)=\epsilon_{K}(e_i,e_j)=Ke_j(v_i)$ so that $He_j=Ke_j$ and so $H=K$ as they agree on a basis.
