Equivalent definitions of quadratic form Let $V$ be an $n$-dimensional vector space over $\mathbb{K}$. $q: V\rightarrow \mathbb{K}$ is a quadratic form if:


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*Definition 1: there exists $\phi: V \times V \rightarrow \mathbb{K}$ bilinear and symmetric such that $q(v)=\phi(v,v)$, $\forall v \in V$

*Definition 2: there exist $f_1...f_N$, $g_1...g_N: V \rightarrow \mathbb{K}$, linear, such that $q(v)=\sum_{i=1}^{N}f_i(v)g_i(v)$
I'd like to know when they're equivalent:


*

*$(1) \implies (2)$: choose a basis $\{ v_1,...,v_n\}$ of $V$ and let $x=(x_1,...,x_n)$ be the coordinates of $v$. Then $q(v)=\sum_{i=1}^{n}x_i^2a_{ii}+\sum_{i<j}x_ix_ja_{ij}$, where $a_{ij}=\phi(v_i,v_j)$. Letting $F_i(v)=x_i$ and doing some manipulation on the indices should yield $(2)$

*$(2) \implies (1)$: assuming $2\neq 0$ in $\mathbb{K}$, $\phi(v,w):=2^{-1}(q(v+w)-q(v)-q(w))$
But what if $2=0$? Is there a way to establish the equivalence regardless of the field's characteristic/is there any other more general definition?
 A: The issue in the case $2=0$ is that several different symmertric forms have the same quadratic form. In $K^n$ for example every symmetric form has the representation $\phi(x,y) = \langle Ax,y\rangle$ with a symmetric matrix $A$. Now, $\langle Ax,x\rangle = \sum_i\sum_jx_ix_j\langle Ae_i,e_j\rangle = \sum_ix_i^2\langle Ae_i,e_i\rangle$. Hence, all symmetric matrices with the same diagonal generate the same quadratic form.
In the case $2=0$ the conditions aren't equivalent. As a counterexample, let $n=2$, and $q(x) = f(x)g(x)$, where $f(x) = x_2$ and $g(x) = x_1$. Thus, $q(x) = x_1x_2$. But as you saw above, we must have $q(x) = a_1x_1^2+a_2x_2^2$. Plugging in $x=e_k$ yields $a_k=0$ for $k=1,2$ and hence $q=0$, a contradiction.
In the case $2=0$, what you need to add to the condition in (2) is a certain symmetry condition, namely that $\sum_i f_i(x)g_i(y) = \sum_i f_i(y)g_i(x)$ (which is easily seen to be satisfied if $q$ is indeed a quadratic form). Then $q$ is indeed a quadratic form. To see this, we assume WLOG that $V = K^n$. Then $f_i(x) = \sum_jf_{ij}x_j$ and $g_i(x) = \sum_jg_{ij}x_j$. The symmetry condition means that $\sum_if_{ij}g_{ik} = \sum_if_{ik}g_{ij}$ for $j\neq k$. Set $a_{jk}:=\sum_if_{ij}g_{ik}$. Then $A = (a_{jk})$ is symmetric and
\begin{align}
q(x)
&= \sum_if_i(x)g_i(x) = \sum_i\sum_j\sum_k f_{ij}g_{ik}x_jx_k = \sum_{j,k}a_{jk}x_kx_j = \langle Ax,x\rangle.
\end{align}
Hence, $q$ is a quadratic form.
Old answer:
Your idea for the proof from (1) to (2) to start with a basis of $V$ is fine. But then it's not clear how you go on. If $v = \sum x_iv_i$ and $w = \sum y_iv_i$, then $\phi(v,w) = \sum_i\sum_j\phi(v_i,v_j)x_iy_j = \langle Ax,y\rangle$, where $A$ is the symmetric matrix with entries $\phi(v_i,v_j)$. Now,
$$
Ax = A\sum_i\langle x,e_i\rangle e_i = \sum_i \langle x,e_i\rangle z_i,
$$
where $z_i := Ae_i$. Hence,
$$
\phi(v,w) = \langle Ax,y\rangle = \sum_i \langle x,e_i\rangle\langle z_i,y\rangle.
$$
Now, set $f_i(v) = \langle x,e_i\rangle$ and $g_i(w):=\langle z_i,y\rangle$. Then $\phi(v,w) = \sum_if_i(v)g_i(w)$, and, in particular, $\phi(v,v) = \sum_if_i(v)g_i(v)$.
