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The following definition is given in Scahum's Ouline of Linear Algebra:

\begin{array}{l}{\text { DEFINTION A: A mapping } q: V \rightarrow K \text { is a quadratic form if } q(v)=f(v, v) \text { for some symmetric }} \\ {\text { bilinear form } f \text { on } V \text { . }} \\ {\text {If } 1+1 \neq 0\text{ in } K,}\\{\text {then the bilinear form } f \text { can be obtained from the quadratic form } q \text { by the following } polar form }\\{\text{ of }f\text{:}} \\ {\qquad f(u, v)=\frac{1}{2}[q(u+v)-q(u)-q(v)]}\end{array}

Where does the last equation appear from, and what does the book mean by polar form?

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You have to compute $q(u+v)$. Using bilinearity, $$ q(u+v)=f(u+v,u+v)=f(u,u)+f(u,v)+f(v,u)+f(v,v)=f(u,v)+f(v,u)+q(u)+q(v). $$ Now, if your field has characteristic $>2$, $f(u,v)+f(v,u)=2f(u,v)$ and you just solve for $f(u,v)$. This identity is called the polarization identity, and it is related to the concept of polar in projective geometry.

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  • $\begingroup$ What do you mean by a field having a characteristic $\gt 2$? $\endgroup$ – David Nov 8 '19 at 16:04
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    $\begingroup$ The characteristic is the minimum number of $1$'s you need to add to get $0$ (respectively, the multiplicative and the additive neutrals in your field). For example, the field $\mathbb{Z}_3$ has characteristic $=3$ since $1+1+1=0$. In the case of the real field, you never get $0$ by adding $1$'s, and in such cases the characteristic is defined to be zero (or infinity, depending on the author). If your characteristic is $>2$, $1+1\neq 0$, and you call $1+1$ simply $2$. The important thing is that $2\neq 0$ so you can divide by it. $\endgroup$ – GReyes Nov 8 '19 at 16:11

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