# Bilinear form obtained from Quadratic form

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?

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.
• What do you mean by a field having a characteristic $\gt 2$? – David Nov 8 '19 at 16:04
• 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. – GReyes Nov 8 '19 at 16:11