# Obtaining bilinear form from Quadratic form

So I have the quadratic form $$Q(x,y,z)=x^2+y^2-z^2$$.
I now would like to obtain its corresponding bilinear map.
Now I've studied Sylvesters' theorem which says that any symmetric bilinear map $$B$$ is determined uniquely by its quadratic form. It also says that you can obtain the bilinear map from the polarization identity.

I am not sure how to obtain the bilinear map from here. I do know that, according to my notes,
$$B(v,v)=Q(v)$$ where $$Q:V \to R^3$$ and $$v ∈ V$$.

How do I obtain the bilinear form from $$B(v,w):=\frac{1}{2}(Q(v+w)-Q(v)-Q(w))$$? (using this polarization equation is given as a hint)

• Well, you just wrote the formula, why don't you just replace $Q$ by its definition and see what happens? Mar 15, 2022 at 21:20
• $B$ itself from your last paragraph is the bilinear form. Mar 15, 2022 at 21:28
• Okay so I substitute $Q(x,y,z)$ for $Q(v)$? And an arbitrary $Q(a,b,c,)$ for $Q(w)$. Then $B(v,w)=B(v,w):=\frac{1}{2}(Q(v+w)-Q(v)-Q(w))$ = $\frac{1}{2}((x+a)^2+(y+b)^2-(z+c)^2)-(x^2+y^2-z^2)-(a^2+b^2-c^2)$. So now I have to work this out and that's the bilinear form? Mar 16, 2022 at 7:40
• @CaptainLama Ah yes I think I see it now, since when I work my previous comment out, I am left with $B(v,w)=ax+by-cz$ which seems okay? Mar 16, 2022 at 8:25
• If $v=(x,y,z)$ and $w=(a,b,c)$ then yes, it's the correct formula. Mar 16, 2022 at 8:35