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)

Thank you in advance.

  • 3
    $\begingroup$ Well, you just wrote the formula, why don't you just replace $Q$ by its definition and see what happens? $\endgroup$ Mar 15, 2022 at 21:20
  • $\begingroup$ $B$ itself from your last paragraph is the bilinear form. $\endgroup$
    – Berci
    Mar 15, 2022 at 21:28
  • $\begingroup$ 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? $\endgroup$
    – Math420
    Mar 16, 2022 at 7:40
  • $\begingroup$ @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? $\endgroup$
    – Math420
    Mar 16, 2022 at 8:25
  • 1
    $\begingroup$ If $v=(x,y,z)$ and $w=(a,b,c)$ then yes, it's the correct formula. $\endgroup$ Mar 16, 2022 at 8:35


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