# How to use Sylvester's law of inertia to show positive definiteness

Let

$G = (<e_i,e_j>)_{i,j=1,...3} :=\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right)$

be a symmetric bilinear form $< \cdot, \cdot>$ with regard to the canonical basis of $\mathbb R^3$,

$H:= \left\{ x \in \mathbb R^3 | <x,x> = -1 \right\}$

and for $a \in H$ name the orthogonal complement of $\mathbb Ra$ with regard to the upper bilinear form: $T_aH$.

I need to show that the restriction $< \cdot, \cdot>|_{T_aH \times T_aH}$ is positiv definite with the help of Sylvester's law of inertia.

How can I do this? I know that we have $\mathbb R^3 = V = V_0 \oplus V_+ \oplus V_-$ and in this case $V_0 = 0$, $V_+= <\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}>$ and $V_-= <\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}>$ as these are the eigenvectors corresponding to the eigenvalues 1 and -1. But that's not the orthogonal complement or $H$, so how can I use this? (Besides I do not really know how to compute the orthogonal complement in this case. If that's necessary, I would be pleased if someone gave me a hint.)

Thanks for help!

By Sylvester's law, you know that $\Bbb Ra$ Is a maximal negative definite subspace. If $\langle b,b \rangle<0$ For some $yb$ in the orthogonal complement to $a$, check to see how the metric would behave in the subspace generated by $a$ and $b$.

With a little more argument, conclude the restriction of the metric to the complement is positive definite.

• Thanks for answering. I'll try;) But just one question - what do you mean by "maximal negative definite"? – xxx May 24 '14 at 20:07
• I mean it is a negative definite subspace which is not properly contained in any other negative definite subspace. This is what Sylvester's law is about :) – rschwieb May 24 '14 at 21:11
• That's what I thought, too, but $\mathbb Ra$, as it's not <(0,0,1)>, is not the negative definite part of the Sylvester decomposition, so I'm a bit confused how to use that. – xxx May 24 '14 at 21:52
• @xxx But Sylvester's law says all maximal negative definite subspaces have equal dimension. It also says the same thing for positive definite subspaces and for isotropic subspaces. If a Sylvester decomposition means decomposing the space into an orthogonal sum of definite and isotropic components, then you should be aware that this isn't unique, ie there is no "the" decomposition. – rschwieb May 25 '14 at 11:44
• @xxx so $\Bbb R a \oplus (\Bbb R a)^\perp$ is another Sylvester decomposition. – rschwieb May 25 '14 at 11:45