# About scalar products on $\mathbb{R}^n$

In what follows, $\mathbb{L}^{n}$ is the Lorentz space (Euclidean space $\mathbb{E}^{n}$ with the Lorentz scalar product).

Theorem

Let $P$ be a $k$-dimensional subspace of $\mathbb{L}^{n}$. Then exactly one of the following statements about $P$ is true:

(i) $P$ = $\mathbb{L}^{k}$, and $< , >_p$ is nondegenerate;

(ii) $P$ = $\mathbb{E}^{k}$, and $< , >_p$ is nondegenerate;

(iii) $< , >_p$ is degenerate, and in this case (and only in this case), $P = E^{k-1}$ $\oplus$ $<c>$, where $c$ satisfies $<c,c> = 0$ and is orthogonal to $E^{n-1}$;

My idea was to use an that $ind$ $\mathbb{L}^{n}$ = $ind$ $P$ + $ind$ $P_{perp}$ (in the cases (i) and (ii)), where ind indicates the index of the subspace. But $ind$ $\mathbb{L}^{n}$ = 1, so ind P is either 0 or 1, which gives (i) and (ii). If < , >\P is degenerate, we have that P $\cap$ $P_{perp}$ is different from $0$, so it's at least a subspace of dimension 1 (and it's here that we find this $c$ such that $<c,c>$ = 0). How can I prove that this subspace has dimension 1? And that it's orthogonal complement is isomorphic to $\mathbb{E}^{k-1}$? Thanks in advance.

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Please add \$ signs to get LaTeX formatting, and also please check my edit to ensure this is what you meant. – nbubis Sep 3 '13 at 0:04
Done! :) Thanks – Br09 Sep 3 '13 at 3:35