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Define a Pseudo-Riemannian Metric $g$ in $\mathbb{R}^{n+1}$ by $g(u,v)=-u_0v_0+u_1v_1+...+u_nv_n$, where $u=(u_0,u_1,...u_n)$. Let $\eta\in\mathbb{R}^{n+1}$ be a vector such that $g(\eta,\eta)=-1$. Is it possible to find a basis $(\eta,w_1,...,w_n)$ of $\mathbb{R}^{n+1}$ such that $g(w_i,w_j)=\delta_{ij}$ and $g(\eta,w_i)=0$ for $i,j=1,..,n$


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What happens when you try the standard basis of $\mathbb{R}^{n+1}$? – Neal Nov 27 '12 at 21:24
Note that i fixed $\eta$ @Neal. If for example $\eta=(-1,0,...,0)$ then we can use the standard basis. – Tomás Nov 27 '12 at 21:25
What if $\eta = (1,0,\cdots,0)$? What is $g(\eta,\eta)$ then? – Neal Nov 27 '12 at 21:27
$g(\eta,\eta)=-1$ – Tomás Nov 27 '12 at 21:28
Misread your question - sorry about that. – Neal Nov 28 '12 at 0:08
up vote 2 down vote accepted

Your desired basis exists. Let $\eta^T=(a,v^T)$. Then $a\not=0$ as $g(\eta,\eta)=-1$. You are essentially asking if there exist a real matrix $C$ and a real vector $u$ such that the following equality holds: $$ (\ast):\ \begin{pmatrix}a&v^T\\u&C^T\end{pmatrix} \begin{pmatrix}-1\\&I_n\end{pmatrix} \underbrace{\begin{pmatrix}a&u^T\\v&C\end{pmatrix}}_{M} =\begin{pmatrix}-1\\&I_n\end{pmatrix}. $$ If so, then the columns of $M$ are the desired basis vectors. Note that the LHS of $(\ast)$ is equal to $$ \begin{pmatrix}a&v^T\\u&C^T\end{pmatrix} \begin{pmatrix}-a&-u^T\\v&C\end{pmatrix} =\begin{pmatrix}-1&-au^T+v^TC\\-au+C^Tv&C^TC-uu^T\end{pmatrix}. $$ So, $(\ast)$ is equivalent to the system of equations \begin{align} (1):\ & C^TC-uu^T=I_n,\\ (2):\ & u = \frac{1}{a}C^Tv. \end{align} Substituting (2) into (1), we get $$(3):\ C^T(I_n - \frac{1}{a^2}vv^T)C=I_n.$$ Now, as $g(\eta,\eta)=-1$, we have $-a^2+\|v\|^2=-1$, i.e. $1-\frac{\|v\|^2}{a^2}=\frac{1}{a^2}$. Therefore $I_n-\frac{1}{a^2}vv^T$ is positive definite. Hence we can write it as $QDQ^T$ for some real orthogonal matrix $Q$ and some positive diagonal matrix $D$. Now, take $C=QD^{-1/2}Q^T$ and (3) holds.

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Thanks you, nice calculation. – Tomás Nov 28 '12 at 10:17

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