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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 3 down vote accepted

Your desired basis exists. Let $\eta^T=(a,v^T)$. You are essentially asking if there exist a real matrix $C$ and a real vector $u$ such that the following equality holds: $$ \begin{pmatrix}a&v^T\\u&C^T\end{pmatrix} \begin{pmatrix}-1\\&I_n\end{pmatrix} \underbrace{\begin{pmatrix}a&u^T\\v&C\end{pmatrix}}_{\color{red}{M}} =\begin{pmatrix}-1\\&I_n\end{pmatrix}.\tag{$\ast$} $$ If so, then the columns of $M$ are the desired basis vectors. Note that the LHS of $(\ast)$ is equal to $$ \begin{pmatrix}a&v^T\\u&C^T\end{pmatrix} \begin{pmatrix}-a&-u^T\\v&C\end{pmatrix} =\begin{pmatrix}-1&-au^T+v^TC\\-au+C^Tv&C^TC-uu^T\end{pmatrix}. $$ Since $g(\eta,\eta)=-a^2+\|v\|^2=-1$, we see that $a\ge1$. Hence $(\ast)$ is equivalent to the system of equations \begin{align} & C^TC-uu^T=I_n,\tag{1}\\ & u = \frac{1}{a}C^Tv.\tag{2} \end{align} Substitute (2) into (1), we get $$ C^T\left(I_n - \frac{1}{a^2}vv^T\right)C=I_n. $$ Now, from $g(\eta,\eta)=-a^2+\|v\|^2=-1$, we also get $1-\frac{\|v\|^2}{a^2}=\frac{1}{a^2}$. Therefore $I_n-\frac{1}{a^2}vv^T$ is positive definite and it has a positive definite square root. Hence we may simply take $$ C=C^T=\left(I_n - \frac{1}{a^2}vv^T\right)^{-1/2} =I+\left[\left(1-\frac{\|v\|^2}{a^2}\right)^{-1/2}-1\right]\frac{vv^T}{\|v\|^2} $$ (with $C=I$ when $v=0$). Using the equality $\|v\|^2=a^2-1$, we further obtain $C=I+(|a|-1)\frac{vv^T}{\|v\|^2}=I+\frac{vv^T}{|a|+1}$ and also $u=\frac{1}{a}C^Tv=\frac{|a|}{a}v$. So, you may obtain the desired basis vectors from the columns of $$ M=\begin{pmatrix}a&\frac{|a|}{a}v^T\\v&I+\frac{vv^T}{|a|+1}\end{pmatrix}. $$

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

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