Invertible submatrix of matrices in $O(n,1)$

Question

Let $O(n,1)$ ={$T$ : $\mathbb R^{n,1} \to \mathbb R^{n,1}$| $q(Tx,Ty) = q(x,y)$} Where $q(x,y)= (\sum_{i=1}^n x_iy_i) - x_{n+1}y_{n+1}$. Now let $e_i$ denote the standard basis for $\mathbb R^{n+1}$. The $e_i$'s also form a $q$-orthonormal basis for $\mathbb R^{n,1}$. Now, let A denote the matrix of T w.r.t. the $e_i$'s, and write A as

A =$\begin{pmatrix} X & \beta \\ \gamma & \delta \end{pmatrix}$

where X is an n by n matrix, $\beta$ is and n by 1 column vector, $\gamma$ is a 1 by n row vector, and $\delta$ is a nonzero real.

Question:

Show X is invertible.

What I've tried:

The obvious first step was the compute $q$ on pairs of $e_i$'s for matching and differing $i$'s and that told me that the columns of A form an orthonomal set. but I can't seem to determine much else. Any solutions or hints are appreciated thank you

Answer 1

Relationship $q(TX,TY)=q(X,Y)$ can be written under a matricial form

$$\forall X,Y \ \ (TX)^tQ(TY)=X^tY \ \iff \ X^t(T^tQT)Y=X^tQY$$

which means that $T^tQT=Q.$

Written in terms of block matrices, it means that:

$$\underbrace{\begin{pmatrix} X^t & \gamma \\ \beta^t & d \end{pmatrix}\begin{pmatrix} I_{n} & 0 \\ 0 & -1 \end{pmatrix}}_{}\begin{pmatrix} X & \beta \\ \gamma^t & d \end{pmatrix}=\begin{pmatrix} I_{n} & 0 \\ 0 & -1 \end{pmatrix} \ \ \iff$$

$$\begin{pmatrix} X^t & -\gamma \\ \beta^t & -d \end{pmatrix}\begin{pmatrix} X & \beta \\ \gamma^t & d \end{pmatrix}=\begin{pmatrix} I_{n} & 0 \\ 0 & -1 \end{pmatrix} \ \ \ \iff$$ $$\tag{1}\iff \ \ \ \begin{cases}X^tX -\gamma \gamma^t &=& I_{n}& \ \ \ \text{matrices}\\ X^t\beta - d \gamma &=&0 & \ \ \ \text{vectors}\\ \beta^t\beta-d^2&=&-1& \ \ \ \text{scalars}\end{cases}$$

We will only consider the first of these equations, under the form:

$$X^tX = I_{n}+\gamma \gamma^t$$

and take the determinants on both sides.

On the LHS, we get

$$\tag{2}det(X^tX)=det(X^t)det(X)=det(X)^2.$$

For the RHS, we are going to use a (rather classical) formula for the determinant of such a matrix ("identity matrix perturbated by a rank-one matrix") (https://en.wikipedia.org/wiki/Matrix_determinant_lemma) or (Determinant of rank-one perturbations of (invertible) matrices)

which gives (see the order of multiplication of the matrices)

$$\tag{3}det(I_{n}+\gamma \gamma^t)=1+\gamma^t\gamma=1+\|\gamma\|^2>0.$$

Comparing (2) and (3), we have necessarily:

$$det(X)\neq0.$$

Remark: The structure of "isometries" preserving the quadratic form is encapsulated into formulas (1). The case $n=1$ is especially simple. In this case, all of $X,\beta,\gamma$ and of course $d$ are reals; thus, we can write:

$$\begin{cases}X^2 &=&\gamma^2+1\\ X\beta &=& d\gamma\\ d^2&=&\beta^2+1\end{cases}$$

which is equivalent to $X=\pm d = \cosh(u)$ and $\gamma = \pm \beta = \sinh(u)$ for a certain $u$. Thus, we have two types of "hyperbolic isometries":

$$\begin{pmatrix}\cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \ \ \ \text{and} \ \ \ \begin{pmatrix}\cosh(u) & -\sinh(u) \\ -\sinh(u) & \cosh(u) \end{pmatrix}.$$

Answer 2

Also, in addition to @JeanMarie's discussion of the matrix viewpoint, one can argue that any endomorphism $T$ of a finite-dimensional vector-space $V$ with a non-degenerate bilinear form $\langle,\rangle$ that preserves that form must be invertible: if $Tv=0$ for $v\not=0$, let $v'\in V$ be such that $\langle v,v'\rangle=1$ using the non-degeneracy, and consider $$ 0\;=\;\langle 0,Tv'\rangle\;=\;\langle Tv,Tv'\rangle\;=\; \langle v,v'\rangle \;=\; 1 $$ Contradiction.