Let \(T\) be a linear operator on \(C^2\) which preserves the quadratic form \(x_1^2 - x_2^2\). Show that
\(\det(T) = \pm 1\).
If \(M\) is the matrix of \(T\) in the standard basis, then \(M_{22} = \pm M_{11}\), \(M_{21} = \pm M_{12}\), \(M_{11}^2 - M_{12}^2 = 1\).
If \(\det M = 1\), then there is a non-zero complex number \(c\) such that \[ M = \frac{1}{2}\begin{bmatrix} c + \frac{1}{c} & c - \frac{1}{c} \\ c - \frac{1}{c} & c + \frac{1}{c} \end{bmatrix}. \]
If \(\det M = -1\) then there is a non-zero complex number \(c\) such that \[ M = \frac{1}{2}\begin{bmatrix} c + \frac{1}{c} & c - \frac{1}{c} \\ -c + \frac{1}{c} & -c - \frac{1}{c} \end{bmatrix}. \]
In the printed source, part (d) says only “there is a complex number \(c\),” but the displayed formula contains \(1/c\). The parameter must be nonzero, as stated here.
Solution
Let \[ J=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad M=\begin{pmatrix}a&b\\ d&e\end{pmatrix}. \] Preservation of \(x_1^2-x_2^2\) is equivalent to \[ M^tJM=J. \]
1. The determinant of \(M\) is \(\pm1\).
Taking determinants in \[ M^tJM=J \] gives \[ (\det M)^2\det J=\det J. \] Since \(\det J=-1\ne0\), \[ (\det M)^2=1, \] so \[ \boxed{\det M=\pm1.} \]
2. Relate the second row of \(M\) to the first row.
Put \[ \Delta=\det M\in\{1,-1\}. \] Because \(J^{-1}=J\), the preservation equation can be rewritten as \[ JM^tJ=M^{-1}. \] Now \[ JM^tJ =\begin{pmatrix}a&-d\\-b&e\end{pmatrix}, \] while \[ M^{-1} =\frac1\Delta \begin{pmatrix}e&-b\\-d&a\end{pmatrix}. \] Equating entries gives \[ e=\Delta a, \qquad d=\Delta b. \] Thus \[ \boxed{M_{22}=\Delta M_{11},\qquad M_{21}=\Delta M_{12}.} \] In particular each is equal to the corresponding entry up to the same sign.
The \((1,1)\) entry of \(M^tJM=J\) is \[ a^2-d^2=1. \] Since \(d=\Delta b\) and \(\Delta^2=1\), \[ \boxed{a^2-b^2=1,} \] which is the asserted relation \(M_{11}^2-M_{12}^2=1\).
3. Parametrize the component with determinant \(1\).
If \(\Delta=1\), step 2 gives \[ M=\begin{pmatrix}a&b\\b&a\end{pmatrix}, \qquad a^2-b^2=1. \] Thus \[ (a+b)(a-b)=1. \] Set \[ c=a+b. \] Then \(c\ne0\) and \[ \frac1c=a-b. \] Adding and subtracting these equations gives \[ a=\frac12\left(c+\frac1c\right), \qquad b=\frac12\left(c-\frac1c\right). \] Therefore \[ \boxed{ M=\frac12 \begin{pmatrix} c+\frac1c&c-\frac1c\\ c-\frac1c&c+\frac1c \end{pmatrix}.} \]
4. Parametrize the component with determinant \(-1\).
If \(\Delta=-1\), step 2 gives \[ M=\begin{pmatrix}a&b\\-b&-a\end{pmatrix}, \qquad a^2-b^2=1. \] Again set \[ c=a+b\ne0, \qquad \frac1c=a-b. \] Then the first row is as in step 3, while \[ -b=\frac12\left(-c+\frac1c\right), \qquad -a=\frac12\left(-c-\frac1c\right). \] Hence \[ \boxed{ M=\frac12 \begin{pmatrix} c+\frac1c&c-\frac1c\\ -c+\frac1c&-c-\frac1c \end{pmatrix}.} \]