Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 8A (Vectors and Matrices)

Preserving the Euclidean scalar product means $(Ax{)}^T(Ay)=x^{T}y$ for every pair of real vectors $x,y$. Since these bilinear expressions agree on all pairs of basis vectors, this is equivalent to $A^{T}A=I$: the columns of $A$ form an orthonormal basis.

The first column is a unit vector, so it is $(\cos \theta ,\sin \theta {)}^T$ for some real $\theta$. A perpendicular unit vector is one of the two choices $\epsilon (-\sin \theta ,\cos \theta {)}^T$, where $\epsilon \in \{1,-1\}$. Hence all the required orthogonal matrices areHere $R(\theta )$ is a planar rotation, and $\theta$ is determined modulo $2\pi$. To check the group laws explicitly, $D_{\epsilon }R(\varphi )=R(\epsilon \varphi )D_{\epsilon }$, soThis proves closure. The identity is $R(0)D_1=I$, the inverse of $R(\theta )D_{\epsilon }$ is $R(-\epsilon \theta )D_{\epsilon }$, and associativity is inherited from matrix multiplication. Thus the set is the orthogonal group $O(2)$.

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

A member of both groups satisfies $A^{T}A=I$ and $A^{T}JA=J$. Multiplication of the second identity by $A$ on the left, using $A{A}^T=I$, gives $JA=AJ$. A real two-by-two matrix commuting with $J=\mathrm{diag}(1,-1)$ must be diagonal, since the two off-diagonal entries change sign in this equation. Its diagonal entries are $\pm 1$ by the orthogonal matrix identity, and the restriction $a>0$ forces the first to be $1$. Therefore the intersection is the two-element group