Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 3

For orthogonal matrices $Q_1,\dots ,Q_m$ of the same size, transpose reverses the order of multiplication. Thereforeby canceling the adjacent pairs $Q_j^T{Q}_j=I$ successively. Hence every finite product is an orthogonal matrix.

The given symmetric matrices satisfy $E_j^T{E}_j=E_j^2=I$, so both are orthogonal matrices. Put $J=E_1{E}_2$, $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. Anticommutation givesThus $A=cI+sJ$ and $A^T=cI-sJ$, giving $A{A}^T=(c^2+s^2)I=I$. These are orthogonal rotations generated by anticommuting involutions. Also $A-A^T=2sJ$. Since $J^2=-I$ makes $J$ invertible, $A$ is symmetric exactly when $\sin (\theta /2)=0$.

Use $J{E}_1=-E_2$, $E_{1}J=E_2$ and $J{E}_{1}J=E_1$ to expand the matrix product:The double-angle identities therefore giveSimilarly $J{E}_2=E_1$, $E_{2}J=-E_1$ and $J{E}_{2}J=E_2$, soThis conjugation rotates the two-dimensional span of the two matrices through the double angle described by orthogonal rotations generated by anticommuting involutions.

Write the general real symmetric matrix as $E_2=\left(\begin{array}{cc}r& t\\ t& u\end{array}\right)$. With the specified $E_1$, anticommutation readsHence $t=0$ and $u=-r$. The remaining condition $E_2^2=I$ gives $r^2=1$, leaving exactlyBoth matrices directly satisfy all the conditions.