Solution
= Solution
The given <symmetric matrices> satisfy $E_j^TE_j=E_j^2=I$, so both are <orthogonal matrices>. Put $J=E_1E_2$, $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$. Anticommutation gives
$$
J^T=E_2E_1=-J,\qquad J^2=E_1E_2E_1E_2=-I.
$$
Thus $A=cI+sJ$ and $A^T=cI-sJ$, giving $AA^T=(c^2+s^2)I=I$. These are <orthogonal rotations generated by anticommuting involutions>. Also $A-A^T=2sJ$. Since $J^2=-I$ makes $J$ invertible, \b[$A$ is symmetric exactly when $\sin(\theta/2)=0$].