Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 5

Yes. The matrix is a real symmetric matrix, so the spectral theorem for real symmetric matrices provides an orthonormal basis of real eigenvectors and hence a real diagonal form. Explicitly its characteristic polynomial is ${\lambda }^2-7\lambda -4$, with distinct real eigenvalues

The characteristic polynomial is ${\lambda }^2-2\lambda \cos \theta +1$, with eigenvalues $\cos \theta \pm i\sin \theta$. A real diagonal form can have only real eigenvalues, so it requires $\sin \theta =0$. Conversely, when $\theta =k\pi$, the rotation matrix is already $(-1{)}^{k}I$. Thus