Past exam of the mathematical methods I course of the University of Oxford / 2017 / 8 / b

In the quadratic form, the two equal off-diagonal contributions together supply the mixed term. Comparing coefficients givesFor a nontrivial physical moment of inertia scale take $I_0>0$; then this inertia tensor is positive definite.

The normalized vectors $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(-1,1{)}^T/\sqrt{2}$ give eigenvalues $I_0$ and $3I_0$. ThereforeThe columns are orthonormal, so $U^{-1}=U^T=U^{\ast }$ and direct multiplication verifies $U^{-1}MU=D$. Signs of individual eigenvectors are free; the displayed signs also make $\det U=+1$.

The matrix has the standard rotation matrix form $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$, with $\cos \theta =\sin \theta =1/\sqrt{2}$. ThusWith $\omega =U{\omega }^{\prime }$, the new first axis points along the original $(1,1)$ direction and is counterclockwise from the original first axis. Specifying this coordinate convention avoids confusing rotation of the axes with the inverse transformation of component columns.

Substituting $\omega =U{\omega }^{\prime }$ and using the diagonalization givesFor $I_0>0$ and energy level one in the chosen units, its ellipse equation isThe major semiaxis has length $a=\sqrt{2/I_0}$ along $u_1=(1,1)/\sqrt{2}$; the minor semiaxis has length $b=\sqrt{2/(3I_0)}$ along $u_2=(-1,1)/\sqrt{2}$. The full axis lengths are $2a$ and $2b$. The original-coordinate curve is the same ellipse tilted by $45^{\circ }$, elongated along ${\omega }_y={\omega }_x$; an orthogonal coordinate change preserves lengths and does not change the ellipse itself.