Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 5

The columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.

The $z$ direction is fixed, giving the eigenvalue one. The planar block has characteristic polynomialThus the full list of eigenvalues isThe last two are real precisely when $\sin \varphi =0$, so all three are real exactly for $\boxed{\varphi \in \pi \mathbb{Z}}$. At even multiples of $\pi$ the matrix is the identity and every vector is fixed. At odd multiples it is a half-turn: the axial eigenvalue remains $+1$, while the whole perpendicular plane has eigenvalue $-1$.

For other angles, the only real eigenvector direction is the rotation axis; the perpendicular plane is rotated rather than preserving an individual real direction. Its complex eigenvectors can be taken as $(1,-i,0{)}^T$ and $(1,i,0{)}^T$ for $e^{i\varphi }$ and $e^{-i\varphi }$, respectively. These represent the two complex conjugate planar eigenvalues.

With the same right-handed active rotation convention as $R_z$, the $y$ direction is fixed, ${\mathbf{e}}_z$ rotates towards ${\mathbf{e}}_x$, and ${\mathbf{e}}_x$ towards $-{\mathbf{e}}_z$. ThereforeIts columns give the images of the three Cartesian unit vectors, which also makes the signs of this rotation matrix explicit.

A product of orthogonal matrices is orthogonal. In fact,Thus $\boxed{R_1\text{ is orthogonal for every }\theta ,\varphi .}$ Its determinant is also one, since each factor has determinant one, so it remains a proper rotation matrix.

Both factors and their product are real, so the Hermitian conjugate equals the transpose. Part (i) therefore givesHence $\boxed{R_1\text{ is unitary for every }\theta ,\varphi .}$ A real orthogonal matrix is always a unitary matrix.

A real matrix is a Hermitian matrix exactly when it is a symmetric matrix. Since $R_1$ is orthogonal, this is equivalent to $R_1=R_1^{-1}$, or $R_1^2=I$. A generic rotation does not satisfy this; for example $\theta =0$, $\varphi =\pi /2$ gives a nonsymmetric quarter-turn. Thus $\boxed{R_1\text{ is not Hermitian in general.}}$

The exceptions are the identity and half-turns, as characterized by real rotation matrices that are Hermitian. For this product, a half-turn occurs when either $\theta$ or $\varphi$ is an odd multiple of $\pi$; the identity occurs when both are even multiples. For example, the half-turn condition follows from $\mathrm{tr}{R}_1=(1+\cos \theta )(1+\cos \varphi )-1=-1$.

The Hermitian conjugate of a real matrix is its transpose, soTherefore $\boxed{M\text{ is anti-Hermitian for all }\theta ,\varphi .}$ It is also a real skew-symmetric matrix. It is Hermitian as well only in the exceptional case $M=0$, precisely when $R_1$ is symmetric. This includes the identity and half-turn cases from part (d), so the zero-matrix overlap should not be excluded.

Applying the rightmost rotation matrix first givesEquality with ${\mathbf{e}}_y$ requires $\cos \theta =0$, so $\theta =\pi /2$ in the specified range. Then $\sin \varphi =1$ and $\cos \varphi =0$ giveIn the reverse order, $R_z$ first fixes ${\mathbf{e}}_z$, so $R_2{\mathbf{e}}_z=R_y(\pi /2){\mathbf{e}}_z={\mathbf{e}}_x$. Therefore $\boxed{R_2\mathbf{v}\ne {\mathbf{v}}^{\prime }}$. This explicitly demonstrates that rotations about different axes do not generally commute.

At the angles in part (f),Let the two diagonal matrices be $D=\mathrm{diag}(d_1,d_2,d_3)$ and $D^{\prime }=\mathrm{diag}(e_1,e_2,e_3)$. Multiplication givesThe vector condition is exactly $c{e}_2{d}_3=b$. For $c\ne 0$, the general form is thereforeEvery pair $p,q$ is attainable, for example with $D=I$ and $D^{\prime }=\mathrm{diag}(-p,b/c,-q)$. The transformation need not be orthogonal or invertible; the question asks only for diagonal rescalings with this one vector image. If $c=0$ and $b\ne 0$, no linear map can satisfy the condition. If $c=b=0$, the $(2,3)$ entry is arbitrary as well.