Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 5

A unitary matrix satisfies $U^{†}U=I$ over the complex numbers; a real orthogonal matrix satisfies $Q^{T}Q=I$. Their inverses are their conjugate transposes and transposes, respectively.

With $d=-a$, the coordinate $e_2=(0,1,0)$ has eigenvalue $c$. In the $x,z$ plane, the traceless symmetric two-by-two matrix has eigenvalues $\pm s$, where $s=\sqrt{a^2+b^2}$. If $s>0$, choose $\theta$ with $\cos 2\theta =a/s$ and $\sin 2\theta =b/s$. Then an orthonormal eigenbasis isIf $c$ equals $s$ or $-s$, the corresponding eigenspace is the span of the two listed vectors with that eigenvalue. If $s=0$, then $a=b=0$, the $x,z$ plane is the zero eigenspace, and $e_2$ retains eigenvalue $c$; if also $c=0$, every vector belongs to the zero eigenspace. This includes the degenerate cases that a division by $b$ would miss.

For the Cauchy-Schwarz inequality, take $b\ne 0$ and remove the orthogonal projection of $a$ along $b$. Positivity of the squared norm givesMultiplication by $\Vert b{\Vert }^2$ and square roots prove $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$; the case $b=0$ is immediate. Equality means linear dependence. ThereforeTaking the nonnegative square root proves the triangle inequality.