= Solution
A <unitary matrix> satisfies $U^\dagger U=I$ over the complex numbers; a real <orthogonal matrix> satisfies $Q^TQ=I$. 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 $\cos2\theta=a/s$ and $\sin2\theta=b/s$. Then an <orthonormal eigenbasis> is
$$
\boxed{\lambda_+=s:\ v_+=(\cos\theta,0,\sin\theta),\quad\lambda_-=-s:\ v_-=(-\sin\theta,0,\cos\theta),\quad\lambda_2=c:\ e_2.}
$$
If $c$ 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 $b\ne0$ and remove the <orthogonal projection> of $a$ along $b$. Positivity of the squared norm gives
$$
0\leq\left\|a-b\frac{\langle b,a\rangle}{\|b\|^2}\right\|^2=\|a\|^2-\frac{|\langle b,a\rangle|^2}{\|b\|^2}.
$$
Multiplication by $\|b\|^2$ and square roots prove $|\langle a,b\rangle|\leq\|a\|\|b\|$; the case $b=0$ is immediate. Equality means linear dependence. Therefore
$$
\|a+b\|^2=\|a\|^2+\|b\|^2+2\operatorname{Re}\langle a,b\rangle\leq\|a\|^2+\|b\|^2+2\|a\|\|b\|=(\|a\|+\|b\|)^2.
$$
Taking the nonnegative square root proves the <triangle inequality>.
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