No continuous choice of a unit eigenvector

Idea. The eigenvector of the largest eigenvalue is determined up to sign, and following it continuously around the loop returns it with the opposite sign: a "holonomy" of $-1$, like a Möbius band.

Step 1: eigenvalues. $A(\theta)$ is symmetric with $\operatorname{tr}A=0$ and $\det A=-\cos^2\theta-\sin^2\theta=-1$, so the eigenvalues are $\pm1$ and the top eigenvalue is $1$, always simple.

Step 2: eigenvectors. Let $w(\theta)=(\cos\frac\theta2,\sin\frac\theta2)$. Then
$$A(\theta)w=\big(\cos\theta\cos\tfrac\theta2+\sin\theta\sin\tfrac\theta2,\ \sin\theta\cos\tfrac\theta2-\cos\theta\sin\tfrac\theta2\big)=\big(\cos\tfrac\theta2,\ \sin\tfrac\theta2\big)=w(\theta),$$
by the subtraction formulas. So $w(\theta)$ is a unit eigenvector for $1$, and since the eigenspace is a line, the unit eigenvectors are exactly $\pm w(\theta)$.

Step 3: any continuous choice is $\pm w$. Let $v$ be a continuous choice. Then $v(\theta)=\varepsilon(\theta)w(\theta)$ with $\varepsilon(\theta)=\langle v(\theta),w(\theta)\rangle\in\{\pm1\}$. The function $\varepsilon$ is continuous on the connected interval $[0,2\pi]$ with values in $\{\pm1\}$, hence constant.

Step 4: contradiction. If $v(0)=v(2\pi)$ then $\varepsilon w(0)=\varepsilon w(2\pi)$, i.e. $w(0)=w(2\pi)$. But $w(0)=(1,0)$ while $w(2\pi)=(\cos\pi,\sin\pi)=(-1,0)\ne w(0)$. Contradiction.

Remark.
- The *eigenvalue* $1$ and the *eigenprojection* $v v^T=w w^T$ do depend continuously (and periodically) on $\theta$; only the *eigenvector* does not. This is why eigenvectors must be defined up to sign when matrices vary along loops.
- Geometrically: the set $\{(A,v):Av=v,|v|=1\}$ is a double cover of the circle $\theta\in S^1$, and it is connected (a single circle traversed twice); a double cover of a circle with a continuous section would be trivial.
- The same obstruction shows there is no continuous orientation of eigenlines of $2\times2$ symmetric matrices with distinct eigenvalues.