Proof that $\frac{Ax}{\|Ax\|}$ has fixed points.

Question

Let $A\in \operatorname{Mat}_{2\times 2}(\Bbb{R})$ with eigenvalues $\lambda\in (1,\infty)$ and $\mu\in (0,1)$. Define $$T:S^1\rightarrow S^1;~~x\mapsto \frac{Ax}{\|Ax\|}$$ I need to show that $T$ has 4 fixed points.

My idea was the following. Since $\mu, \lambda$ are eigenvalues there exists, $x,y\neq 0$ such that $Ax=\mu x$ and $Ay=\lambda y$.Then I claim that $x,y$ are fixed points:$$Tx=\frac{Ax}{\|Ax\|}=\frac{\mu x}{\|\mu x\|}=\operatorname{sign}(\mu)\frac{x}{\|x\|}=\frac{x}{\|x\|}=x$$ similarly one can show that $Ty=y$. But then I don't see where the other two fixed points should come from?

Answer 1

The other two fixed points are $-x$ and $-y$, when $x$ and $y$ lie on $S^1$ (they can always be chosen to lie on $S^1$ since eigenvectors can be scaled), since $S^1$ is invariant under negation: $$-x\mapsto\frac{-Ax}{\|Ax\|}=\frac{-\mu x}{\|\mu x\|}=-x$$ Note that the problem statement still holds, and this proof still works (if "$4$ fixed points" means at least $4$), as long as $A$ has linearly independent eigenvectors and positive eigenvalues.

Answer 2

Your two eigenvectors $x,y$ are only assumed to be non-zero, so they are not in $S^1.$

A point $z\in S^1$ is fixed by $T$ iff $Az=\|Az\|z,$ i.e. iff $z$ is an eigenvector for some positive eigenvalue.

Therefore, the fixed points of $T$ are exactly the unit eigenvectors of $A,$ i.e. the $4$ vectors $$\frac x{\|x\|},\;-\frac x{\|x\|},\; \frac y{\|y\|},\;-\frac y{\|y\|}.$$