For $\theta\in\mathbb{R}$ and $0<\lambda<\frac{1}{2}$ we define $$A_{\theta}:=\begin{pmatrix}2&0&0\\ -\pi\sin(2\pi\theta)&\lambda&0\\ \pi\cos(2\pi\theta)&0&\lambda\end{pmatrix}.$$ For $\alpha>0$ we define $$K_{\alpha}:=\{v\in\mathbb{R}^{3}:v_{2}^{2}+v_{3}^{2}\leq\alpha^{2}v_{1}^{2}\}.$$ Can we find $\alpha>0$ such that the following holds?
- $A_{\theta}v\in K_{\alpha}$ for all $v\in K_{\alpha}$
- $\|A_{\theta}v\|>\|v\|$ for all non-zero $v\in K_{\alpha}$.
What I tried so far: Fix a vector $v\in K_{\alpha}$ and write $w:=A_{\theta} v$. For 1. we want to find restrictions for $\alpha>0$ such that $w_{1}^{2}+w_{2}^{2}\leq\alpha^{2}w_{3}^{2}$. I can estimate $$w_{1}^{2}+w_{2}^{2}\leq\frac{1}{4}(\pi^{2}+\alpha^{2}\lambda^{2}+4\pi\lambda\alpha)w_{3}^{2}.$$ So we have $w_{1}^{2}+w_{2}^{2}\leq\alpha^{2}w_{3}^{2}$ if $$\frac{1}{4}(\pi^{2}+\alpha^{2}\lambda^{2}+4\pi\lambda\alpha)\leq\alpha^{2}.$$ However, I cannot forge 2. into some useful restrictions on $\alpha$. Any suggestions will be greatly appreciated.
EDIT: I made a mistake in the definition of $K_{\alpha}$. It should be $v_{2}^{2}+v_{3}^{2}\leq\alpha^{2}v_{1}^{2}$ instead of $v_{1}^{2}+v_{2}^{2}\leq\alpha^{2}v_{3}^{2}$. I'm sorry! Many thanks to QC_QAOA for giving a counterexample which allowed me to spot my typo.
EDIT: The original post was edited to reflect an error that led to the following answer. While correct for the post, it is no longer relevant. OP had defined
$$K_\alpha=\{v\in\mathbb{R}^3:v_1^2+v_2^2\leq \alpha^2 v_3^2\}$$
when they had meant to define it
$$K_\alpha=\{v\in\mathbb{R}^3:v_2^2+v_3^2\leq \alpha^2 v_1^2\}.$$
As such, the following answer is no longer relevant and I provide a new answer below that answer OP's question instead.
We shall show that the first condition does not hold for all vectors in $K_\alpha$ for any $\alpha>0$.
Consider the vector $v=(\alpha,0,1)^T$. Clearly, $v\in K_\alpha$ as
$$v_1^2+v_2^2=\alpha^2+0^2=\alpha^2=\alpha^2\cdot 1^2=\alpha^2 v_3^2.$$
However, if we consider $A_\theta v$ at $\theta=3/4$, we get
$$w=A_\theta(3/4)v=(2\alpha,\pi \alpha,\lambda)^T$$
This implies
$$w_1^2+w_2^2=\pi ^2 \alpha ^2+4 \alpha ^2=\alpha^2(4+\pi^2)>\alpha^2 \frac{1}{2^2}\geq \alpha^2 \lambda^2=\alpha^2w_3^2$$
Thus, $A_\theta(3/4)v\not\in K_\alpha$ and the condition is impossible to fulfill.
EDIT: New answer enclosed below
With the new conditions for $K_\alpha$ it is in fact possible to find an $\alpha$ which satisfies both conditions. In fact, I will show that any $\alpha$ of the form
$$\alpha\in\left[\frac{2\pi}{3}, \frac{1}{2} \left(\sqrt{12+5 \pi ^2}-\pi \right)\right]=[2.0944,2.34545]$$
works (I do not believe this is a strict bound, but I do not have a proof of this fact).
Before starting the proof in earnest, we shall prove the following lemma:
For $x,y,\theta\in\mathbb{R}$
$$f(\theta)=x\cos(2\pi\theta)-y\sin(2\pi\theta)$$
has maximum and minimum values of
$$\pm\sqrt{x^2+y^2}$$
This is easily shown as the derivative of this function is
$$f'(\theta)=-2 \pi x \sin (2 \pi \theta )-2 \pi y \cos (2 \pi \theta )$$
which has zeros at
$$\frac{\tan ^{-1}\left(-\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)}{2 \pi }$$
$$\frac{\tan ^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}}\right)}{2 \pi }$$
Plugging these into $f(\theta)$, we get
$$f\left(\frac{\tan ^{-1}\left(\pm\frac{x}{\sqrt{x^2+y^2}},\mp\frac{y}{\sqrt{x^2+y^2}}\right)}{2 \pi }\right)=\pm\sqrt{x^2+y^2}$$
Having proved this lemma, we shall proceed to the main proof. First, we shall use the notation provided above and define $w=A_\theta v$. Then the first condition is equivalent to
$$w_2^2+w_3^2-\alpha^2 w^1\leq 0$$
when
$$v_2^2+v_3^2\leq \alpha^2 v_1^2.$$
We can easily compute the first equation to get
$$w_2^2+w_3^2-\alpha^2 w_1^2=\left(\pi ^2-4 \alpha ^2\right) v_1^2+2 \pi \lambda v_1 (v_3 \cos (2 \pi \theta )-v_2 \sin (2 \pi \theta ))+\lambda ^2 \left(v_2^2+v_3^2\right)$$
$$\leq \left(\pi ^2-4 \alpha ^2\right) v_1^2+2 \pi \lambda v_1 (v_3 \cos (2 \pi \theta )-v_2 \sin (2 \pi \theta ))+\lambda ^2 \alpha^2 v_1^2$$
Without loss of generality, we may as well assume $v_1\geq 0$. If $v_1=0$, then clearly $v$ is the zero vector and the condition holds no matter what $\alpha$ is chosen. If not, using the lemma we get
$$\leq \left(\pi ^2-4 \alpha ^2\right) v_1^2+2 \pi \lambda v_1 \sqrt{v_2^2+v_3^2}+\lambda ^2 \alpha^2 v_1^2$$
$$\leq \left(\pi ^2-4 \alpha ^2\right) v_1^2+2 \pi \lambda v_1^2 \alpha+\lambda ^2 \alpha^2 v_1^2$$
$$=v_1^2(\pi ^2-4 \alpha ^2+2\pi\alpha\lambda+\alpha^2\lambda^2)$$
In order to make this expression less than or equal $0$, we need
$$\pi ^2-4 \alpha ^2+2\pi\alpha\lambda+\alpha^2\lambda^2\leq 0$$
Using the bounds on $\lambda$, we get
$$\pi ^2-4 \alpha ^2+2\pi\alpha\lambda+\alpha^2\lambda^2\leq \frac{1}{4}(4\pi^2+4\pi\alpha-15\alpha^2)$$
This quadratic is easily solved to get that the condition holds when
$$\alpha\geq \frac{2\pi}{3}.$$
Now, the second condition holds when
$$w_1^2+w_2^2+w_3^2-v_1^2-v_2^2-v_3^2>0$$
We can explicit write this out as
$$w_1^2+w_2^2+w_3^2-v_1^2-v_2^2-v_3^2$$
$$=\left(3+\pi ^2\right) v_1^2+2 \pi \lambda v_1 (v_3 \cos (2 \pi \theta )-v_2 \sin (2 \pi \theta ))+\left(\lambda ^2-1\right) \left(v_2^2+v_3^2\right)$$
Using the condition on $v$ and the lemma (with the minimum of the function this time) we get
$$\geq \left(3+\pi ^2\right) v_1^2-2 \pi \lambda \alpha v_1^2 -\left(1-\lambda ^2\right) \left(\alpha^2v_1^2\right)$$
$$=v_1^2\left(3+\pi^2-2\pi\lambda\alpha-\alpha^2+\lambda^2\alpha^2\right)$$
With the bounds on $\lambda$, this is
$$>v_1^2\left(3+\pi^2-\pi\alpha-\alpha^2\right)$$
Again, this is a simple quadratic to solve, and we get
$$\alpha\leq \frac{1}{2} \left(\sqrt{12+5 \pi ^2}-\pi \right).$$
It is easy to see that this upper bound is larger than the previous lower bound we found
$$\pi^2<4^2=16<27$$
$$\frac{4}{9}\pi^2<12$$
$$\frac{49}{9}\pi^2<12+5\pi^2$$
$$\frac{7}{3}\pi<\sqrt{12+5\pi^2}$$
$$\frac{4}{3}\pi<\sqrt{12+5\pi^2}-\pi$$
$$\frac{2}{3}\pi<\frac{1}{2} \left(\sqrt{12+5 \pi ^2}-\pi \right)$$
Thus, our final bounds on $\alpha$ do define a proper upper and lower bound, which gives us an interval of
$$\alpha\in\left[\frac{2\pi}{3}, \frac{1}{2} \left(\sqrt{12+5 \pi ^2}-\pi \right)\right]=[2.0944,2.34545]$$