Showing existence of a vector $v$ such that $\|Av\|=\|v\|$ given two eigenvalues of the linear transformation $A$ Let $A : \mathbb R ^2 \rightarrow \mathbb R^ 2$ be a linear transformation with eigenvalues $\frac{2}{3}$
and $\frac{9}{5}$.  
Then, show that there exists a non-zero
vector $v \in \mathbb  R^ 2$ such that $\|Av\|=\|v\|$.
My try:
$\exists v_1 ,v_2$ (non-zero) such that $Av_1=\frac{2}{3}v_1$ and $Av_2=\frac{9}{5}v_2$
Will I have to construct some vector $v$ using $v_1,v_2$ such that $\|Av\|=\|v\|$?
I am not so sure. But I could not construct such a vector yet.
Any help?
 A: Let the eigenvectors be $v_1$ and $v_2$ as you described.
Define an application $\Bbb S^1\to \Bbb R$ by 
$$f(a,b)=\frac{\|A(av_1+bv_2)\|}{\|av_1+bv_2\|}.$$
This is a continuous function defined on a compact, hence it takes all possible values between its $\inf$ and $\sup$.
We know that $f(1,0) = \frac 23$ and $f(0,1)=\frac 95$, therefore $\sup f\ge \frac 95$ and $\inf f\le \frac 23$ hence there exists a point $(a,b)$ on $\Bbb S^1$ such that $f(a,b)=1$, which gives you the desired result. 
A: Hint Consider the function $$f: v \mapsto ||Av|| - ||v||$$ on $\mathbb{R}^2$ and a path $\gamma$ in $\mathbb{R}^2$ from $v_1$ to $v_2$.
Note that, as stated, you need not actually construct a vector $v$ for which $||Av|| = ||v||$, only show that such a vector exists.
Remark More generally, using the same idea one can show that given any linear transformation $B: \mathbb{R}^n \to \mathbb{R}^n$ with eigenvalues $\lambda_1, \lambda_2$, say, $|\lambda_1| \leq |\lambda_2|$, and any $\mu \in [\lambda_1, \lambda_2]$, there is a vector $v \in \mathbb{R}^n$ such that $||A v|| = \mu ||v||$ (where $||v||$ is the usual norm, say).
