Linear map with Distinct eigenvalues

Question

Let $A: \Bbb{R}^2 \rightarrow \Bbb{R}^2$ be a linear transformation with eigenvalues $\frac{2}{3}$ and $\frac{9}{5}$.

Then how to prove there exist a non- zero vector $v \in \Bbb{R}^2$ such that $\vert \vert Av \vert \vert=\vert \vert v \vert \vert$ ?

I know A has distinct eigenvalues, so diagonalizable. How to proceed further?

Answer 1

By continuity. There is a vector $v$ with norm $1$ such that $Av=\frac95v$ and there is a vector $w$ with norm $1$ such that $Aw=\frac23w$. Consider the map$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&t&\mapsto&\dfrac{\left\|A\bigl((1-t)v+tw\bigr)\right\|}{\|(1-t)v+tw\|},\end{array}$$Since $f(0)=\frac95$ and $f(1)=\frac23$, $f(t)=1$ for some $t\in(0,1)$.

Answer 2

Let $v:=\lambda e_{2/3}+\mu e_{9/5}$ where $e_a$ denotes the Eigenvectors, which are orthogonal.

Then enforce

$$\|v\|^2=\lambda^2+\mu^2=1$$ and $$\|Av\|^2=\frac49\lambda^2+\frac{81}{25}\mu^2=1.$$

This system has positive solutions in $\lambda^2,\mu^2$.