Let a matrix $A\in\Bbb M_n(R) $, $n>2$ for which exists a number $a\in\Bbb [-2,2 ]$ so that :
$A^2 -aA + I_n = O_n$
Prove that for any natural number $m\in\Bbb N$ there exists a unique $a_m\in\Bbb [-2 , 2 ]$ for which $A^{2m}-a_mA^m + I_n = O_n$.
How I tried solving it : I've written $A^2 = aA – I_n$ from which I generalized $A^n$ as the following series :
$A^n = x_nA – x_{n-1}I_n$ where $x_1 = 1$ and $x_2 = a$ and where $x_{n+1} = ax_n – x_{n-1}$ after which i applied the characteristic equation and got :
$r^2-ar=1=0$ so naturally $r_{1,2}=\frac{a\pm i\sqrt{4-a^2}}{2}$
We can write $r_{1,2} = \cos{t} \pm i\sin{t}$ where :
$\sin{t} = \frac{\sqrt{4-a^2}}{2}$ , $\cos{t} = \frac{a}{2}$ and $t\in [0,\pi]$
so $x_n = C_1\cos{nt} + C_2\sin{nt}$ where if we replace $n$ with $1$ and $2$ and solve the system we get
$C_1= 0$ and $C_2= \frac{2}{\sqrt{4-a_2} }$ so we easily get $x_n = \frac{2}{\sqrt{4-a_2} }\sin{nt}$
so $A^n = \frac{2}{\sqrt{4-a_2} }$ $[\sin{nt} A – \sin{(n-1)t} I_n]$
Furthermore we have that $A^2 -aA = I_n$ so $A(aI_n-A) = I_n$ so $\det{A}$ can't be equal to $0$
so we pretty much know the value of $A^{-n}$, can someone help me proceed ?
I've tried to calculate $A^m + A^{-m}$ and got $a_mI_n$ as an answer, if this enough to prove the statement correct ? And if it isn't i could really use a hand here.
Best Answer
Because $A$ is invertible, you can likewise conclude from the recurrence relation that
$$A^{-m} = \frac{ \sin -mt \times A - \sin (-m-1)t }{\sin t }\times I_n.$$
Thus, $ A^m + A^{-m} = \frac{\sin (m+1)t - \sin (m-1) t}{\sin t} \times I_n = 2 \sin mt \times I_n. $
So, for $ a_m = 2 \sin mt \in [2, -2]$, we have $A^{2m} - a_m A^m + I_n = 0. $
Uniqueness follows since $ A^m \neq 0$ (as $ \det A \neq 0 $).