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$V$ is a real vector space where $x = (x_{1}, x_{2}, ...)$ which fulfills the recursion equation $x_{n}=3x_{n-1}-2x_{n-2}, \forall n > 2$

We have two basis vectors in $V$:

$v_{1}=(1,0,-2,-6,-14,...)$

$v_{2}=(0,1,3,7,15,...)$

We consider the endomorphism $\phi:V \rightarrow V$ , $(x_{1},x_{2},x_{3},...) \mapsto(x_{2},x_{3},x_{4},...)$

What is $\phi(v_1)= \square \cdot v_{1} + \square \cdot v_{2}$ ,$\phi(v_2)= \square \cdot v_{1} + \square \cdot v_{2}$

and the minimalpolynom of $\phi$ is $c_{\phi}=x^2+\square x + \square$, What are the eigenvalues?

I have discovered the closed formula of the Fibonacci sequence can by proved by using linear algebra which is amazing to me. Related to this problem I have found the problem above but I'm still struggling how to solve it. It's clear the we define the endomorphism by this "shifting" but I still don't know how to compute the values in the $\square$-s. Thank you in advance.

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1 Answer 1

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$v_{1}=(1,0,-2,-6,-14,...)$

$\phi (v_{1})=(1,0,-2,-6,-14,...)$

$v_{2}=(0,1,3,7,15,...)$

$\phi (v_{2})=(1,3,7,15,...)$

$\implies$

$0 = 0 \cdot v_{1} + \lambda \cdot v_2$

$-2 = 0 \cdot v_{1} + -2 \cdot v_2$

$-6 = 0 \cdot v_{1} + -2 \cdot v_2$

on the other side:

$1 = 1 \cdot v_{1} + \lambda \cdot v_2$

$3 = 1 \cdot v_{1} + 3 \cdot v_2$

$7 = 1 \cdot v_{1} + 3 \cdot v_2$

which implies that

$c(ϕ):= \begin{vmatrix} 0 & -2 \\ 1 & 3 \end{vmatrix} = (x-2)(x-1) = x^2-3x+2$. The eigenvalues are: $1,2$

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