# Minimal polynomial of an endomorphism

$$V$$ 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.

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

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

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

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

$$\implies$$

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

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

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

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: $$1,2$$