Eigenvalues of $A^n$, $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$ 
Let $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$ and $\alpha_n$ and $\beta_n$ denote the two eigenvalues of $A^n$ such that $|\alpha_n|\geq |\beta_n|$. Then

*

*$\alpha_n\rightarrow \infty$ as $n\to\infty$


*$\beta_n\to 0$ as $n\to\infty$


*$\beta_n$ is positive if $n$ is even.


*$\beta_n$ is negative if $n$ is odd.

$F_n$ is $n$-th Fibonacci sequence, with $F_{-1}=0,F_0=1,F_1=1$
I found $A^n=\begin{bmatrix}F_n&F_{n-1}\\F_{n-1}&F_{n-2}\end{bmatrix}$
Eigenvalues are $\dfrac{F_n+F_{n-2}\pm\sqrt{(F_n-F_{n-2})^2+4F^2_{n-1}}}{2}$
Is there any result I need to know, because it is a MSQ(Multiple selection question) and meant to solve within 4-5 minutes.
One other thing, I found (which may not be important here):
$\begin{bmatrix}F_n&F_{n-1}\\F_{n-1}&F_{n-2}\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}=\begin{bmatrix}F_{n+1}&0\\0&F_n\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$
 A: The eigenvalues of $A$ are $\varphi=\frac{1+\sqrt5}2$ (the golden ratio) and $-\varphi^{-1}=\frac{1-\sqrt5}2$. Therefore, the eigenvalues of $A^n$ are $\varphi^n$ and $(-1)^n\varphi^{-n}$. So, $\alpha_n=\varphi^n$ and $\beta_n=(-1)^n\varphi^{-n}$. Note that $\varphi>1$ and that $-\varphi^{-1}\in(-1,0)$.
A: It can be easily noticed that for all $n\geq 3$, $A^n=A^{n-1}+A^{n-2}$ and so $\alpha_n=\alpha_{n-1}+\alpha_{n-2}$ and $\beta_n=\beta_{n-1}+\beta_{n-2}$. Also note that 
$$\alpha_1=\dfrac{1+\sqrt{5}}{2} \quad \alpha_2=\dfrac{3+\sqrt{5}}{2} \qquad  \beta_1=\dfrac{1-\sqrt{5}}{2} \quad \beta_2=\dfrac{3-\sqrt{5}}{2} $$ Now characteristic equation of recurring relation (See: Wikipedia) $\alpha_n=\alpha_{n-1}+\alpha_{n-2}$ is $x^2-x-1=0$, which gives us 
$$\alpha_n = a\left( \frac{1+\sqrt{5}}{2} \right)^n +b\left( \frac{1-\sqrt{5}}{2} \right)^n. $$
Taking $n=1$ we have $a=1, b=0$ and so  we get $$ \alpha_n = \left( \frac{1+\sqrt{5}}{2} \right)^n=\alpha_1^n $$ 
Similarly we have $$\beta_n = \left( \frac{1-\sqrt{5}}{2} \right)^n = \beta_1^n$$. 
As $\alpha_1>2$, we have $\alpha_n\to\infty$, as $n\to\infty$. Again that $\beta_1\in(-1,0)$, so we have $\beta_n\to 0$, as $n\to \infty$.  Also $\beta_n$ is positive if $n$ is even and $\beta_n$ is negative if $n$ is odd.
Answer:  $(1), (2), (3), (4)$.
A: Hint:


*

*diagonalize the matrix A.

*Prove that eigenvalues of $A^n$ are precisely $(\text{eigenvalues of A})^n$.

