How to find the (1,1) entry in this dynamical system? Here's the question:
Consider the dynamical system $V_{k+1}$ = $AV_k$ where
$$ 
A=\begin{pmatrix}
0 & 1 \\
-13 & 4\\
\end{pmatrix}
\quad \text{and} \quad 
V_0=\begin{pmatrix}
 1 \\
3 
\end{pmatrix}
$$
Find a formula in terms of k for the $(1,1)$-entry $x_k$ of $V_k$.
Can anyone help point me in the right direction for this? my first though to Diagonalize A and use the formula $V_k = PD^kP^-1V_0$, but all I got wax this insanely janky looking vector that I know is wrong. How are you supposed to find the formula for k in this instance?
 A: As explained in the comments,
$$x_k={\rm Re}((a+ib)(2+3i)^k)$$
with $a,b\in\Bbb R$ determined by $1=x_0=a$ and $3=x_1=2a-3b,$ i.e. $a=1$ and $b=-\frac13.$
A: Diagonalization seems to work:
$$D=\begin {pmatrix}2+3i\quad 0\\0\quad 2-3i\end {pmatrix}$$, with the eigenvalues on the diagonal.
For $P$ we need a basis of eigenvectors.   I got $$P=\begin{pmatrix}1\quad 1\\2+3i\quad 2-3i\end {pmatrix}$$.
So, we should have $A^k=PD^kP^{-1}$.
$P^{-1}=\dfrac i6\begin{pmatrix}2-3i\quad-2-3i\\-1\quad 1\end {pmatrix}$,
and the rest is a straight forward calculation.
A: Obviously,
$ V_k = A^k V_0 $
Now,
$ A^k = \alpha_0 I + \alpha_1 A \hspace{30pt} (*)$
To find $\alpha_0$ and $\alpha_1$, we use the eigenvalues of $A$.
The characteristic polynomial of $A$ is $ \lambda^2 - 4 \lambda + 13 = 0 $
whose roots (the eigenvalues) are $\lambda_1= 2 - 3 i $ and $ \lambda_2 = 2 + 3 i $
Substitute this in the above equation, i.e.
$ \lambda_1^k = \alpha_0 + \alpha_1 \lambda_1 $
$ \lambda_2^k = \alpha_0 + \alpha_1 \lambda_2 $
Solving this linear system in $\alpha_0, \alpha_1 $ we get
$\alpha_0 = \dfrac{ \lambda_2 \lambda_1^{k} - \lambda_1 \lambda_2^k }{ \lambda_2 - \lambda_1 } $
$\alpha_1 = \dfrac{ \lambda_2^k - \lambda_1^k } {\lambda_2 - \lambda_1} $
Now, we have
$ \lambda_2 - \lambda_1 = 6 i $
$ \lambda_1^k = \sqrt{13} e^{-i k \phi} , \lambda_2^k = \sqrt{13} e^{i k \phi} $
where $ \phi = \tan^{-1} \dfrac{3}{2} $
Substituting this, and simplifying,
$ \alpha_0 = - \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) $
$ \alpha_1 = \dfrac{13^{k/2}}{3} \sin (k \phi ) $
Substituting these in equation (*) above, then
$ A^k = \begin{bmatrix} - \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) && \dfrac{13^{k/2}}{3} \sin (k \phi )  \\  -13 \dfrac{13^{k/2}}{3} \sin (k \phi ) &&  - \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) + 4 \dfrac{13^{k/2}}{3} \sin (k \phi ) \end{bmatrix} $
Hence,
$ V_k = A^k V_0 = \begin{bmatrix}- \dfrac{13^{(k+1)/2}}{3} \sin( ( k - 1) \phi ) + 13^{k/2} \sin (k \phi ) \\  - \dfrac{13^{k/2}}{3} \sin (k \phi )  - 13^{(k+1)/2} \sin( ( k - 1) \phi ) \end{bmatrix} $
