Pattern with $\begin{bmatrix}0&1\\-1&3\end{bmatrix}^k$. How can I prove what corner element will be? The matrix
$$B= \begin{bmatrix}0&1\\-1&3\end{bmatrix}$$
Is inspired by this matrix which generates the Fibonacci sequence.
Upon calculating a few powers of $B$, I noticed that $({B^k})_{2,2}$ seems to be ${F_{2(k+1)}} - {F_{2k}}$ but I lack an idea of how to prove it.
What would be a fruitful approach to do it?
 A: I think a good idea to start is with the characteristic equation of $B$ which is
$$
\lambda(\lambda-3)+1=0\iff \lambda^2=3\lambda-1\implies B^2=3B+1.
$$
It is reasonable to write that
$$
B^k=a_kB+b_k,
$$hence
$$
B^{k+1}=BB^k=a_kB^2+b_kB=(3a_k+b_k)B+a_k=a_{k+1}B+b_{k+1},
$$
which yields
$$
{\\a_{k+1}=3a_k+b_k
\\b_{k+1}=a_k,
}
$$
hence
$$
{
a_{k+1}=3a_k+a_{k-1}
\\
b_k=a_{k-1}.
}
$$
From here, a bit of substitution and mathematical calculation would yield the final result.
A: Empirically:
Let us have a look at a few successive powers:
$$\begin{pmatrix}0&1\\-1&3\end{pmatrix}$$
$$\begin{pmatrix}-1&3\\-3&8\end{pmatrix}$$
$$\begin{pmatrix}-3&8\\-8&21\end{pmatrix}$$
$$\begin{pmatrix}-8&21\\-21&55\end{pmatrix}$$
$$\begin{pmatrix}-21&55\\-55&144\end{pmatrix}$$
We see that the elements are each time shifted north-west and a new corner element appears, which is $(-1,3)$ times the right column. Hence the recurrence
$$c_{n+2}=3c_{n+1}-c_n,$$ starting with $1,3$.
A: Hint: Prove by induction that
$$
\begin{bmatrix}\hphantom-0&1\\-1&3\end{bmatrix}^k
=
\begin{bmatrix}-F_{2k-2} & F_{2k} \\ -F_{2k} & F_{2k+2} \end{bmatrix}
$$
You'll need that $F_{n+2} = 3F_n-F_{n-2}$. For a proof, adapt this answer. See also this question.
