The Fibonacci-sequence has been solved using diagonalization.
$1,1,2,3,5,8, \ldots$.
Starting with $\left(1,1 \right)^T$, the next number in the sequence is given by the last entry of
\begin{align*}
\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}
\end{align*}
where the vector $\left(a,b \right)^T$ contains the previous two numbers of the sequence.
Note that the last entry of the previous vector becomes the new first entry.
Now, clearly, you can repeat this procedure to find the $n+2$-th number of the sequence by calculating
\begin{align*}
& \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^n \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\
&= A^n \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\end{align*}
This is where diagnoalization comes into play. If you diagonalize $A$, i.e. $A = P D P^{-1}$ with $D$ diagonal, it is easily seen that
\begin{align*}
&A^n \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\
& = P D^n P^{-1} \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\end{align*}
$D^n$ is calculated easily just by raising each diagonal entry to the power of $n$.
You can determine $P$ and $D$ explicitly to derive a formula for the $n+2$-th number in the sequence.
Note that you can replace the starting vector $(1,1)^T$ by any vector of your choice.