\subsection{Hamilton-Cayley}
Data $A = \left[
\arraycolsep=2.0pt\def\arraystretch{1.0}
\begin{array}{@{}ccc@{}}
1 & 2 \\
3 & 4 \\
\end{array}
\right]$ calcolare $A^4$.
$p(\lambda) = \lambda^2-5\lambda-2$, quindi $A^2=5A+2I$.
$A^4 = A^2A^2 = (5A+2I)(5A+2I) = 25A^2+20A+4I = 25(5A+2I)+20A+4I = 145A + 54I$
