0.10 Cayley-hamilton theorem

Take the following matrix and its characteristic polynomial. $$A=\begin{pmatrix}2 & 1\\ 1 & 1\\ \end{pmatrix}$$ $$x_A()=^2-3+1$$ Plugging $A$ into the characteristic polynomial, we can find an expression for $A^2$ in terms of $A$ and the identity matrix: $$A^2-3A+I=0$$

To compute $A^2$ , we could actually perform the matrix multiplication, as below: $$A^2=\begin{pmatrix}2 & 1\\ 1 & 1\\ \end{pmatrix}\begin{pmatrix}2 & 1\\ 1 & 1\\ \end{pmatrix}=\begin{pmatrix}5 & 3\\ 3 & 2\\ \end{pmatrix}$$ Or taking equation of characteristic polynomial to heart, we can compute (with fewer operations) by scaling the elements of $A$ by $3$ and then subtracting $1$ from the elements on the diagonal. $$A^2=\begin{pmatrix}6 & 3\\ 3 & 3\\ \end{pmatrix}-I=\begin{pmatrix}5 & 3\\ 3 & 2\\ \end{pmatrix}$$