I'm given the following matrix $$ A = \begin{pmatrix} 2.5 & -1 & -1.5\\ 1.5 & 0.5 & -1.5\\ 0.5 & -1 & 0.5 \end{pmatrix}. $$ Its characteristic equation is $λ^3 - 3.5λ^2 +3.5λ -1$ (e.g. eigenvalues of $2$, $1$, and $0.5$).
How can I find $A^{10}$ using the known facts and a system of equations?
I know that I have to use the Cayley-Hamilton method.
\begin{align} f(v) &= v^{10} = a + bv + cv^2 \\ f'(v) &= 10v^9 = b + 2cv \end{align}
So that: \begin{align} f(2) &= a + 2b + 4c \\ f'(1) &= b + 2c \\ f(0.5) &= a + 0.5b + 0.25c \end{align}
I solved for $a$, $b$, and $c$. Then I used the following formula to find $A^{10}$: $$ A^{10} = a*I + b*A + c*A^2 $$
However, when I compare my answer with Matlab, my answer doesn't seem right.
You know that $A^{10}=aI+bA+cA^2$ for some $a,b,c$.
If $v$ is an eigenvector relative to $2$, then $$ A^{10}v=2^{10}v=av+2bv+4cv $$ so $$ a+2b+4c=2^{10} $$ Similarly, with the eigenvalue $1$, you get $$ a+b+c=1 $$ and, with the eigenvalue $1/2$, you get $$ a+\frac{1}{2}b+\frac{1}{4}c=\frac{1}{2^{10}} $$