[Math] Find the $n$-th power of a $3{\times}3$ matrix using the Cayley-Hamilton theorem.

Question

I need to find $A^n$ of the matrix $A=\begin{pmatrix}
2&0 & 2\\
0& 2 & 1\\
0& 0 & 3
\end{pmatrix}$ using Cayley-Hamilton theorem.

I found the characteristic polynomial $P(A)=(2-A)^2(3-A)$ from which I got $A^3=7A^2-16A+12$. How to continue?

Answer 1

We can compute $A^2$ directly: $$ A^2 = \left( \begin{array}{ccc} 2 & 0 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{array} \right)\left( \begin{array}{ccc} 2 & 0 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{array} \right) = \left( \begin{array}{ccc} 4 & 0 & 10 \\ 0 & 4 & 5 \\ 0 & 0 & 9 \\ \end{array} \right). $$ From the Cayley-Hamilton theorem, it follows that \begin{align} A^3 &= 7A^2 -16A + 12I\\ &= 7\left( \begin{array}{ccc} 4 & 0 & 10 \\ 0 & 4 & 5 \\ 0 & 0 & 9 \\ \end{array} \right) - 16\left( \begin{array}{ccc} 2 & 0 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{array} \right) + 12\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\\ &= \left( \begin{array}{ccc} 8 & 0 & 38 \\ 0 & 8 & 19 \\ 0 & 0 & 27 \\ \end{array} \right). \end{align} Observe the pattern $$ A^n = \left( \begin{array}{ccc} 2^n & 0 & -2 \left(2^n-3^n\right) \\ 0 & 2^n & -2^n+3^n \\ 0 & 0 & 3^n \\ \end{array} \right). $$ Clearly this holds for $n=1$. Assume that it holds for some $n\geqslant 1$, then \begin{align} A^{n+1} &= AA^{n}\\ &= \left( \begin{array}{ccc} 2 & 0 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{array} \right)\left( \begin{array}{ccc} 2^n & 0 & -2 \left(2^n-3^n\right) \\ 0 & 2^n & -2^n+3^n \\ 0 & 0 & 3^n \\ \end{array} \right)\\ &= \left( \begin{array}{ccc} 2^{n+1} & 0 & -2 \left(2^{n+1}-3^{n+1}\right) \\ 0 & 2^{n+1} & -2^{n+1}+3^{n+1} \\ 0 & 0 & 3^{n+1}. \\ \end{array} \right) \end{align} So by induction, this formula holds for all positive integers $n$.