Diagonalizing the matrix $A$ then finding $A^{10}$

Question

Diagonalize the matrix
$$A= \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}$$
Then find $A^{10}.$
We have the characteristic polynomial of $A:$
$$\left | A- \lambda I \right |= \left | \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}- \lambda\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \right |= \begin{vmatrix} -3- \lambda & -14 & -10\\ 2 & 13- \lambda & 10\\ -2 & -7 & -4- \lambda \end{vmatrix}=$$
$$= \left ( 6- \lambda \right )\left ( 1+ \lambda \right )\left ( 1- \lambda \right )$$
which has the roots $\lambda_{1}= 6, \lambda_{2}= -1, \lambda_{3}= 1.$ These roots are the diagonal elements as well as the eigenvalues of $A.$ These three eigenvalues correspond to the eigenvectors $v_{1}= \left ( 2, -2, 1 \right ),$
$v_{2}= \left ( 2, -1, 1 \right ), v_{3}= \left ( 1, -1, 1 \right ).$ Hence, $A= SJS^{-1}$ with $S= \begin{pmatrix} 2 & 1 & 2\\ -1 & -1 & -2\\ 1 & 1 & 1 \end{pmatrix},$ $J= \begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 6 \end{pmatrix}, S^{-1}= \begin{pmatrix} 1 & 1 & 0\\ -1 & 0 & -2\\ 0 & -1 & -1 \end{pmatrix}.$ I have no plans to find $A^{10}.$ I need to the helps

Answer 1

Observe that if $A=SJS^{-1}$ we also have $A^n = SJ^n S^{-1}$ for all $n \in \mathbb N$. We can see this by induction with $A^2= SJS^{-1}SJS^{-1} = SJ^2S^{-1}$ and $A^n = SJS^{-1} SJ^{n-1}S^{-1} = SJ^nS^{-1}$

Now $J^{10} = \text{diag}(1,1, 6^{10})$ and $A^{10} = SJ^{10}S^{-1}$.