<s>[INST] Eigenvalues of $A^{2018}$ Find the eigenvalues and eigenvectors of $A^{2018}$.
$$
A=\begin{bmatrix} 1 & 3 & 4\\ 3 & 1 & 4\\ 0 & 0 & 4\end{bmatrix}
$$

My solution:
First, by substracting first row times three from second row we get:
$$
A\approx \begin{bmatrix} 1 & 3 & 4\\ 0 & -8 & -8\\ 0 & 0 & 4\end{bmatrix}
$$
We achieved the upper triangular matrix so the characteristic polynomial is:
$$
\chi_{A^{2018}}(\lambda)=det (\begin{bmatrix} 1 & 3 & 4\\ 0 & -8 & -8\\ 0 & 0 & 4\end{bmatrix}^{2018}-\lambda I)=(1^{2018}-\lambda)((-8)^{2018}-\lambda)(4^{2018}-\lambda)
$$
Therefore the set of eigevalues is $\{1,4^{2018},8^{2018},\}$.

Please verify if this the correct solution, and in case it isn't, help me find the correct one.
 [/INST] Hint:  The characteristic polynomial is $c_A(x)=(x-4)((x-1)^2-9)=(x-4)^2(x+2)$.
You can check that the Jordan normal form  is $B=\begin{pmatrix} 4&1&0\\0&4&0\\0&0&-2\end{pmatrix}$.
Then $A^n$ is similar to $B^n$.
But $B^n=\begin{pmatrix}4^n&4+2\cdot4^n&0\\0&4^n&0\\0&0&(-2)^n\end{pmatrix}$.
Hence we get $\{4^n,(-2)^n\}$ as the eigenvalues.
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