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.
 A: 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.
A: Your matirx is the sum of a relatively easy one (symmetric) and a nilpotent matrix, and these commute. $N$ is zero except for the pair of $4$ in positions $(1,3)$  and $(2,3).$  Call the symmetric one $S$ which can be nicely diagonalized. because they commute,
$$  (S + N)^{2018} = S^{2018} + 2018 S^{2017} N. $$
It stops there since $N^2 = 0.$
I just guessed this, all that is needed is explicit $P^{-1} S P = D.$  It is not really necessary that $P$ be orthogonal. In fact, what I do is
$$ P^T S P = D  $$
$$\left( 
\begin{array}{rrr} 
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\ 
-\frac{1}{\sqrt 2}  & \frac{1}{\sqrt 2} & 0 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 & 3 & 0 \\ 
3 & 1 & 0 \\ 
0 & 0 & 4 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}  & 0 \\ 
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
4 & 0 & 0 \\ 
0 &  - 2 & 0 \\ 
0 & 0 & 4 \\ 
\end{array}
\right) 
$$
where $PP^T = P^TP = I$
$ P^T S P = D  $  so that $S = PDP^T.$  Thus $S^{n} = P D^n P^T$ for all $n \geq 1$
the $S+N$ is called a Jordan Chevalley decomposition, sometimes much easier than Jordan canonical form. This time, not much different
