Let $A= \begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}$. Find the entry in the first row and second column of $A^{2014}$ I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=\begin{pmatrix}  1&1 \\   -4&-1  \end{pmatrix}$
$D=\begin{pmatrix}  0&0 \\   0&6  \end{pmatrix}$
$P^{-1}=\frac{1}{3}\begin{pmatrix}  -1&-1 \\   4&1  \end{pmatrix}$.  
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.  
 A: Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$\begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}\begin{pmatrix} a&b \\ -a&-b \end{pmatrix}=6\begin{pmatrix} a&b \\ -a&-b \end{pmatrix}$$
but this is overkill.
A: By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2\cdot6^{2013}$.  
Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.
A: Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=\begin{bmatrix}1\\-1\end{bmatrix}\begin{bmatrix}8&2\end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=\left(\begin{bmatrix}1\\-1\end{bmatrix}\begin{bmatrix}8&2\end{bmatrix}\right)^{2014}=\begin{bmatrix}1\\-1\end{bmatrix}\left(\begin{bmatrix}8&2\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}\right)^{2013}\begin{bmatrix}8&2\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}6^{2013}\begin{bmatrix}8&2\end{bmatrix}=6^{2013}A$$
