compute $A^2$ and $A^6$. $A= \begin{bmatrix} -1 & 1 & 1 & -1\\ 1 & -1 & -1 & 1\\ 1 & -1 & -1 & 1\\ -1 & 1 & 1 & -1 \end{bmatrix} $ . Fuzhen Zhang's linear algebra, problem 3.11
$A= \begin{bmatrix}
-1 & 1 & 1 & -1\\ 
1 & -1 & -1 & 1\\ 
1 & -1 & -1 & 1\\ 
-1 & 1 & 1 & -1
\end{bmatrix}   $ 
compute $A^2$ and $A^6$.
The answers are $-4A$ and $-2^{10} A$, respectively.
I have no clue how to calculate higher powers. 
Thanks!
 A: $A^2$ you can calculate directly. There are tricks you can do, but in this case I don't think they will save you any significant amount of time. Especially once you notice that every single one of the 16 entries of $A^2$ is the result of basically the same calculation (well, there is your trick, I guess).
Once you know that $A^2=-4A$, you can calculate any higher power rather efficiently:
$$
A^6=A^2\cdot A^2\cdot A^2\\
=(-4A)\cdot (-4A)\cdot (-4A)\\
=-64A^3\\
=-64A^2\cdot A\\
=-64(-4A)\cdot A\\
=256A^2=-1024 A
$$
A: We can observe $$A_4 = \begin{bmatrix}A_2&-A_2\\-A_2&A_2\end{bmatrix}$$
where
$$A_2 = -\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$
Which we see follow the same pattern (except minus sign). So this is nothing but Kronecker product of $A_2$ on $A_2$. Since it is smaller, we can investigate this $A_2$ more readily and see that it has eigenvalues $\lambda(A_2) = \{0,2\}$ and eigenvector (of course) $[1,-1]^T$.
Then the Kronecker laws of eigenvalue propagation tells us the eigenvalues for the $A_4$ shall be all possible products of $\{0,2\}$ on itself, these are the four combinations : $$\lambda(A_4) = \lambda(A_2 \otimes A_2) = \lambda(A_2) \otimes \lambda(A_2) = \{0,2\} \otimes \{0,2\} = \\\phantom{a}\\= \{0\cdot 0, 0\cdot 2, 2\cdot 0, 2\cdot 2\} = \{0,0,0,4\}$$So the eigenvalue $4$ is the only we need to worry about.
Now we can directly calculate what it will be. Multiply by $4$ times exponent (minus 1).
$$4\cdot (2 - 1) = 4$$
$$4\cdot (6 - 1) = 4\cdot 5 = 2 \cdot 10$$
And by law of exponents we know $4^{5} = 2^{10}$
