Relation between the common divisor of eigenvalues and the power of a matrix

Question

I was given the following question:

$A$ = \begin{bmatrix}1&2&3&4\\2&4&6&8\\0&0&2&4\\0&0&4&8\end{bmatrix}

Without calculating $A^4$, show that every element of $A^4$ is an integer that's divisible by five.

I'm not sure on how to approach this, but I will say that I noticed an interesting fact - the eigenvalues of $A$ are all divisible by five (0, 0, 5 and 10) and it is also diagonalizable. I'm pretty sure this is the way to go, but I'm not sure on how to take this forward. Can anyone point me in a generally useful direction? Also, could this idea be expanded into a more general theorem regarding common divisors of eigenvalues?

Thanks in advance!

Answer 1

You're on the right track. In particular, you have found that the characteristic polynomial of $A$ is $$ p(x) = x^2(x-5)(x-10) $$ If we consider the entries of the matrix to be elements of $\Bbb Z_5$ (the integers modulo 5), then this can be rewritten as $$ p(x) = x^2(x-0)(x-0) = x^4 $$ By the Cayley-Hamilton theorem, $p(A) = 0$ when the entries are taken modulo 5. In other words, the entries of $A^4$ are multiples of $5$.