When powers of matrices are represented as a sum of integral matrices There is a ring $R$ and its subring $K$ with unit. We have a matrix $A$ of order $n$ over $R$. Someone said, that if $A^m$ for $m=1,...,n$ can be represented as a sum of matrices over $R$ which a integral over $K$, than $A$ is integral over $K$. Why is that?
 A: A matrix is integral iff its eigenvalues are integral. Thus A is integral if and only if A^m is integral.
A: --- This should go as a comment at @guest 's answer. ---- 
Let $$ E = \small \begin{bmatrix} 
 1 & . & . & . & . & . & . & . \\\ 
 0 & 1 & . & . & . & . & . & . \\\ 
 0 & 1/2 & 1 & . & . & . & . & . \\\ 
 0 & 1/6 & 1 & 1 & . & . & . & . \\\ 
 0 & 1/24 & 7/12 & 3/2 & 1 & . & . & . \\\ 
 0 & 1/120 & 1/4 & 5/4 & 2 & 1 & . & . \\\ 
 0 & 1/720 & 31/360 & 3/4 & 13/6 & 5/2 & 1 & . \\\ 
 0 & 1/5040 & 1/40 & 43/120 & 5/3 & 10/3 & 3 & 1
 \end{bmatrix} $$ 
(which is the factorially scaled matrix of Stirling-numbers 2nd kind), then let's define
$$ M = E \cdot diag(1,2,4,8,16,2^5,2^6,2^7) \cdot E^{-1} $$ 
Then M has integral eigenvalues but has fractional entries:
$$ M= \small \begin{bmatrix} 
 1 & . & . & . & . & . & . & . \\\ 
 0 & 2 & . & . & . & . & . & . \\\ 
 0 & -1 & 4 & . & . & . & . & . \\\ 
 0 & 1 & -4 & 8 & . & . & . & . \\\ 
 0 & -13/12 & 5 & -12 & 16 & . & . & . \\\ 
 0 & 5/4 & -19/3 & 18 & -32 & 32 & . & . \\\ 
 0 & -541/360 & 49/6 & -26 & 56 & -80 & 64 & . \\\ 
 0 & 223/120 & -961/90 & 37 & -272/3 & 160 & -192 & 128
 \end{bmatrix} $$ 
So is there something that I misunderstood in the question/your answer?
