How can I construct matrices with positive integer values and distinct integer eigenvalues (not necessarily positive, but 0 should not be an eigenvalue). The standard-method to construct matrices with given eigenvalues is a transformation $A\rightarrow B^{-1}AB$, but the problem is that such transformations do not tend to produce positive matrices. A few examples are (the eigenvalues are shown in brackets)
$[4, 2; 3, 5] \ \ \ \ [2, 7]~$
$[2, 5, 2; 5, 2, 6; 6, 6, 5] \ \ \ [-3, -1, 13]~$
$[1, 5, 4, 3; 4, 6, 1, 2; 5, 1, 4, 3; 2, 5, 4, 2] \ \ \ [-3, -1, 4, 13]~$
$[1, 1, 5, 3, 5; 4, 6, 3, 4, 5; 6, 6, 5, 5, 4; 6, 3, 5, 5, 4; 5, 6, 4, 5, 4] $
$[-3, -1, 1, 2, 22]~$
I do not know a 6x6-matrix or a bigger one with the desired property.
For a $6 \times 6$ example, try $$ \left[ \begin {array}{cccccc} 6&1&1&1&1&1\\ 1&6&1&1 &1&1\\ 1&2&5&1&1&1\\ 1&2&2&3&1&2 \\ 1&2&2&1&3&2\\ 1&2&2&1&2&3 \end {array} \right] $$ with eigenvalues $1,2,3,4,5,11$.
EDIT: More generally, start with an $n \times n$ diagonal matrix $D$ with positive integer diagonal entries $d_1 > d_2 > \ldots > d_n$. Let $B$ be the $n \times n$ matrix with first row $[d_1, d_2, \ldots, d_n$ and all else $0$. Thus $B + D$ is upper triangular with eigenvalues $2 d_1, d_2, \ldots d_n$ which are distinct positive integers. Take $A = L (B + D) L^{-1}$ where $L$ is the lower triangular matrix with all entries $1$ on and below the main diagonal. Of course, this has the same eigenvalues as $B+D$. Note that $L^{-1}$ has $1$ on the main diagonal, $-1$ on the diagonal below that, and everything else $0$. Then $$A_{ij} = \cases{d_j - d_{j+1} & if $ i < j < n$\cr 2 d_j - d_{j+1} & if $ i = j < n$\cr 2 d_j - 2 d_{j+1} & if $i > j$\cr d_n & if $i < j = n$\cr 2 d_n & if $i=j=n$\cr }$$ which are all positive integers.
EDIT: Still more generally, you don't need to use the same positive integers in $B$ and $D$...