Proof wanted that there is no positive integer matrix with positive integer eigenvalues u,v,w, if $0<u<v$ and $1\le w-v\le 2$

Question

I have the following conjecture :

If u,v,w are integers with $0<u<v<w$, then there is a POSITIVE INTEGER 3x3 - matrix A with eigenvalues u,v,w if and only if $w-v\ge 3$. I approved the conjecture upto w = 9 with brute force.

Is it true that for $w-v\le 2$, there is no solution ?

I found out that the matrix [ [u+1 , v , 1] , [1 , u + v , 1] , [1 , 1 , u + v + 2] ] has the eigenvalues u,u+v and u+v+3, therefore the case $w-v=3$ is solved.

I extended the result to the eigenvalues u, u+v and u+v+w+3 to solve the
case $w-v>3$

[ [ 1 + u , v , 1 ] , [1 , u + v , 1] , [w + 1 , 1 , 2 + u + v + w] ]

So, with the proof for the impossibility in the case $w-v\le 2$, the problem would be solved completely.

Answer 1

Are you trying to find a matrix whose eigenvalues are $u$, $u+v$, and $u+v+w+3$? The straightforward way to do that would be $$\begin{pmatrix} u & 0 & 0 \\ 0 & u+v & 0 \\ 0 & 0 & u+v+w+3 \end{pmatrix}$$

If you want something that looks less boring than that, conjugate it with an arbitrary invertible matrix.

If that does not answer your question, you need to express clearer what it is you're trying to achieve.