Given a matrix $M$ with integer coordinates in general we know that each eigenvalue is an algebraic number.
Is it true that for every algebraic number $\alpha$ exists a matrix $M$ with integer entries such that $\alpha$ is an eigenvalue of $M$?
In case of yes a proof would be very nice.
In case of no, it would be nice to know what we do know about the possible eigenvalues.
thanks Till
You need to add the condition that $\alpha$ is an algebraic integer. Note that if $M$ has integer entries, then the characteristic polynomial of $M$ is a monic polynomia with integer coefficients, and hence the only eigenvalues are algebraic integers.
Now let $\alpha$ be an arbitrary algebraic integer satisfying some $f(x)=x^n+a_{n-1}x^{n-1}+\dots+a_0$. Then the companion matrix satisfies what you want: $$\begin{pmatrix}0&0&\dots&0&-a_{0}\\1&0&\dots&0&-a_1\\0&1&\dots&0&-a_2\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\dots&1&-a_{n-1}\end{pmatrix}$$