I'll do an example, with I hope is general enough to demonstrate the general procedure.
Let $\alpha$ have minimum polynomial $x^2+x+1$ and $\beta$ have minimal polynomial $x^3-x-1$. I'll construct a matrix $M$ with integer entries having $\alpha+\beta$ as an eigenvalue.
This will show that $\alpha+\beta$ is a zero of the characteristic equation of $M$,
and as that has integer coefficients, meaning that $\alpha+\beta$ is an algebraic integer.
I will construct $M$ such that
$$M\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
=(\alpha+\beta)\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
.$$
Then $\alpha+\beta$ will be an eigenvalue of $M$.
Using $\alpha^2=-\alpha-1$ and $\beta^3=\beta+1$ we can express all entries
in the vector in the right as integer linear combinations of the purported eigenvector.
In detail,
$$(\alpha+\beta)1=\alpha+\beta,$$
$$(\alpha+\beta)\alpha=-1-\alpha+\alpha\beta,$$
$$(\alpha+\beta)\beta=\alpha\beta+\beta^2,$$
$$(\alpha+\beta)\alpha\beta=-\beta-\alpha\beta+\alpha\beta^2,$$
$$(\alpha+\beta)\beta^2=1+\beta+\alpha\beta^2,$$
$$(\alpha+\beta)\alpha\beta^2=\alpha+\alpha\beta-\beta^2-\alpha\beta^2$$
Then (E&OE)
$$M=\pmatrix{0&1&1&0&0&0\\
-1&-1&0&1&0&0\\
0&0&0&1&1&0\\
0&0&-1&-1&0&1\\
1&0&0&0&0&1\\
0&1&0&1&-1&-1}$$
and $\alpha+\beta$ is an eigenvalue of $M$.
As an exercise, find $N$ with integer entries and
$$N\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
=\alpha\beta\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
.$$
In general, with $\alpha$ and $\beta$ having degrees $m$ and $n$,
take the vector whose entries are $\alpha^j\beta^k$ with $0\le j<m$
and $0\le k<n$.
