Both matrices and Algebraic integers have minimal polynomials. But I struggle to get how these two types of minimal polynomials are related. Well, they are two different rings, but I think we can find some connection between them
For example, if an algebraic number $\alpha$ can be expressed as a combination of the linear independent basis $1,\sqrt{3}, \sqrt{5},\sqrt{15}$, then $\alpha^n$ can be expressed as a combination of this basis as well. The equation $\sum_{i=0}^4 c_i\alpha^i=0$ is essentially four linear equations with four unknowns $c_i$. So, in general, the values of $c_i$ are uniquely determined, and we therefore find the minimal polynomial.
The process above is essentially linear algebra. However, I find that the power $\alpha^n$ cannot be easily represented by the power of a matrix. Is there an easy way to embed the algebraic integers into a set of matrices so that the minimal polynomial of algebraic integers can be found easily by finding the minimal polynomial of matrices?
Best Answer
Let $\alpha$ be an algebraic integer, let $K={\bf Q}(\alpha)$. Then multiplication by $\alpha$ is a $\bf Q$-linear transformation on $K$. Let $B$ be a basis for $K$ as a $\bf Q$-vector space, and let $M$ be the matrix representing multiplication by $\alpha$ with respect to $B$. Then the minimal polynomial for $\alpha$ over $\bf Q$ is the minimal polynomial for $M$.
Let the degree of $\alpha$ over $\bf Q$ be $n$. Then you can choose the basis $\{\,1,\alpha,\alpha^2,\dots,\alpha^{n-1}\,\}$, and the matrix $M$ that you get has a particularly simple form; it's called the companion matrix for $\alpha$. But it requires that you know how to express $\alpha^n$ in terms of the basis, which is the same as knowing the minimal polynomial of $\alpha$.