# Algebraic Number is an Eigenvalue of Matrix with Rational Entries

## Problem 88

A complex number $z$ is called algebraic number (respectively, algebraic integer) if $z$ is a root of a monic polynomial with rational (respectively, integer) coefficients.

Prove that $z \in \C$ is an algebraic number (resp. algebraic integer) if and only if $z$ is an eigenvalue of a matrix with rational (resp. integer) entries.

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## Hint.

Use the companion matrix.

Recall that the characteristic polynomial of the companion matrix of a polynomial is the polynomial.

See the post Companion matrix for a polynomial for the definition of the companion matrix and the proof of the above fact.

## Proof.

$(\implies)$
Suppose that $z$ is algebraic number (resp. algebraic integer). Then $z$ is a root of a monic polynomial
$p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$ where $a_i$ are rational numbers (resp. integers).

Then consider the matrix
$A=\begin{bmatrix} 0 & 0 & \dots & 0 &-a_0 \\ 1 & 0 & \dots & 0 & -a_1 \\ 0 & 1 & \dots & 0 & -a_2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \dots & 1 & -a_{n-1} \end{bmatrix}.$

Note that the matrix $A$ has rational (resp. integer) entries. Then the characteristic polynomial $\det(xI-A)$ of $A$ is the polynomial $p(x)$.
Hence $z$ is an eigenvalue of the matrix $A$.

$(\impliedby)$
Suppose that $z$ is an eigenvalue of a matrix $A$ with rational (resp. integer) entries.
Then $z$ is a root of the characteristic polynomial of $A$.

The characteristic polynomial of $A$ is a monic polynomial with rational (resp. integer) coefficients. Thus $z$ is an algebraic number (resp. integer).

Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$....