Cayley-Hamilton | Arithmetic variety

A complex number is algebraic if it is the root of some polynomial $P(x)$ with rational coefficients. $\sqrt{2}$ is algebraic (e.g. the polynomial $x^2 -2$ ); $i$ is algebraic (e.g. the polynomial $x^2 + 1$ ); $\pi$ and $e$ are not. (A complex number that is not algebraic is called transcendental)

Previously, I wrote some blog posts (see here and here) which sketched a proof of the fact that the sum and product of algebraic numbers is also algebraic (and more). This is not an obvious fact, and to prove this requires some amount of field theory and linear algebra. Nevertheless, the ideas in the proof lead the way to a better understanding of the structure of the algebraic numbers and towards the theorems of Galois theory. In that post, I tried to introduce the minimum algebraic machinery necessary in order to state and prove the main result; I don’t think I entirely succeeded.

However, there is a more direct approach, one which also allows us find a polynomial that has $\alpha + \beta$ (or $\alpha\beta$ ) as a root, for algebraic numbers $\alpha$ and $\beta$ . That is the subject of this post. Instead of trying to formally prove the result, I will illustrate the approach for a specific example: showing $\sqrt{2} + \sqrt{3}$ is algebraic.

This post will assume familiarity with the characteristic polynomial of a matrix, and not much more. (In particular, none of the algebra from the previous posts)

A case study

Define the set $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \ | \ a, b, c, d \in \mathbb{Q} \}$ . We will think of this as a four-dimensional vector space, where the scalars are elements of $\mathbb{Q}$ , and the basis is $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ . Every element can be uniquely expressed as $a(1) + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$ , for $a, b, c, d \in \mathbb{Q}$ .

We’re trying to prove $\sqrt{2} + \sqrt{3}$ is algebraic. Consider the linear transformation $T$ on $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ defined as “multiply by $\sqrt{2} + \sqrt{3}$ “. In other words, consider the linear map $T: \mathbb{Q}(\sqrt{2}, \sqrt{3}) \to \mathbb{Q}(\sqrt{2}, \sqrt{3})$ which maps $v \mapsto (\sqrt{2} + \sqrt{3})v$ . This is definitely a linear map, since it satisfies $T(v + w) = Tv + Tw$ and $T(cv) = c(Tv)$ . In particular, we should be able to represent it by a matrix.

What is the matrix of $T$ ? Well, $T(1) = \sqrt{2} + \sqrt{3}$ , $T(\sqrt{2}) = 2 + \sqrt{6}$ , $T(\sqrt{3}) = 3 + \sqrt{6}$ , and $T(\sqrt{6}) = 3\sqrt{2} + 2\sqrt{3}$ . Thus we can represent $T$ by the matrix

$\begin{bmatrix}0 & 2 & 3 & 0 \\1 & 0 & 0 & 3 \\1 & 0 & 0 & 2 \\0 & 1 & 1 & 0\end{bmatrix}$ .

Now, the characteristic polynomial $\chi_T(x)$ of this matrix, which is defined as $\text{det}(T-xI)$ , is $x^4 - 10x^2 + 1$ , which has $\sqrt{2} + \sqrt{3}$ as a root. Thus $\sqrt{2} + \sqrt{3}$ is indeed algebraic.

Why it works

The basic reason is the Cayley-Hamilton theorem. It tells us that $T$ should satisfy the characteristic polynomial: $T^4 - 10T^2 + I$ is the zero matrix. But the matrix we get when plugging $T$ into $\chi_T(x)$ should correspond to multiplication by $\chi_T(\sqrt{2} + \sqrt{3})$ ; thus $\chi_T(\sqrt{2} + \sqrt{3}) = 0$ .

Note that I chose $\sqrt{2} + \sqrt{3}$ randomly. I could have chosen any element of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ and used this method to find a polynomial with rational coefficients having that element as a root.

At the end of the day, to prove that such a method always works requires the field theory we have glossed over: what is $\mathbb{Q}(\alpha, \beta)$ in general, why is it finite-dimensional, etc. This constructive method, which assumes the Cayley-Hamilton theorem, only replaces the non-constructive “linear dependence” argument in Proposition 4 of the original post.

Here is a sketch of proof of the Cayley-Hamilton theorem via classical algebraic geometry.

The set of n x n matrices over an algebraically closed field can be identified with the affine space $\mathbb{A}^{n^2}$ . Let $V$ be the subset of matrices that satisfy their own characteristic polynomial. We will prove that $V$ is in fact all of $\mathbb{A}^{n^2}$ . Since affine space is irreducible, it suffices to show that $V$ is closed and $V$ contains a non-empty open set.

Fix a matrix $M$ . First, observe that the coefficients of the characteristic polynomial are polynomials in the entries in $M$ . In particular, the condition that a matrix satisfy its own characteristic polynomial amounts to a collection of polynomials in the entries of $M$ vanishing. This establishes that $V$ is closed.

Let $U$ be the set of matrices that have $n$ distinct eigenvalues. A matrix has $n$ distinct eigenvalues if and only if its characteristic polynomial has no double roots when it splits. This occurs if and only if the discriminant of the characteristic polynomial is nonzero. The discriminant is a polynomial in the coefficients of the characteristic polynomial. Thus the condition that a matrix have $n$ distinct eigenvalues amounts to a polynomial in the entries of $M$ not vanishing. Thus $U$ is open.

Finally, we have to show $U \subseteq V$ . It is easy to check this for $U$ a diagonal matrix. The general result follows from the fact that the determinant and thus the characteristic polynomial is basis-invariant.

I learned this from https://aergus.net/blog/posts/using-zariski-topology-to-prove-the-cayley-hamilton-theorem.html/

Arithmetic variety

Tag Archives: Cayley-Hamilton

More on algebraic numbers

A case study

Why it works

Algebro-geometric proof of Cayley-Hamilton