More on algebraic numbers

Posted on July 16, 2019 by rohanjoshi

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.

Two proofs complex matrices have eigenvalues

Posted on July 16, 2019 by rohanjoshi

Today I will briefly discuss two proofs that every matrix $T$ over the complex numbers (or more generally, over an algebraically closed field) has an eigenvalue. Notice that this is equivalent to finding a complex number $\lambda$ such that $T - \lambda I$ has nontrivial kernel. The first proof uses facts about “linear dependence” and the second uses determinants and the characteristic polynomial. The first proof is drawn from Axler’s textbook [1]; the second is the standard proof.

Proof by linear dependence

Let $p(x) = a_nx^n + \dots a_1x + a_0$ be a polynomial with complex coefficients. If $T$ is a linear map, $p(T) = a_nT^n + \dots a_n T + a_0I$ . We think of this as “ $p$ evaluated at $T$ ”.

Exercise: Show $(pq)(T) = p(T)q(T)$ .

Proof: Pick a random vector $v \in V$ . Consider the sequence of vectors $v, Tv, T^2v, \dots T^nv.$ This is a set of $n+1$ vectors, so they must be linearly dependent. Thus there exist constants $a_0, \dots a_n \in \C$ such that $a_nT^nv + a_{n-1}T^{n-1}v + \dots a_1Tv + a_0v$ $= (a_nT^n + a_{n-1}T^{n-1} + \dots a_1T + a_0I)v = 0$ .

Define $p(x) = a_nx^n + \dots a_1x + a_0$ . Then, we can factor $p(x) = a_n(x-\lambda_1)\dots (x-\lambda_n).$ By the Exercise, this implies $a_n(T - \lambda_1I)\dots (T-\lambda_nI)v = 0$ . So, at least one of the maps $T - \lambda_iI$ has a nontrivial kernel, so $T$ has an eigenvalue. $\square$

Proof by the characteristic polynomial

Proof: We want to show that there exists some $\lambda$ such that $T - \lambda I$ has nontrivial kernel: in other words, that $T - \lambda I$ is singular. A matrix is singular if and only if its determinant is nonzero. So, let $\chi_T(x) = \det(T - xI)$ ; this is a polynomial in $x$ , called the characteristic polynomial of $T$ . Now, every polynomial has a complex root, say $\lambda$ . This implies $T - \lambda I$ , so $T$ has an eigenvalue. $\square$

Thoughts

To me, it seems like the determinant based proof is more straightforward, although it requires more machinery. Also, the determinant based proof is “constructive”, in that we can actually find all the eigenvalues by factoring the characteristic polynomial. On subject of determinant-based vs determinant-free approaches to linear algebra, see Axler’s article “Down With Determinants!” [3].

There is a similar situation for the problem of showing that the sum (or product) of two algebraic numbers is algebraic. Here there is a non-constructive proof using “linear dependence” (which I attempted to describe in a previous post) and a constructive proof using the characteristic polynomial (which will hopefully be the subject of a future blog post). A further advantage of the determinant-based proof is that it can be used more generally to show that the sum and product of integral elements over a ring are integral. In this more general context, we no longer have linear dependence available.

References

Sheldon Axler, Linear algebra done right. Springer 2017
Evan Chen, An Infinitely Large Napkin, available online
Sheldon Axler. Down with Determinants! The American Mathematical Monthly, 102(2), 139, 1995. doi:10.2307/2975348, available online

Hilbert’s Basis Theorem

Posted on July 10, 2019 by rohanjoshi

Here is a proof of Hilbert’s Basis Theorem I thought of last night.

Let $R$ be a noetherian ring. Consider an ideal $J$ in $R[X]$ . Let $I_i$ be the ideal in $R$ generated by the leading coefficients of the polynomials of degree $i$ in $I$ . Notice that $I_i \subseteq I_{i+1}$ , since if $P \in I_i$ , $xP \in I_{i+1}$ , and it has the same leading coefficient. Thus we have an ascending chain $\dots \subseteq I_{i-1} \subseteq I_i \subseteq I_{i+1} \subseteq \dots$ , which must terminate, since $R$ is noetherian. Suppose it terminates at $i = n$ , so $I_n = I_{n+1} = \dots$ .

Now for each $I_i$ choose a finite set of generators (which we can do since $R$ is noetherian). For each generator, choose a representative polynomial in $J$ with that leading coefficient. This gives us a finite collection polynomials: define $J_i$ to be the ideal of $R[x]$ generated by these polynomials.

Let $J' = J_0 + J_1 + \dots + J_n$ . I claim $J = J'$ . Assume for the sake of contradiction that there is a polynomial $P$ of minimal degree (say $i$ ) which is in $J$ but $J'$ . If $i \leq n$ , there is an element of $P' \in J'$ with the same leading coefficient, so $P - P'$ is not in $J'$ but has degree smaller than $i$ : contradiction. If $P$ is of degree $i > n$ , then there is an element of $P'$ of $J_n$ which has the same leading coefficient as $P$ . Thus $P - x^{i-n}P'$ is of degree smaller than $i$ but is not in $J'$ : contradiction.

Thus $J = J'$ . Since $J$ is therefore finitely generated, this proves $R[x]$ is noetherian.

Algebro-geometric proof of Cayley-Hamilton

Posted on July 10, 2019 by rohanjoshi

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

Monthly Archives: July 2019