# Sylvester’s law of inertia, from the spectral theorem

Here we prove Sylvester’s law of inertia. Let ${A}$ be a real symmetric matrix, and assume the spectral theorem.

Take an arbitrary orthogonal basis with respect to the form ${A(v,w) = X^tAY}$, and scale (and optionally reorder) so that there are ${(n_+,n_-,n_0)}$ entries of ${+1,-1,0}$. (This can also be shown using the spectral theorem, but I think it’s overkill and conceptually messier (we’re “mixing up” linear transformation matrices and bilinear form matrices).)

So first we show ${n_+,n_-,n_0}$ are unique (i.e. invariant under change of basis). This is not too hard, since our basis ${v_1,\ldots,v_n}$ is quite nice. The maximum dimension of a positive-definite set is ${n_+}$, or else we would get a linearly independent set of at least ${(n_+ + 1) + n_-+n_0 = n+1}$ vectors. (More precisely, this would force a nontrivial intersection between a dimension-(${\ge n_+ + 1}$) positive-definite space and a dimension-${(n_-+n_0)}$ negative-semi-definite space, which is clearly absurd.) Similarly, the maximum dimension of a negative-definite set is ${n_-}$.

Now that we have uniqueness, we move on to the eigenvalues and principal minor determinant interpretations. By the spectral theorem (for real symmetric matrices), the fact that symmetric matrices have real eigenvalues, and uniqueness of Sylvester form, ${A}$ has ${n_+}$ positive eigenvalues, ${n_-}$ negative eigenvalues, and ${n_0}$ zero eigenvalues.

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# Non-inductive proof of the spectral theorem (for normal matrices)

Here we present a non-inductive proof of the spectral theorem for normal matrices (which doesn’t use, for instance, proposition 8.6.4 in Artin’s Algebra). (But it does seem to be the same as the proof in Herstein’s Topics in algebra.) It is motivated by a similar direct proof (presented in my class) for Hermitian operators with ${n}$ distinct eigenvalues.

We work in matrix form, so we need to prove that there exists an orthonormal basis of ${\mathbb{C}^n}$ consisting of eigenvectors of a normal matrix ${A}$.

# Galois theory basics, part 2

From the previous post, we know that (in characteristic ${0}$), if ${[K:F]}$ is finite, then (by the primitive element theorem to get some ${\alpha}$ to satisfy hypotheses of previous explanation) ${G(K/F) = G(K/K^{G(K/F)})}$ and

$\displaystyle [K:K^{G(K/F)}] = \lvert{G(K/F)}\rvert = \lvert{G(K/K^{G(K/F)})}\rvert$

(so ${K}$ is a splitting field over/Galois over ${K^{G(K/F)}}$) equals the degree of the “splitting part” of min poly of ${\alpha}$.

So naturally, we wonder if we can say anything starting from the perspective of groups, i.e. “can ${G(K/F)}$ be replaced by an arbitrary (finite) group of automorphisms (of ${K}$) ${H}$?”

Perhaps not surprisingly, yes!

# Galois theory basics, part 1

I was recently looking at algebraic proofs of FTA, and decided to review some Galois theory. So here’s some notes/an attempt to motivate Galois theory (perhaps up to or a bit before fundamental theorem of Galois theory), from the perspective of someone who much prefers field theory/polynomials to group theory/automorphisms (although maybe I should just stop being stubborn).

(Random side note: in the context of abstract algebra, polynomials are naturally motivated by “minimal (free) ring extensions”: if we have a commutative ring ${R}$, then ${R[x]}$ is the smallest ring containing ${R}$ with the additional element ${x}$ satisfying no relations. On the other hand, any constraints/extra relations would be polynomials, so at least for monic polynomials ${f}$ we get the quotient construction ${R[x]/(f)}$.)

Suppose ${K = F(\alpha)}$ is a singly-generated field extension (by primitive element theorem this is broader than it seems). If ${f}$ is the minimal polynomial of ${\alpha}$ of degree ${d}$, then let’s look at how it splits/factors in ${K[x]}$. If ${f}$ has some set of roots ${r\le n}$ of roots lying in ${K}$ (the “splitting part”/linear factors of ${f}$ in ${K[x]}$), say ${z_1 = \alpha, z_2, \ldots, z_r}$.

Generally, the main object of Galois theory seems to be splitting fields (or if we’re lazy, algebraic closures), but I’m still fighting through the material myself so I don’t fully appreciate/communicate this myself. Perhaps the point is just that it’s much easier to work concretely with roots (in algebraic closures) than directly with irreducible polynomials. (For example, we can then work with symmetric sums of the roots, and generally draw lots of intuition from number fields.)

We’ll work in characteristic ${0}$ for convenience, so e.g. the ${z_i}$ are pairwise distinct.

1. “Symmetry” of the ${z_i}$: crux of Galois theory, and introducing Galois groups

The key is that ${K = F(z_1) = F(z_2) = \cdots = F(z_r)}$ (recall ${\alpha = z_1}$ by definition), since the ${z_i}$ share minimal polynomials ${f \in F[x]}$.

To make this symmetry precise, we phrase things in terms of ${F}$-automorphisms of ${K}$; each ${F}$-automorphism fixes coefficients of ${f}$, hence is uniquely determined by sending ${\alpha \rightarrow z_i}$. Thus they form a Galois group ${G := G(K/F)}$ of size ${r}$, since we easily check the automorphisms to be bijections of ${K}$.