### Where Does The Spectrum Come From?

#### Posted by Tom Leinster

Perhaps you, like me, are going to spend some of this semester teaching
students about eigenvalues. At some point in our lives, we absorbed the
lesson that eigenvalues are important, and we came to appreciate that the
invariant *par excellence* of a linear operator on a finite-dimensional
vector space is its spectrum: the set-with-multiplicities of
eigenvalues. We duly transmit this to our students.

There are lots of good ways to motivate the concept of eigenvalue, from lots of points of view (geometric, algebraic, etc). But one might also seek a categorical explanation. In this post, I’ll address the following two related questions:

If you’d never heard of eigenvalues and knew no linear algebra, and someone handed you the category $\mathbf{FDVect}$ of finite-dimensional vector spaces, what would lead you to identify the spectrum as an interesting invariant of endomorphisms in $\mathbf{FDVect}$?

What is the analogue of the spectrum in other categories?

I’ll give a fairly complete answer to question 1, and, with the help of that answer, speculate on question 2.

*(New, simplified version posted at 22:55 UTC, 2015-09-14.)*

Famously, trace has a kind of cyclicity property: given maps

$X \stackrel{f}{\to} Y \stackrel{g}{\to} X$

in $\mathbf{FDVect}$, we have

$tr(g \circ f) = tr(f \circ g).$

I call this “cyclicity” because it implies the more general property that for any cycle

$X_0 \stackrel{f_1}{\to} X_1 \stackrel{f_2}{\to} \,\, \cdots\,\, \stackrel{f_{n-1}}{\to} X_{n - 1} \stackrel{f_n}{\to} X_0$

of linear maps, the scalar

$tr(f_i \circ \cdots \circ f_1 \circ f_n \circ \cdots \circ f_{i + 1})$

is independent of $i$.

A slightly less famous fact is that the same cyclicity property is enjoyed by a finer invariant than trace: the set-with-multiplicities of nonzero eigenvalues. In other words, the operators $g\circ f$ and $f\circ g$ have the same nonzero eigenvalues, with the same (algebraic) multiplicities. Zero has to be excluded to make this true: for instance, if we take $f$ and $g$ to be the projection and inclusion associated with a direct sum decomposition, then one composite operator has $0$ as an eigenvalue and the other does not.

I’ll write $Spec(T)$ for the set-with-multiplicities of eigenvalues of a
linear operator $T$, and $Spec'(T)$ for the set-with-multiplicities of
*nonzero* eigenvalues. Everything we’ll do is on finite-dimensional
vector spaces over an algebraically closed field $k$. Thus, $Spec(T)$ is a
finite subset-with-multiplicity of $k$ and $Spec'(T)$ is a finite
subset-with-multiplicity of $k^\times = k \setminus \{0\}$.

I’ll call $Spec'(T)$ the **invertible spectrum** of $T$. Why? Because
every operator $T$ decomposes uniquely as a direct sum of operators
$T_{nil} \oplus T_{inv}$, where every eigenvalue of $T_{nil}$ is $0$ (or
equivalently, $T_{nil}$ is nilpotent) and no eigenvalue of $T_{inv}$ is $0$
(or equivalently, $T_{inv}$ is invertible). Then the invertible spectrum
of $T$ is the spectrum of its invertible part $T_{inv}$.

If excluding zero seems forced or unnatural, perhaps it helps to consider the “reciprocal spectrum”

$RecSpec(T) = \{\lambda \in k : ker(\lambda T - I) \,\,\text{ is nontrivial} \}.$

There’s a canonical bijection between $Spec'(T)$ and $RecSpec(T)$ given by $\lambda \leftrightarrow 1/\lambda$. So the invariants $Spec'$ and $RecSpec$ carry the same information, and if $RecSpec$ seems natural to you then $Spec'$ should too.

Moreover, if you know the space $X$ that your operator $T$ is acting on, then to know the invertible spectrum $Spec'(T)$ is to know the full spectrum $Spec(T)$. That’s because the multiplicities of the eigenvalues of $T$ sum to $dim(X)$, and so the multiplicity of $0$ in $Spec(T)$ is $dim(X)$ minus the sum of the multiplicities of the nonzero eigenvalues.

The cyclicity equation

$Spec'(g\circ f) = Spec'(f\circ g)$

is a very strong property of $Spec'$. A second, seemingly more mundane, property is that for any operators $T_1$ and $T_2$ on the same space, and any scalar $\lambda$,

$Spec'(T_1) = Spec'(T_2) \implies Spec'(T_1 + \lambda I) = Spec'(T_2 + \lambda I).$

In other words, for an operator $T$, if you know $Spec'(T)$ and you know the space that $T$ acts on, then you know $Spec'(T + \lambda I)$ for each scalar $\lambda$. Why? Well, we noted above that if you know the invertible spectrum of an operator and you know the space it acts on, then you know the full spectrum. So $Spec'(T)$ determines $Spec(T)$, which determines $Spec(T + \lambda I)$ (as $Spec(T) + \lambda$), which in turn determines $Spec'(T + \lambda I)$.

I claim that the invariant $Spec'$ is universal with these two properties, in the following sense.

TheoremLet $\Omega$ be a set and let $\Phi : \{ \text{linear operators} \} \to \Omega$ be a function satisfying:

$\Phi(g \circ f) = \Phi(f \circ g)$ for all $X \stackrel{f}{\to} Y \stackrel{g}{\to} X$

$\Phi(T_1) = \Phi(T_2)$ $\implies$ $\Phi(T_1 + \lambda I) = \Phi(T_2 + \lambda I)$ for all operators $T_1, T_2$ on the same space, and all scalars $\lambda$.

Then $\Phi$ is a specialization of $Spec'$, that is, $Spec'(T_1) = Spec'(T_2) \implies \Phi(T_1) = \Phi(T_2)$ for all $T_1, T_2$. Equivalently, there is a unique function $\bar{\Phi} : \{ \text{finite subsets-with-multiplicity of }\,\, k^\times\} \to \Omega$ such that $\Phi(T) = \bar{\Phi}(Spec'(T))$ for all operators $T$.

For example, take $\Phi$ to be trace. Then conditions 1 and 2 are satisfied, so the theorem implies that trace is a specialization of $Spec'$. That’s clear anyway, since the trace of an operator is the sum-with-multiplicities of the nonzero eigenvalues.

I’ll say just a little about the proof.

The invertible spectrum of a nilpotent operator is empty. Now, the Jordan normal form theorem invites us to pay special attention to the special nilpotent operators $P_n$ on $k^n$ defined as follows: writing $e_1, \ldots, e_n$ for the standard basis of $k^n$, the operator $P_n$ is given by

$e_n \mapsto e_{n - 1} \mapsto \cdots \mapsto e_1 \mapsto 0.$

So if the theorem is to be true then, in particular, $\Phi(P_n)$ must be independent of $n$.

But it’s not hard to cook up maps $f: k^n \to k^{n - 1}$ and $g: k^{n - 1} \to k^n$ such that $g\circ f = P_n$ and $f \circ g = P_{n - 1}$. Thus, condition 1 implies that $\Phi(P_n) = \Phi(P_{n - 1})$. It follows that $\Phi(P_n)$ is independent of $n$, as claimed.

Of course, that doesn’t prove the theorem. But the rest of the proof is straightforward, given the Jordan normal form theorem and condition 2, and in this way, we arrive at the conclusion of the theorem:

$Spec'(T_1) = Spec'(T_2) \implies \Phi(T_1) = \Phi(T_2)$

for any operators $T_1$ and $T_2$.

One way to interpret the theorem is as follows. Let $\sim$ be the smallest equivalence relation on $\{\text{linear operators}\}$ such that:

$g\circ f \sim f \circ g$

$T_1 \sim T_2$ $\implies$ $T_1 + \lambda I \sim T_2 + \lambda I$

(where $f$, $g$, etc. are quantified as in the theorem). Then the natural surjection

$\{ \text{linear operators} \} \longrightarrow \{ \text{linear operators} \}/\sim$

is isomorphic to

$Spec': \{ \text{linear operators} \} \longrightarrow \{ \text{finite subsets-with-multiplicity of }\,\, k^\times\}.$

That is, there is a bijection between $\{ \text{linear operators} \}/\sim$ and $\{ \text{finite subsets-with-multiplicity of }\,\, k^\times\}$ making the evident triangle commute.

So, we’ve characterized the invariant $Spec'$ in terms of conditions 1 and 2. These conditions seem reasonably natural, and don’t depend on any prior concepts such as “eigenvalue”.

Condition 2 *does* appear to refer to some special features of the category
$\mathbf{FDVect}$ of finite-dimensional vector spaces. But let’s now think
about how it could be interpreted in other categories. That is, for a
category $\mathcal{E}$ (in place of $\mathbf{FDVect}$) and a function

$\Phi: \{ \text{endomorphisms in }\,\, \mathcal{E} \} \to \Omega$

into some set $\Omega$, how can we make sense of condition 2?

Write $\mathbf{Endo}(\mathcal{E})$ for the category of endomorphisms in $\mathcal{E}$, with maps preserving those endomorphisms in the sense that the evident square commutes. (It’s the category of functors from the additive monoid $\mathbb{N}$, seen as a one-object category, into $\mathcal{E}$.)

For any scalars $\kappa \neq 0$ and $\lambda$, there’s an automorphism $F_{\kappa, \lambda}$ of the category $\mathbf{Endo}(\mathbf{FDVect})$ given by

$F_{\kappa, \lambda}(T) = \kappa T + \lambda I.$

I guess, but haven’t proved, that these are the *only* automorphisms of
$\mathbf{Endo}(\mathbf{FDVect})$ that leave the underlying vector space
unchanged. In what follows, I’ll assume this guess is right.

Now, condition 2 says that $\Phi(T)$ determines $\Phi(T + \lambda I)$ for
each $\lambda$, for operators $T$ on a known space. That’s weaker than the
statement that $\Phi(T)$ determines $\Phi(\kappa T + \lambda I)$ for each
$\kappa \neq 0$ and $\lambda$ — but $Spec'(T)$ *does* determine
$Spec'(\kappa T + \lambda I)$. So the theorem remains true if we replace
condition 2 with the statement that $\Phi(T)$ determines $\Phi(F(T))$ for
each automorphism $F$ of $\mathbf{Endo}(\mathbf{FDVect})$ “over
$\mathbf{FDVect}$” (that is, leaving the underlying vector space
unchanged).

This suggests the following definition:

DefinitionLet $\mathcal{E}$ be a category. Let $\sim$ be the equivalence relation on $\{ \text{endomorphisms in }\,\, \mathcal{E} \}$ generated by:

$g\circ f \sim f \circ g$ for all $X \stackrel{f}{\to} Y \stackrel{g}{\to} X$ in $\mathcal{E}$

$T_1 \sim T_2$ $\implies$ $F(T_1) \sim F(T_2)$ for all endomorphisms $T_1, T_2$ on the same object of $\mathcal{E}$ and all automorphisms $F$ of $\mathbf{Endo}(\mathcal{E})$ over $\mathcal{E}$.

Call $\{ \text{endomorphisms in }\,\, \mathcal{E}\}/\sim$ the set of

invertible spectral valuesof $\mathcal{E}$. Write $Spec': \{ \text{endomorphisms in }\,\, \mathcal{E} \} \to \{ \text{invertible spectral values of}\,\, \mathcal{E} \}$ for the natural surjection. Theinvertible spectrumof an endomorphism $T$ in $\mathcal{E}$ is $Spec'(T)$.

In the case $\mathcal{E} = \mathbf{FDVect}$, the invertible spectral values are the finite subsets-with-multiplicity of $k^\times$, and the invertible spectrum $Spec'(T)$ is as defined at the start of this post — namely, the set of nonzero eigenvalues with their algebraic multiplicities.

AsideAt least, that’s the case up to isomorphism. You might feel that we’ve lost something, though. After all, the spectrum of a linear operator is a subset-with-multiplicities of the base field, not just an element of some abstract set.But the theorem

doesgive us some structure on the set of invertible spectral values. This remark of mine below (written after I wrote a first version of this post, but before I wrote the revised version you’re now reading) shows that if $\mathcal{E}$ has finite coproducts then $\sim$ is a congruence for them; that is, if $S_1 \sim S_2$ and $T_1 \sim T_2$ then $S_1 + T_1 \sim S_2 + T_2$. (Here $+$ is the coproduct in $\mathbf{Endo}(\mathcal{E})$, which comes from the coproduct in $\mathcal{E}$ in the obvious way.) So the coproduct structure on endomorphisms induces a binary operation $\vee$ on the set of invertible spectral values, satisfying$Spec'(S \oplus T) = Spec'(S) \vee Spec'(T).$

In the case $\mathcal{E} = \mathbf{FDVect}$, this is the union of finite subsets-with-multiplicity of $k^\times$ (adding multiplicities). And in general, the algebraic properties of coproduct imply that $\vee$ gives the set of invertible spectral values the structure of a commutative monoid.

Similarly, condition 2 implies that the automorphism group of $\mathbf{Endo}(\mathcal{E})$ acts on the set of invertible spectral values; and since automorphisms preserve coproducts (if they exist), it acts by monoid homomorphisms.

We can now ask what this general definition produces for other categories. I’ve only just begun to think about this, and only in one particular case: when $\mathcal{E}$ is $\mathbf{FinSet}$, the category of finite sets.

I believe the category of endomorphisms in $\mathbf{FinSet}$ has no nontrivial automorphisms over $\mathbf{FinSet}$. After all, given an endomorphism $T$ of a finite set $X$, what natural ways are there of producing another endomorphism of $X$? There are only the powers $T^n$, I think, and the process $T \mapsto T^n$ is only invertible when $n = 1$.

So, condition 2 is trivial. We’re therefore looking for the smallest equivalence relation on $\{ \text{endomorphisms of finite sets} \}$ such that $g \circ f \sim f \circ g$ for all maps $f$ and $g$ pointing in opposite directions. I believe, but haven’t proved, that $T_1 \sim T_2$ if and only if $T_1$ and $T_2$ have the same number of cycles

$x_1 \mapsto x_2 \mapsto \cdots \mapsto x_p \mapsto x_1$

of each period $p$. Thus, the invertible spectral values of $\mathbf{FinSet}$ are the finite sets-with-multiplicity of positive integers, and if $T$ is an endomorphism of a finite set then $Spec'(T)$ is the set-with-multiplicities of periods of cycles of $T$.

All of the above is a record of thoughts I had in spare moments at this workshop I just attended in Louvain-la-Neuve, so I haven’t had much time to reflect. I’ve noted where I’m not sure of the facts, but I’m also not sure of the aesthetics:

In other words, do the theorem and definition above represent the best approach? Here are three quite specific reservations:

I’m not altogether satisfied with the fact that it’s the invertible spectrum, rather than the full spectrum, that comes out. Perhaps there’s something to be done with the observation that if you know the invertible spectrum, then knowing the full spectrum is equivalent to knowing (the dimension of) the space that your operator acts on.

Condition 2 of the theorem states that $Spec'(T)$ determines $Spec'(T + \lambda I)$ for an operator $T$ on a known space (and, of course, for known $\lambda)$. That was enough to prove the theorem. But there’s also a much stronger true statement: $Spec'(T)$ determines $Spec'(p(T))$ for any polynomial $p$ over $k$ (again, for an operator $T$ on a known space). Any polynomial $p$ gives an endomorphism $T \mapsto p(T)$ of $\mathbf{Endo}(\mathbf{FDVect})$ over $\mathbf{FDVect}$, and I guess these are the only endomorphisms. So, we could generalize condition 2 by using

*endomorphisms*rather than*automorphisms*of $\mathbf{Endo}(\mathcal{E})$. Should we?

## Re: Where Does The Spectrum Come From?

The “slightly less famous fact” is essentially the Sylvester determinant theorem: https://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem . (Sorry, I can’t figure out how to make HTML links with this parser.)