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 of finite-dimensional vector spaces, what would lead you to identify the spectrum as an interesting invariant of endomorphisms in ?
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
in , we have
I call this “cyclicity” because it implies the more general property that for any cycle
of linear maps, the scalar
is independent of .
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 and have the same nonzero eigenvalues, with the same (algebraic) multiplicities. Zero has to be excluded to make this true: for instance, if we take and to be the projection and inclusion associated with a direct sum decomposition, then one composite operator has as an eigenvalue and the other does not.
I’ll write for the set-with-multiplicities of eigenvalues of a linear operator , and for the set-with-multiplicities of nonzero eigenvalues. Everything we’ll do is on finite-dimensional vector spaces over an algebraically closed field . Thus, is a finite subset-with-multiplicity of and is a finite subset-with-multiplicity of .
I’ll call the invertible spectrum of . Why? Because every operator decomposes uniquely as a direct sum of operators , where every eigenvalue of is (or equivalently, is nilpotent) and no eigenvalue of is (or equivalently, is invertible). Then the invertible spectrum of is the spectrum of its invertible part .
If excluding zero seems forced or unnatural, perhaps it helps to consider the “reciprocal spectrum”
There’s a canonical bijection between and given by . So the invariants and carry the same information, and if seems natural to you then should too.
Moreover, if you know the space that your operator is acting on, then to know the invertible spectrum is to know the full spectrum . That’s because the multiplicities of the eigenvalues of sum to , and so the multiplicity of in is minus the sum of the multiplicities of the nonzero eigenvalues.
The cyclicity equation
is a very strong property of . A second, seemingly more mundane, property is that for any operators and on the same space, and any scalar ,
In other words, for an operator , if you know and you know the space that acts on, then you know for each scalar . 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 determines , which determines (as ), which in turn determines .
I claim that the invariant is universal with these two properties, in the following sense.
Theorem Let be a set and let be a function satisfying:
for all
for all operators on the same space, and all scalars .
Then is a specialization of , that is, for all . Equivalently, there is a unique function such that for all operators .
For example, take to be trace. Then conditions 1 and 2 are satisfied, so the theorem implies that trace is a specialization of . 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 on defined as follows: writing for the standard basis of , the operator is given by
So if the theorem is to be true then, in particular, must be independent of .
But it’s not hard to cook up maps and such that and . Thus, condition 1 implies that . It follows that is independent of , 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:
for any operators and .
One way to interpret the theorem is as follows. Let be the smallest equivalence relation on such that:
(where , , etc. are quantified as in the theorem). Then the natural surjection
is isomorphic to
That is, there is a bijection between and making the evident triangle commute.
So, we’ve characterized the invariant 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 of finite-dimensional vector spaces. But let’s now think about how it could be interpreted in other categories. That is, for a category (in place of ) and a function
into some set , how can we make sense of condition 2?
Write for the category of endomorphisms in , with maps preserving those endomorphisms in the sense that the evident square commutes. (It’s the category of functors from the additive monoid , seen as a one-object category, into .)
For any scalars and , there’s an automorphism of the category given by
I guess, but haven’t proved, that these are the only automorphisms of that leave the underlying vector space unchanged. In what follows, I’ll assume this guess is right.
Now, condition 2 says that determines for each , for operators on a known space. That’s weaker than the statement that determines for each and — but does determine . So the theorem remains true if we replace condition 2 with the statement that determines for each automorphism of “over ” (that is, leaving the underlying vector space unchanged).
This suggests the following definition:
Definition Let be a category. Let be the equivalence relation on generated by:
for all in
for all endomorphisms on the same object of and all automorphisms of over .
Call the set of invertible spectral values of . Write for the natural surjection. The invertible spectrum of an endomorphism in is .
In the case , the invertible spectral values are the finite subsets-with-multiplicity of , and the invertible spectrum is as defined at the start of this post — namely, the set of nonzero eigenvalues with their algebraic multiplicities.
Aside At 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 does give 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 has finite coproducts then is a congruence for them; that is, if and then . (Here is the coproduct in , which comes from the coproduct in in the obvious way.) So the coproduct structure on endomorphisms induces a binary operation on the set of invertible spectral values, satisfying
In the case , this is the union of finite subsets-with-multiplicity of (adding multiplicities). And in general, the algebraic properties of coproduct imply that gives the set of invertible spectral values the structure of a commutative monoid.
Similarly, condition 2 implies that the automorphism group of 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 is , the category of finite sets.
I believe the category of endomorphisms in has no nontrivial automorphisms over . After all, given an endomorphism of a finite set , what natural ways are there of producing another endomorphism of ? There are only the powers , I think, and the process is only invertible when .
So, condition 2 is trivial. We’re therefore looking for the smallest equivalence relation on such that for all maps and pointing in opposite directions. I believe, but haven’t proved, that if and only if and have the same number of cycles
of each period . Thus, the invertible spectral values of are the finite sets-with-multiplicity of positive integers, and if is an endomorphism of a finite set then is the set-with-multiplicities of periods of cycles of .
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 determines for an operator on a known space (and, of course, for known . That was enough to prove the theorem. But there’s also a much stronger true statement: determines for any polynomial over (again, for an operator on a known space). Any polynomial gives an endomorphism of over , and I guess these are the only endomorphisms. So, we could generalize condition 2 by using endomorphisms rather than automorphisms of . 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.)