### Spectra of Operators and Rings

#### Posted by Tom Leinster

There are many uses of the word ‘spectrum’ in mathematics, and most of them are related. (The main exception that I’m aware of is spectra in the sense of homotopy theory.) In particular, the spectrum of a linear operator on a finite-dimensional vector space — that is, its set of eigenvalues — can be seen as the spectrum of an associated commutative ring.

I’ll explain how that story goes. But let me state up front the question that I want to ask. Although it was motivated by my desire to link up the various notions of spectra, it doesn’t actually mention spectra at all.

Let $V$ be a finite-dimensional vector space over a field $k$. Let $T$ be an operator on $V$, in other words, a linear map $V \to V$. Write $\chi(x) \in k[x]$ for its characteristic polynomial. What is the significance of the ring

$k[x]/(\chi(x))?$

Let me explain the kind of answer I’m hoping for. Suppose we replace the characteristic polynomial $\chi$ by the minimal polynomial $m$. Then the ring

$k[x]/(m(x))$

can be seen as the ring of polynomials in $T$. Precisely: write $End(V)$ for the $k$-algebra of operators (endomorphisms) on $V$, with composition as its product. The homomorphism $k[x] \to End(V)$ sending $x$ to $T$ has $(m(x))$ as its kernel, and the image is the subalgebra of $End(V)$ generated by $T$. Hence $k[x]/(m(x))$ is isomorphic to this subalgebra, which consists of the polynomials in $T$.

But is there a nice way to think about $k[x]/(\chi(x))$?

I said that the **spectrum** of a linear operator $T$ was its set of
eigenvalues, but it should really be thought of as a set *with
multiplicity*. The multiplicities here are the *algebraic*
multiplicities. I’ll write $Spec(T)$ for the spectrum of $T$.

Usually people define the algebraic multiplicity of an
eigenvalue $\lambda$ as the power of $(x - \lambda)$ appearing in the
characteristic polynomial. In the first post I ever wrote for the
Café, *Linear Algebra Done Right*,
I explained a more conceptual way to think about the algebraic multiplicity.
It’s the dimension of the **eventual kernel** of $T - \lambda$, defined as

$evKer(T - \lambda) = \bigcup_{i \in \mathbb{N}} Ker((T - \lambda)^i).$

(I’m assuming that we’re over an algebraically closed field.) For that reason, I wanted to use the term ‘dynamic multiplicity’ instead. But I’ll just say ‘multiplicity’.

In any case, we can write the characteristic polynomial of $T$ as

$\chi(x) = (x - \lambda_1)^{\alpha_1} \cdots (x - \lambda_k)^{\alpha_k}$

where $\lambda_1, \ldots, \lambda_k$ are the distinct eigenvalues of $T$ and $\alpha_1, \ldots, \alpha_k$ are their multiplicities. We can form the ring

$R(T) = k[x]/(\chi(x)),$

which is commutative. So, starting from a linear operator $T$, we’ve defined a commutative ring $R(T)$.

Every commutative ring $R$ has a spectrum, $Spec(R)$. At its most basic, the spectrum of a ring is the set of prime ideals; but it also carries a topology and a sheaf of rings. In the jargon, $Spec(R)$ is a ‘ringed space’.

So: given a linear operator $T$, we get two things called ‘spectra’: the set-with-multiplicities $Spec(T)$ of eigenvalues, and the ringed space $Spec(R(T))$. How are they related?

The answer was worked out long ago, and I guess all algebraic geometers know it. Nevertheless, I think there are plenty of people who know both uses of the word ‘spectrum’, but aren’t aware that there’s a close connection between them. So here goes.

For our ring $R(T)$, the (mere) set $Spec(R(T))$ of prime ideals is

$Spec(R(T)) = \{ (x - \lambda_1), \ldots, (x - \lambda_k) \}.$

The points of $Spec(R(T))$ are, therefore, in natural, one-to-one correspondence with the points of $Spec(T)$. That’s an encouraging start.

As a topological space, $Spec(R(T))$ is discrete: every subset is open. A sheaf of widgets on a discrete space $S$ is just an $S$-indexed family of widgets (namely, the stalks of the sheaf). Here, ‘widget’ is ‘ring’. So, to describe the sheaf $O$ with which $Spec(R(T))$ comes equipped, we just have to describe one ring $O_p$ for each point $p$ of $Spec(R(T))$. And in fact, the ring associated to the point $(x - \lambda_i)$ of $Spec(R(T))$ is just

$k[x]/((x - \lambda_i)^{\alpha_i}).$

The point now is that you can recover the multiplicity $\alpha_i$ from this ring. At least, you can do so if you’re allowed to know its $k$-algebra structure: for $\alpha_i$ is just its dimension as a vector space over $k$.

Thus, even if you don’t know $T$, knowing $Spec(T)$ tells you what $Spec(R(T))$ is, and vice versa. In that sense, $Spec(T)$ and $Spec(R(T))$ contain equivalent information about the operator $T$.

This is satisfying, but one question remains. How do we understand the ring
$R(T)$? As I said at the beginning, it would be easily understandable if we’d
used $k[x]/(m(x))$ instead of $k[x]/(\chi(x))$; but then the spectrum would tell us the powers of
$(x - \lambda_i)$ in the *minimal* polynomial, rather than the characteristic polynomial, and that’s not such
important information.

Maybe an example will make the problem clearer. In fact, a really trivial example will do: $T = 0$. Say $dim(V) = N$. Then

$m(x) = x, \quad \chi(x) = x^N.$

Hence

$k[x]/(m(x)) = k, \quad R(T) = k[x]/(\chi(x)) = k[x]/(x^N).$

Conceptually, what is the process that turns the zero operator on $k^N$ into the ring $k[x]/(x^N)$?

## Re: Spectra of Operators and Rings

Deformation? Given a linear operator you can consider the moduli space of all linear operators with the same characteristic polynomial. Generically they should have minimal polynomial the characteristic polynomial, so an arbitrarily small deformation of a given operator (preserving its characteristic polynomial) will generate an algebra isomorphic to the quotient by the characteristic polynomial. I don’t know how satisfying this is, though.