A adic PerronFrobenius Theorem
Abstract.
We prove that if an matrix defined over (or more generally an arbitrary complete, discretelyvalued, nonArchimedean field) satisfies a certain congruence property, then it has a strictly maximal eigenvalue in , and that iteration of the (normalized) matrix converges to a projection operator onto the corresponding eigenspace. This result may be viewed as a adic analogue of the PerronFrobenius theorem for positive real matrices.
Key words and phrases:
PerronFrobenius theorem, maximal eigenvalue, adic and nonArchimedean fields, iteration of matrices2010 Mathematics Subject Classification:
15B48, 15B51, 11S99, 37P201. Introduction
The PerronFrobenius theorem gives information about the eigenvalues and eigenvectors of certain matrices with nonnegative real entries. In 1907 Perron proved the simplest version of the result, for positive matrices, and in 1912 Frobenius extended the result to (irreducible) nonnegative matrices. (Recall that a matrix is said to be positive (resp. nonnegative) if all of its entries are positive (resp. nonnegative) real numbers.) Perron’s version of the theorem can be stated as follows.
Theorem 1 ([4], Ch. 16, Thm. 1).
Let be a positive matrix. Then has a positive real eigenvalue , of multiplicity one, such that for all other complex eigenvalues of . Furthermore, there exists a positive eigenvector of .
A useful application of the PerronFrobenius theorem is the following wellknown result. Following [4], an matrix is said to be stochastic if it is nonnegative and the sum along each column of the matrix is equal to .
Theorem 2 ([4], Ch. 16, Thm. 3).
Let be a positive stochastic matrix. Then the maximal eigenvalue of is equal to , and for any nonnegative vector , the sequence converges to a positive eigenvector of .
The PerronFrobenius theorem and its application to the iteration of stochastic matrices have implications in many areas of mathematics, including graph theory, probability theory, and symbolic dynamical systems. Realworld applications of these results include the modeling of changes in atomic nuclei and populations in ecological systems, and the PageRank algorithm used by Google as a basis for its internet search strategy. See [4] and [5] for further background on the PerronFrobenius theorem and its applications.
In this paper we give an analogue of (Perron’s form of) the PerronFrobenius theorem for matrices defined over the field of adic numbers. Denote by the adic absolute value on , normalized as usual so that , and for each and real , denote by the closed adic disc with center and radius . As usual, denote by the completion of the algebraic closure of .
To formulate a adic analogue of Perron’s theorem, we take the adic disc as our analogue of the positive real numbers. The motivation for this choice is simply the observation that is the largest subinterval of containing but not containing , and is the largest disc in containing but not containing . This leads to a fairly close parallel with Perron’s theorem, and as we explain in Remark 1, the choice of the disc is not as limiting as it may at first appear. The simplest nontrivial case of our main result is the following.
Theorem 3.
Let be a prime number and let be a positive integer such that . Let be an matrix over with all entries in the disc . Then the following conditions hold.

has a simple eigenvalue such that , and for all other eigenvalues of .

The maximal eigenvalue is an element of the disc .

There exists a eigenvector of , all of whose components are elements of the disc .

The (adic) limit
exists and is a projection operator onto the eigenspace of .
Remark 1.
Theorem 3, and its generalization Theorem 6, can be used to study a broader class of matrices than just those whose entries are near . For example, suppose that is an matrix over , all of whose entries lie in a adic disc such that . Then and for all . Therefore the matrix satisfies the hypotheses of Theorem 3, and the conclusions pertaining to easily imply corresponding statements about the matrix .
Example 1.
The matrix
over satisfies the hypotheses of Theorem 3. The characteristic polynomial of is
Since and , we have . The eigenvector of has both entries in , and the adic limit
exists and is equal to a projection operator onto the eigenspace of . The evaluation of this limit, as shown, follows from the diagonalization argument given in the proof of Theorem 3; see especially .
Example 2.
Example 3.
While , the nonmaximal eigenvalues are not necessarily in . Consider
over , which satisfies the hypotheses of Theorem 3. Its characteristic polynomial is . Thus , but is irreducible over . Indeed, is not a square in because it is not a square in the residue field . Therefore is not a square in .
Example 4.
This example illustrates that, if the requirement is omitted, then the theorem is false as written. Let and consider the matrix
over . All of the entries are elements of the disc , and the characteristic polynomial of is
A Newton polygon argument shows that the two eigenvalues of both have absolute value . Thus has no strictly maximal eigenvalue in , and no eigenvalues at all in .
Despite Example 4, a version of Theorem 3 can be recovered, even when , provided we strengthen the hypothesis that all entries of are near . Precisely, if we require that all entries of lie in the smaller disc for a positive integer , then (a slightly modified version of) Theorem 3 continues to hold. We also prove, in 4, that this result is sharp by giving a counterexample if the hypothesis is weakened.
Since it requires no extra work, we prove our results in the setting of an arbitrary complete, discretelyvalued, nonArchimedean field. Thus in 2 we briefly review some notation and basic facts about such fields, and we prove two crucial lemmas. In 3 we prove our main result, Theorem 6, of which Theorem 3 is a special case. In 4 we give a counterexample to our main result if the main hypothesis is weakened.
This work was done during the Summer 2015 REU program in Mathematics at Oregon State University, with support by National Science Foundation Grant DMS1359173.
2. Notation and Preliminary Lemmas
Throughout this paper, denotes a field which is complete with respect to a nontrivial, nonArchimedean absolute value , and which has a discrete value group . This means that, in addition to the usual axioms ([3] 2.1) satisfied by an absolute value, also satisfies the strong triangle inequality
(1) 
and a standard argument ([3] Prop. 2.3.3) shows that
(2) 
Let denote the ring of integers in and let be a uniformizing paramater. This means that is an element of satisfying
or equivalently, that is the unique maximal ideal of . (Such an element exists by the assumption that is discretely valued.) Denote by the discrete valuation on normalized so that for all ; thus .
Given an element and a real number , denote by
the closed disc in with center and radius .
Given a matrix with entries in , define its norm by
(3) 
This defines a norm on the vector space of matrices over . The strong triangle inequality implies the bound
(4) 
whenever and .
Let be the completion of the algebraic closure of . The field is both complete and algebraically closed, and the absolute value extends uniquely to a nonArchimedean absolute value on [1]. We extend the norm to matrices defined over via the formula .
Lemma 4.
Let be an matrix over with all entries in the disc for some positive integer . Then
(5) 
Proof.
By hypothesis we may write
(6) 
Let denote the group of permutations of the index set , and let denote an arbitrary subset of . We have
(7) 
where for each we define
We will now show that
(8) 
For if , then the set has at least two elements, so there exists a transposition that fixes each element of . Then since the map is a bijection from to itself, we have
which implies . (Of course, the equality does not imply in a field of characteristic . However, one may view the proof of as taking place in the polynomial ring in doubly indexed indeterminates , which implies that the identity holds in arbitrary characteristic.)
Because of the identity (7) becomes
(9) 
When , it follows from the strong triangle inequality that
(10) 
and we conclude from and that . ∎
Lemma 5.
Let be an matrix over with all entries in the disc for some positive integer , and let
be the characteristic polynomial of . Then
(11) 
Moreover, if , then
(12) 
Proof.
First we establish some notation. Given positive integers and , let represent the Kronecker delta function of and . Given a permutation , let denote the set of indices that are fixed by . Given a subset , let .
Again with written as in , we have
(13) 
where
(14) 
The secondtolast equality in follows from the fact that, because of the Kronecker delta factors, the th summand vanishes unless fixes every element of .
Note that if we let denote the matrix formed by removing the th row and th column from for each , then becomes
(15) 
It follows that for , the coefficient of can be calculated by adding and subtracting the determinants of matrices with entries in ; Lemma 4 implies that each of these determinants has an absolute value of at most , and the desired bound follows.
Finally, if , then since is the negative of the trace of , we have
by and the fact that . ∎
3. The main result
Theorem 6.
Let be an matrix over with all entries in the disc for a positive integer . Then the following conditions hold.

has a simple eigenvalue such that for all other eigenvalues of . Moreover, .

The maximal eigenvalue is an element of the disc .

There exists a eigenvector of , all of whose components are elements of the disc .

The limit
exists and is a projection operator onto the eigenspace of .
Remark 3.
It is implicit in the hypothesis that is not divisible by the characteristic of , for if then one has .
Remark 4.
The hypothesis can be written in the form , which implies in particular that
(16) 
The inequalities show that the bounds quoted in parts (b) and (c) of the Theorem are nontrivial and can be made independent of the dimension .
Proof of Theorem 6 (a).
In order to show the existence of a strictly maximal eigenvalue of of multiplicity one, we require an analysis of the characteristic polynomial
of via its Newton polygon. Recall that the Newton polygon of is the lower convex hull of the set of points
in the plane. (See [2] 6.3 or [3] 6.4.) A partial Newton polygon for is depicted in Figure 1.
From Lemma 5 and the hypothesis , we have
(17) 
and
(18) 
Observe that the line segment connecting with has slope
Extending this line segment to the left until it intersects the axis, this line, call it , has equation . For , we bound the coordinate of the point on the line with coordinate equal to :
(19) 
Here we have used , then , and finally the inequality , which follows from and the fact that is a positive integer.
The inequality shows that the line lies strictly below all of the leftmost points , , of the Newton polygon of (see Figure 1). Since the distinct slopes of the segments of the Newton polygon are increasing from left to right, it follows that the slopes of all of the segments of the Newton polygon to the left of the point are strictly less than the slope of the segment from to . Since this final segment has horizontal length , it follows from the Theorem of the Newton Polygon ([2] Thm. 6.3.1) that factors over as for satisfying
(20) 
and that all roots of in have absolute value . ∎
Proof of Theorem 6 (b).
Let be the nonmaximal eigenvalues of . Then for each , by and the fact that . So on the one hand we have
On the other hand, with written as in , using Lemma 5 and the fact that is the negative of the trace of , we have
In this inequality, the evaluation of the maximum as follows from an elementary argument and the hypothesis that , which implies . Combining these last two calculations for we deduce the desired inequality . ∎
Proof of Theorem 6 (c).
Let be an arbitrary eigenvector. Normalizing by a suitable nonzero scalar, we may assume without loss of generality that . Let denote the dimensional column vector, all of whose entries are equal to , and let denote the matrix, all of whose entries are equal to . With written as in , we have
where and . Thus
where . We then have
This last equality follows from , because the largest entry of has absolute value , while all entries of have absolute value at most
using . In particular, is nonzero and .
We now set . Then is a eigenvector of and
as desired.∎
In view of part (a) of Theorem 6, part (d) follows from the following theorem which holds in somewhat greater generality.
Theorem 7.
Let be an matrix with entries in , and suppose that has a simple eigenvalue with the property that for all other eigenvalues of . Then the limit
(21) 
exists and is a projection operator onto the eigenspace of .
Proof.
It suffices to prove that the limit exists in the space endowed with the norm defined in . Plainly this is equivalent to componentwise convergence, and the fact that such a limit must have entries in follows from the completeness of .
We have , where is a nonsingular change of basis matrix, and
is a Jordan canonical form for (written here in block form); here is the maximal eigenvalue guaranteed by Theorem 6 part (a), and each is an Jordan block
Thus the denote the (not necessarily distinct) nonmaximal eigenvalues of . Given a positive integer , we have
and a straightforward induction argument gives the wellknown formula
for the th power of an Jordan block; here we interpret whenever , as is standard. We now divide through by to obtain
and
where for each we define .
Noting that for , and recalling that binomial coefficients are integers and therefore have nonArchimedean absolute value at most , we see that each block converges to the zero matrix as . We conclude that
where
Now define
(22) 
Using we have
as , completing the proof of . We have , and therefore is a projection operator. Finally, given any vector , we have
and thus is a eigenvector; therefore the image of is the eigenspace of . ∎
4. A Counterexample when
In this section we give an example showing that Theorem 6 is sharp, in the sense that the hypothesis cannot be relaxed.
Let be a prime, let , and let be a positive integer such that . Set , and consider the matrix
In other words, all entries of are equal to except the upperleft entry, which is equal to . In particular, all entries of are elements of the disc , as required in the hypotheses of Theorem 6. Since has rank , its characteristic polynomial vanishes to multiplicity at least . We then have
where
(23) 
and we will show that
(24) 
In particular, the eigenvalues of in are , where are the roots of the quadratic polynomial
In fact we have . This can be seen by an elementary argument, showing that the discriminant of is positive; alternatively, all entries of are positive, so the classical PerronFrobenius theorem implies that must have a simple nonzero eigenvalue.
From and we have