T.8. Perron-Frobenius theory of positive matrices From: H.R. Thieme, Mathematics in Population Biology, Princeton University Press, Princeton 2003

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1 T.8. Perron-Frobenius theory of positive matrices From: H.R. Thieme, Mathematics in Population Biology, Princeton University Press, Princeton 2003 A vector x R n is called positive, symbolically x > 0, if all components are nonnegative and at least one is positive. It is called strictly positive, x 0, if all components are positive. A square matrix is called positive if all entries are non-negative numbers and the matrix is not the zero matrix. It is called quasi-positive if it is not the zero matrix and all off-diagonal entries are non-negative numbers. It is called strictly positive if all entries are strictly positive. If n 2, an n n matrix A = (a ik ) is called irreducible if if the following holds: For any proper non-empty subset P of {1,, m} there are k P, j P such that a jk 0. A 1 1 matrix is called irreducible if it is not the 0 matrix. Equivalently A is irreducible if and only if, for all i, k = 1,..., n, there exist numbers j 1,..., j r {1,..., n} such that i = j 1, k = j r and a jl j l+1 0 for all l = 1,... r 1. A non-negative matrix A is irreducible if and only if the matrix exponential e A is strictly positive. A non-negative square matrix A is called primitive if one of its powers, A k, has strictly positive entries. It is easily seen that a non-negative matrix is primitive if it is irreducible and all entries in its main diagonal are strictly positive (Exercise 2). If A is a complex square matrix, a complex number λ is called a spectral value of A if the matrix λ A is singular. The set of spectral values of A is called the spectrum of A and is denoted by σ(a). For a matrix, a spectral value is an eigenvalue and vice versa, i.e., there exists a non-zero vector x, called eigenvector of A, such that Ax = λx. The spectral radius of the matrix A, r(a), is defined as r(a) = max{ λ ; λ σ(a)}, while the spectral bound of the matrix A, s(a) is defined as s(a) = max{rλ; λ σ(a)}. Obviously, s(a) r(a). Theorem 8.1. Let A be a positive matrix. Then its spectral radius, r(a), is an eigenvalue associated both with a positive eigenvector of A and a positive eigenvector of the transposed matrix A. In particular, s(a) = r(a). 29

2 For a proof see Schaefer (1974), Prop. I.2.3. Theorem 8.2. Let A be a positive matrix and x > 0 be a vector and µ 0 such that A q z µz for some natural number q and some vector z = z z with z, z R m +, z R m +. Then the spectral radius of A satisfied r(a) µ 1/q. Proof: This is the finite dimensional special case of Theorem 2.5 by Krasnosel skii (1964). Theorem 8.3. Let A be a quasi-positive matrix. Then its spectral bound (modulus of stability), s(a), is an eigenvalue of A associated both with a positive eigenvector of A and a positive eigenvector of the transposed matrix A. Moreover if x > 0 is a vector and µ R such that Ax µx, there exists some vector z > 0 and some scalar λ µ such that Az = λz and in particular s(a) µ. Proof: Since all off-diagonal elements of A are non-negative, then the matrix A + νi is non-negative for some (and then all) sufficiently large ν > 0. Let λ C be an eigenvalue of A such that Rλ = s(a). Then there exists a (possibly complex) vector x 0 such that Ax = λx. So (A + ν)x = (λ + ν)x. Let x = ( x 1,..., x n ) be the modulus (or absolute value) of the vector x. Since A+νI is a positive matrix, ν +λ x = (A+ν)x (A+ν) x. By Corollary 8.2 and Theorem 8.1, there exists some r ν+λ ν+s(a) and some vector z > 0 such that (A + ν)z = rz. So Az = (r ν)z. By definition of s(a), r ν s(a). Together with our previous inequality, r ν = s(a) and s(a) is an eigenvalue of A associated with a non-negative eigenvector. Since s(a) = s(a ) and A is a quasi-positive matrix, we can conclude that s(a) is also associated with a positive eigenvector of A. Now let Ax µx for some vector x > 0 and some µ R. Then (A + ν)x (ν + µ)x. Since A + νi is a positive matrix, by Corollary 8.2 and Theorem 8.1, there exists some r (ν + µ) and some vector z > 0 such that (A + ν)z = rz. Obviously Az = (r ν)z and r ν µ. So we choose λ = r ν. By definition of s(a), λ s(a) and so µ s(a). Theorem 8.4. Let A and D be positive matrices, D diagonal with all diagonal elements being positive. Then s(a D) and r(d 1 A) 1 have the same sign, i.e., these numbers are simultaneously positive, zero, or negative. 30

3 Proof: Let λ = s(a D). Since the off-diagonal elements of A D are non-negative, by Theorem 8.3 there exists some vector x > 0 such that (A D)x = λx. Reorganizing terms, Ax = Dx + λx. Since D is an invertible matrix, D 1 Ax = x + λd 1 x (1 + λɛ)x with ɛ being the reciprocal of the largest of the diagonal elements in D. So r(d 1 A) 1 + λɛ. Now let r = r(d 1 A) 1. By Theorem 1 there exist some vector x > 0 such that D 1 Ax = rx. Reorganizing terms, Ax = rdx. So (A D)x = (r 1)Dx (r 1)dx with d being the smallest diagonal element of D. By Theorem 8.3, s(a D) (r 1)d. The continuous and discrete dynamical systems associated with irreducible quasipositive or even primitive matrices have a strikingly simple large-time behavior. In the following x, y = m k=1 x ky k is the canonical scalar (or inner) product on R m. Theorem 8.5. Let A be a quasi-positive irreducible matrix. Then s = s(a) is an eigenvalue of both A and A with strictly positive eigenvectors v and v and s(a) is larger than the real parts of all other eigenvalues of A. Further any non-negative solution x of the differential equation x = Ax which is not identical 0 satisfies e s(t r) x(t) x(r), v v, v v, t, r > 0. Proof: If A is a quasi-positive irreducible matrix, then A+ν is a positive irreducible matrix for a sufficiently large ν > 0. So all matrices e ta = e νt e t(a+ν) are strictly positive and so form an irreducible uniformly continuous semigroup of compact operators on the Banach lattice R m. The claim now follows from Theorem 9.11 in Heijmans, de Pagter (1987). Remarks. Theorem 8.5 has significant side effects for an irreducible quasi-positive matrix A: (a) Every subspace that is forward invariant under A and contains a positive vector also contains the eigenvector v associated with s(a). In particular (b) Eigenvalues of A different from s(a) have no positive eigenvector or positive generalized eigenvector. (c) There are no generalized eigenvectors associated with s(a) and the eigenspace associated with s(a) is one-dimensional. In other words, s(a) is a simple eigenvalue. 31

4 (d) s(a) > Rλ for all eigenvalues λ of A that are different from s(a). Proof of (a): Let Y be a subspace of X and x 0 Y positive. Consider the solution x = Ax with x(0) = x 0. Then x(t) = e ta x 0 Y for all t 0 and so is v = lim t v, v x 0, v e s(a)t x(t). (d) Let λ be an eigenvalue of A that is different from s(a), but with Rλ = s(a). Let x be the solution of x = Ax with x(0) = v being the eigenvector associated with λ. Then e s(a)t x(t) = e ı(iλ)t v does not converge, contradicting the statement in Theorem 8.5. Lemma 8.6. Let A be a quasi-positive irreducible matrix and Ax λx or A x λx with λ R and x being a positive vector. Then s(a) λ. Proof: We consider Ax λx, the other case is done similarly. By Theorem 8.5, there exists a strictly positive vector v such that A v = sv with s = s(a). Then λ x, v = λx, v Ax, v = x, A v = x, sv = s x, v. Since the vector x is positive and the vector v strictly positive, x, v > 0 and λ s = s(a) follows by division. Remark. Choosing x with x j = 1 for j = 1,..., m in Lemma 8.6 provides the estimates s(a) max 1 j m k=1 m a jk, s(a) max 1 k m j=1 m a jk. Since the vector x is strictly positive, it is actually sufficient that A is quasi-positive. Theorem 8.7. Let A be a primitive matrix with spectral radius r = r(a) = s(a). Then r k A k x x, v v, v v, k where v and v are strictly positive eigenvectors of A and A associated with the eigenvalue r according to Theorem 8.5. A perhaps more intuitive equivalent formulation is the following one. 32

5 Corollary 8.8. Let A be a primitive matrix with spectral radius r = r(a) = s(a). Then, for every positive vector x and for every norm on R m, A k x A k x v v, k, where v is a positive eigenvector of A associated with the eigenvalue r according to Theorem 8.5. Proof: Corollary 8.8 obviously follows from Theorem 8.7 by the continuity of the norm. The converse follows by choosing the norm x = x, v where x is the vector ( x 1,..., x n ) and v a strictly positive eigenvector of A associated with r. Theorem 8.7 is only valid for primitive matrices. Actually, for non-negative matrices, primitivity is equivalent to the ergodicity statement in Theorem 8.7 (Schaefer, 1974, I.Proposition 7.3). But mean ergodicity still holds for irreducible matrices, which means that the convergence in Theorem 8.7 holds in average (Schaefer, 1974, end of Section I.6). Notice that the convergence statement in Theorem 8.7 implies the convergence statement in Theorem 8.9 (Exercise 2). Theorem 8.9. Let A be an irreducible positive matrix with spectral radius r = r(a). Then 1 k + 1 k j=0 r j A j x x, v v, v v, k where v and v are the strictly positive eigenvectors of A and A associated with the eigenvalue r according to Theorem 8.5. Noticing that A j x, v = r j x, v, Theorem 8.8 can be reformulated as follows. Theorem Let A be an irreducible positive matrix with spectral radius r = r(a) and x a positive vector. Then 1 k + 1 k j=0 A j x A j x, v v v, v, k where v and v are the strictly positive eigenvectors of A and A associated with the eigenvalue r according to Theorem

6 Exercises 1. Show that an irreducible non-negative matrix is primitive if all entries in the main diagonal are strictly positive. 2. Let (z(l)) be a convergent sequence of vectors in a normed vector space. Show that the averages converge to the same limit as l. 1 l + 1 l z(l) l=0 34