Inmatrix theory, thePerron–Frobenius theorem, proved byOskar Perron (1907) andGeorg Frobenius (1912), asserts that areal square matrix with positive entries has a uniqueeigenvalue of largest magnitude and that eigenvalue is real. The correspondingeigenvector can be chosen to have strictly positive components, and also asserts a similar statement for certain classes ofnonnegative matrices. This theorem has important applications to probability theory (ergodicity ofMarkov chains); to the theory ofdynamical systems (subshifts of finite type); to economics (Okishio's theorem,^[1]Hawkins–Simon condition^[2]);to demography (Leslie population age distribution model);^[3] to social networks (DeGroot learning process); to Internet search engines (PageRank);^[4] and even to ranking of American footballteams.^[5] The first to discuss the ordering of players within tournaments using Perron–Frobenius eigenvectors isEdmund Landau.^[6][7]

Letpositive andnon-negative respectively describematrices with exclusivelypositive real numbers as elements and matrices with exclusively non-negative real numbers as elements. Theeigenvalues of a realsquare matrixA arecomplex numbers that make up thespectrum of the matrix. Theexponential growth rate of the matrix powersA^k ask → ∞ is controlled by the eigenvalue ofA with the largestabsolute value (modulus). The Perron–Frobenius theorem describes the properties of the leading eigenvalue and of the corresponding eigenvectors whenA is a non-negative real square matrix. Early results were due toOskar Perron (1907) and concerned positive matrices. Later,Georg Frobenius (1912) found their extension to certain classes of non-negative matrices.

Let${\displaystyle A=(a_{ij})}$ be an${\displaystyle n\times n}$ positive matrix:${\displaystyle a_{ij}>0}$ for${\displaystyle 1\le i,j\le n}$. Then the following statements hold.

1. There is a positive real numberr, called thePerron root or thePerron–Frobenius eigenvalue (also called theleading eigenvalue,principal eigenvalue ordominant eigenvalue), such thatr is an eigenvalue ofA and any other eigenvalueλ (possiblycomplex) inabsolute value is strictly smaller thanr , |λ| <r. Thus, thespectral radius${\displaystyle \rho (A)}$ is equal tor. If the matrix coefficients are algebraic, this implies that the eigenvalue is aPerron number.
2. The Perron–Frobenius eigenvalue is simple:r is a simple root of thecharacteristic polynomial ofA. Consequently, theeigenspace associated tor is one-dimensional. (The same is true for the left eigenspace, i.e., the eigenspace forA^T, the transpose ofA.)
3. There exists an eigenvectorv = (v₁,...,v_n)^T ofA with eigenvaluer such that all components ofv are positive:A v =r v,v_i > 0 for 1 ≤i ≤n. (Respectively, there exists a positive left eigenvectorw :w^T A =w^T r,w_i > 0.) It is known in the literature under many variations as thePerron vector,Perron eigenvector,Perron-Frobenius eigenvector,leading eigenvector,principal eigenvector ordominant eigenvector.
4. There are no other positive (moreover non-negative) eigenvectors except positive multiples ofv (respectively, left eigenvectors except ww'w), i.e., all other eigenvectors must have at least one negative or non-real component.
5. ${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{A}^k/r^k=v{w}^T}$, where the left and right eigenvectors forA are normalized so thatw^Tv = 1. Moreover, the matrixvw^T is theprojection onto the eigenspace corresponding to r. This projection is called thePerron projection.
6. Collatz–Wielandt formula: for all non-negative non-zero vectorsx, letf(x) be the minimum value of [Ax]_i / x_i taken over all thosei such thatx_i ≠ 0. Thenf is a real valued function whosemaximum over all non-negative non-zero vectorsx is the Perron–Frobenius eigenvalue.
7. A "Min-max" Collatz–Wielandt formula takes a form similar to the one above: for all strictly positive vectorsx, letg(x) be the maximum value of [Ax]_i / x_i taken overi. Theng is a real valued function whoseminimum over all strictly positive vectorsx is the Perron–Frobenius eigenvalue.
8. Birkhoff–Varga formula: Letx andy be strictly positive vectors. Then,^[8]$${\displaystyle r=\underset{x>0}{sup}\underset{y>0}{inf}\frac{y^{\mathrm{\top }}Ax}{y^{\mathrm{\top }}x}=\underset{x>0}{inf}\underset{y>0}{sup}\frac{y^{\mathrm{\top }}Ax}{y^{\mathrm{\top }}x}=\underset{x>0}{inf}\underset{y>0}{sup}\sum _{i,j=1}^n{y}_i{a}_{ij}{x}_j/\sum _{i=1}^n{y}_i{x}_i.}$$
9. Donsker–Varadhan–Friedland formula: Letp be a probability vector andx a strictly positive vector. Then,^[9][10]$${\displaystyle r=\underset{p}{sup}\underset{x>0}{inf}\sum _{i=1}^n{p}_i[Ax{]}_i/x_i.}$$
10. Fiedler formula:^[11]$${\displaystyle r=\underset{z>0}{sup}\underset{x>0,y>0,x\circ y=z}{inf}\frac{y^{\mathrm{\top }}Ax}{y^{\mathrm{\top }}x}=\underset{z>0}{sup}\underset{x>0,y>0,x\circ y=z}{inf}\sum _{i,j=1}^n{y}_i{a}_{ij}{x}_j/\sum _{i=1}^n{y}_i{x}_i.}$$
11. The Perron–Frobenius eigenvalue satisfies the inequalities$${\displaystyle \underset{i}{min}\sum _j{a}_{ij}\le r\le \underset{i}{max}\sum _j{a}_{ij}.}$$

All of these properties extend beyond strictly positive matrices toprimitive matrices (see below). Facts 1–7 can be found in Meyer^[12]chapter 8 claims 8.2.11–15 page 667 and exercises 8.2.5,7,9 pages 668–669.

The left and right eigenvectorsw andv are sometimes normalized so that the sum of their components is equal to 1; in this case, they are sometimes calledstochastic eigenvectors. Often they are normalized so that the right eigenvectorv sums to one, while${\displaystyle w^{T}v=1}$.

There is an extension to matrices with non-negative entries. Since any non-negative matrix can be obtained as a limit of positive matrices, one obtains the existence of an eigenvector with non-negative components; the corresponding eigenvalue will be non-negative and greater thanor equal, in absolute value, to all other eigenvalues.^[13][14] However, for the example, the maximum eigenvaluer = 1 has the same absolute value as the other eigenvalue −1; while for, the maximum eigenvalue isr = 0, which is not a simple root of the characteristic polynomial, and the corresponding eigenvector (1, 0) is not strictly positive.

However, Frobenius found a special subclass of non-negative matrices —irreducible matrices — for which a non-trivial generalization is possible. For such a matrix, although the eigenvalues attaining the maximal absolute value might not be unique, their structure is under control: they have the form${\displaystyle \omega r}$, where${\displaystyle r}$ is a real strictly positive eigenvalue, and${\displaystyle \omega }$ ranges over the complexh' throots of 1 for some positive integerh called theperiod of the matrix.The eigenvector corresponding to${\displaystyle r}$ has strictly positive components (in contrast with the general case of non-negative matrices, where components are only non-negative). Also all such eigenvalues are simple roots of the characteristic polynomial. Further properties are described below.

LetA be an × n square matrix overfieldF.The matrixA isirreducible if any of the following equivalent propertiesholds.

Definition 1 :A does not have non-trivial invariantcoordinate subspaces.Here a non-trivial coordinate subspace means alinear subspace spanned by anyproper subset of standard basis vectors ofFⁿ. More explicitly, for any linear subspace spanned by standard basis vectorse_i1, ...,e_ik, 0 <k < n its image under the action ofA is not contained in the same subspace.

Definition 2:A cannot be conjugated into block upper triangular form by apermutation matrixP:

whereE andG are non-trivial (i.e. of size greater than zero) square matrices.

Definition 3: One can associate with a matrixA a certaindirected graphG_A. It hasn vertices labeled 1,...,n, and there is an edge from vertexi to vertexj precisely whena_ij ≠ 0. Then the matrixA is irreducible if and only if its associated graphG_A isstrongly connected.

IfF is the field of real or complex numbers, then we also have the following condition.

Definition 4: Thegroup representation of${\displaystyle (\mathbb{R},+)}$ on${\displaystyle {\mathbb{R}}^n}$ or${\displaystyle (\mathbb{C},+)}$ on${\displaystyle {\mathbb{C}}^n}$ given by${\displaystyle t\mapsto \exp (tA)}$ has no non-trivial invariant coordinate subspaces. (By comparison, this would be anirreducible representation if there were no non-trivial invariant subspaces at all, not only considering coordinate subspaces.)

A matrix isreducible if it is not irreducible.

A real matrixA isprimitive if it is non-negative and itsmth power is positive for some natural numberm (i.e. all entries ofA^m are positive).

LetA be real and non-negative. Fix an indexi and define theperiod of indexi to be thegreatest common divisor of all natural numbersm such that (A^m)_ii > 0. WhenA is irreducible, the period of every index is the same and is called theperiod ofA. In fact, whenA is irreducible, the period can be defined as the greatest common divisor of the lengths of the closed directed paths inG_A (see Kitchens^[15] page 16). The period is also called the index of imprimitivity (Meyer^[12] page 674) or the order of cyclicity. If the period is 1,A isaperiodic. It can be proved that primitive matrices are the same as irreducible aperiodic non-negative matrices.

All statements of the Perron–Frobenius theorem for positive matrices remain true for primitive matrices. The same statements also hold for a non-negative irreducible matrix, except that it may possess several eigenvalues whose absolute value is equal to its spectral radius, so the statements need to be correspondingly modified. In fact the number of such eigenvalues is equal to the period.

Results for non-negative matrices were first obtained by Frobenius in 1912.

Let${\displaystyle A}$ be an irreducible non-negative${\displaystyle N\times N}$ matrix with period${\displaystyle h}$ andspectral radius${\displaystyle \rho (A)=r}$. Then the following statements hold.

• The number${\displaystyle r\in {\mathbb{R}}^{+}}$ is a positive real number and it is an eigenvalue of the matrix${\displaystyle A}$. It is calledPerron–Frobenius eigenvalue.
• The Perron–Frobenius eigenvalue${\displaystyle r}$ issimple. Both right and left eigenspaces associated with${\displaystyle r}$ are one-dimensional.
• ${\displaystyle A}$ has both right and left eigenvectors, respectively${\displaystyle \mathbf{v}}$ and${\displaystyle \mathbf{w}}$, with eigenvalue${\displaystyle r}$ and whose components are all positive. Moreover theonly eigenvectors whose components are all positive are those associated with the eigenvalue${\displaystyle r}$.
• The matrix${\displaystyle A}$ has exactly${\displaystyle h}$ (where${\displaystyle h}$ is theperiod) complex eigenvalues with absolute value${\displaystyle r}$. Each of them is a simple root of the characteristic polynomial and is the product of${\displaystyle r}$ with an${\displaystyle h}$throot of unity.
• Let${\displaystyle \omega =2\pi /h}$. Then the matrix${\displaystyle A}$ issimilar to${\displaystyle e^{i\omega }A}$, consequently the spectrum of${\displaystyle A}$ is invariant under multiplication by${\displaystyle e^{i\omega }}$ (i.e. to rotations of the complex plane by the angle${\displaystyle \omega }$).
• If${\displaystyle h>1}$ then there exists a permutation matrix${\displaystyle P}$ such that

where${\displaystyle O}$ denotes a zero matrix and the blocks along the main diagonal are square matrices.

• Collatz–Wielandt formula: for all non-negative non-zero vectors${\displaystyle \mathbf{x}}$ let${\displaystyle f(\mathbf{x})}$ be the minimum value of${\displaystyle [A\mathbf{x}{]}_i/x_i}$ taken over all those${\displaystyle i}$ such that${\displaystyle x_i\ne 0}$. Then${\displaystyle f}$ is a real valued function whosemaximum is the Perron–Frobenius eigenvalue.

• The Perron–Frobenius eigenvalue satisfies the inequalities

${\displaystyle \underset{i}{min}\sum _j{a}_{ij}\le r\le \underset{i}{max}\sum _j{a}_{ij}.}$

The example shows that the (square) zero-matrices along the diagonal may be of different sizes, the blocksA_j need not be square, andh need not divide n.

LetA be an irreducible non-negative matrix, then:

1. (I+A)ⁿ⁻¹ is a positive matrix. (Meyer^[12]claim 8.3.5 p. 672). For a non-negativeA, this is also a sufficient condition.^[16]
2. Wielandt's theorem.^{[17][clarification needed]} If |B|<A, thenρ(B)≤ρ(A). If equality holds (i.e. ifμ=ρ(A)e^iφ is eigenvalue forB), thenB =e^iφD AD⁻¹ for some diagonal unitary matrixD (i.e. diagonal elements ofD equals toe^iΘ_l, non-diagonal are zero).^[18]
3. If some powerA^q is reducible, then it is completely reducible, i.e. for some permutation matrixP, it is true that:, whereA_i are irreducible matrices having the same maximal eigenvalue. The number of these matricesd is the greatest common divisor ofq andh, whereh is period ofA.^[19]
4. Ifc(x)= xⁿ + c_k1 x^n-k₁ + c_k2 x^n-k₂ + ... + c_ks x^n-k_s is the characteristic polynomial ofA in which only the non-zero terms are listed, then the period ofA equals the greatest common divisor ofk₁, k₂, ... , k_s.^[20]
5. Cesàroaverages:${\displaystyle \underset{k\to \mathrm{\infty }}{lim}1/k\sum _{i=0,...,k}{A}^i/r^i=(v{w}^T),}$ where the left and right eigenvectors forA are normalized so thatw^Tv = 1. Moreover, the matrixv w^T is thespectral projection corresponding tor, the Perron projection.^[21]
6. Letr be the Perron–Frobenius eigenvalue, then the adjoint matrix for (r-A) is positive.^[22]
7. IfA has at least one non-zero diagonal element, thenA is primitive.^[23]
8. If 0 ≤A <B, thenr_A ≤r_B. Moreover, ifB is irreducible, then the inequality is strict:r_A < r_B.

A matrixA is primitive provided it is non-negative andA^m is positive for somem, and henceA^k is positive for allk ≥ m. To check primitivity, one needs a bound on how large the minimal suchm can be, depending on the size ofA:^[24]

• IfA is a non-negative primitive matrix of sizen, thenA^{n2 − 2n + 2} is positive. Moreover, this is the best possible result, since for the matrixM below, the powerM^k is not positive for everyk <n² − 2n + 2, since (M^{n2 − 2n+1})_1,1 = 0.

Numerous books have been written on the subject of non-negative matrices, and Perron–Frobenius theory is invariably a central feature. The following examples given below only scratch the surface of its vast application domain.

The Perron–Frobenius theorem does not apply directly to non-negative matrices. Nevertheless, any reducible square matrixA may be written in upper-triangular block form (known as thenormal form of a reducible matrix)^[25]

PAP⁻¹ =

whereP is a permutation matrix and eachB_i is a square matrix that is either irreducible or zero. Now ifA isnon-negative then so too is each block ofPAP⁻¹, moreover the spectrum ofA is just the union of the spectra of theB_i.

The invertibility ofA can also be studied. The inverse ofPAP⁻¹ (if it exists) must have diagonal blocks of the formB_i⁻¹ so if anyB_i isn't invertible then neither isPAP⁻¹ orA.Conversely letD be the block-diagonal matrix corresponding toPAP⁻¹, in other wordsPAP⁻¹ with theasterisks zeroised. If eachB_i is invertible then so isD andD⁻¹(PAP⁻¹) is equal to theidentity plus a nilpotent matrix. But such a matrix is always invertible (ifN^k = 0 the inverse of 1 −N is1 +N +N² + ... +N^k−1) soPAP⁻¹ andA are both invertible.

Therefore, many of the spectral properties ofA may be deduced by applying the theorem to the irreducibleB_i. For example, the Perron root is the maximum of the ρ(B_i). While there will still be eigenvectors with non-negative components it is quite possiblethat none of these will be positive.

A row (column)stochastic matrix is a square matrix each of whose rows (columns) consists of non-negative real numbers whose sum is unity. The theorem cannot be applied directly to such matrices because they need not be irreducible.

IfA is row-stochastic then the column vector with each entry 1 is an eigenvector corresponding to the eigenvalue 1, which is also ρ(A) by the remark above. It might not be the only eigenvalue on the unit circle: and the associated eigenspace can be multi-dimensional. IfA is row-stochastic and irreducible then the Perron projection is also row-stochastic and all its rows are equal.

The theorem has particular use inalgebraic graph theory. The "underlying graph" of a nonnegativen-square matrix is the graph with vertices numbered 1, ...,n and arcij if and only ifA_ij ≠ 0. If the underlying graph of such a matrix is strongly connected, then the matrix is irreducible, and thus the theorem applies. In particular, theadjacency matrix of astrongly connected graph is irreducible.^[26][27]

The theorem has a natural interpretation in the theory of finiteMarkov chains (where it is the matrix-theoretic equivalent of the convergence of an irreducible finite Markov chain to its stationary distribution, formulated in terms of the transition matrix of the chain; see, for example, the article on thesubshift of finite type).

More generally, it can be extended to the case of non-negativecompact operators, which, in many ways, resemble finite-dimensional matrices. These are commonly studied in physics, under the name oftransfer operators, or sometimesRuelle–Perron–Frobenius operators (afterDavid Ruelle). In this case, the leading eigenvalue corresponds to thethermodynamic equilibrium of adynamical system, and the lesser eigenvalues to the decay modes of a system that is not in equilibrium. Thus, the theory offers a way of discovering thearrow of time in what would otherwise appear to be reversible, deterministic dynamical processes, when examined from the point of view ofpoint-set topology.^[28]

A common thread in many proofs is theBrouwer fixed point theorem. Another popular method is that of Wielandt (1950). He used theCollatz–Wielandt formula described above to extend and clarify Frobenius's work.^[29] Another proof is based on thespectral theory^[30] from which part of the arguments are borrowed.

IfA is a positive (or more generally primitive) matrix, then there exists a real positive eigenvaluer (Perron–Frobenius eigenvalue or Perron root), which is strictly greater in absolute value than all other eigenvalues, hencer is thespectral radius ofA.

This statement does not hold for general non-negative irreducible matrices, which haveh eigenvalues with the same absolute eigenvalue asr, whereh is the period ofA.

LetA be a positive matrix, assume that its spectral radius ρ(A) = 1 (otherwise considerA/ρ(A)). Hence, there exists an eigenvalue λ on the unit circle, and all the other eigenvalues are less or equal 1 in absolute value. Suppose that another eigenvalue λ ≠ 1 also falls on the unit circle. Then there exists a positive integerm such thatA^m is a positive matrix and the real part of λ^m is negative. Let ε be half the smallest diagonal entry ofA^m and setT =A^m − εI which is yet another positive matrix. Moreover, ifAx =λx thenA^mx =λ^mx thusλ^m − ε is an eigenvalue ofT. Because of the choice ofm this point lies outside the unit disk consequentlyρ(T) > 1. On the other hand, all the entries inT are positive and less than or equal to those inA^m so byGelfand's formulaρ(T) ≤ρ(A^m) ≤ρ(A)^m = 1. This contradiction means that λ=1 and there can be no other eigenvalues on the unit circle.

Absolutely the same arguments can be applied to the case of primitive matrices; we just need to mention the following simple lemma, which clarifies the properties of primitive matrices.

Given a non-negativeA, assume there existsm, such thatA^m is positive, thenA^m+1,A^m+2,A^m+3,... are all positive.

A^m+1 =AA^m, so it can have zero element only if some row ofA is entirely zero, but in this case the same row ofA^m will be zero.

Applying the same arguments as above for primitive matrices, prove the main claim.

For a positive (or more generally irreducible non-negative) matrixA the dominanteigenvector is real and strictly positive (for non-negativeA respectively non-negative.)

This can be established using thepower method, which states that for a sufficiently generic (in the sense below) matrixA the sequence of vectorsb_k+1 =Ab_k / |Ab_k | converges to theeigenvector with the maximumeigenvalue. (The initial vectorb₀ can be chosen arbitrarily except for some measure zero set). Starting with a non-negative vectorb₀ produces the sequence of non-negative vectorsb_k. Hence the limiting vector is also non-negative. By the power method this limiting vector is the dominant eigenvector forA, proving the assertion. The corresponding eigenvalue is non-negative.

The proof requires two additional arguments. First, the power method converges for matrices which do not have several eigenvalues of the same absolute value as the maximal one. The previous section's argument guarantees this.

Second, to ensure strict positivity of all of the components of the eigenvector for the case of irreducible matrices. This follows from the following fact, which is of independent interest:

Lemma: given a positive (or more generally irreducible non-negative) matrixA andv as any non-negative eigenvector forA, then it is necessarily strictly positive and the corresponding eigenvalue is also strictly positive.

Proof. One of the definitions of irreducibility for non-negative matrices is that for all indexesi,j there existsm, such that (A^m)_ij is strictly positive. Given a non-negative eigenvectorv, and that at least one of its components sayi-th is strictly positive, the corresponding eigenvalue is strictly positive, indeed, givenn such that (Aⁿ)_ii >0, hence:rⁿv_i =Aⁿv_i ≥(Aⁿ)_iiv_i>0. Hencer is strictly positive. The eigenvector is strict positivity. Then givenm, such that (A^m)_ji >0, hence:r^mv_j =(A^mv)_j ≥(A^m)_jiv_i >0, hencev_j is strictly positive, i.e., the eigenvector is strictly positive.

This section proves that the Perron–Frobenius eigenvalue is a simple root of the characteristic polynomial of the matrix. Hence the eigenspace associated to Perron–Frobenius eigenvaluer is one-dimensional. The arguments here are close to those in Meyer.^[12]

Given a strictly positive eigenvectorv corresponding tor and another eigenvectorw with the same eigenvalue. (The vectorsv andw can be chosen to be real, becauseA andr are both real, so the null space ofA-r has a basis consisting of real vectors.) Assuming at least one of the components ofw is positive (otherwise multiplyw by −1). Given maximal possibleα such thatu=v- α w is non-negative, then one of the components ofu is zero, otherwiseα is not maximum. Vectoru is an eigenvector. It is non-negative, hence by the lemma described in theprevious section non-negativity implies strict positivity for any eigenvector. On the other hand, as above at least one component ofu is zero. The contradiction implies thatw does not exist.

Case: There are no Jordan blocks corresponding to the Perron–Frobenius eigenvaluer and all other eigenvalues which have the same absolute value.

If there is a Jordan block, then theinfinity norm(A/r)^k_∞ tends to infinity fork → ∞,but that contradicts the existence of the positive eigenvector.

Givenr = 1, orA/r. Lettingv be a Perron–Frobenius strictly positive eigenvector, soAv=v, then:

${\displaystyle \Vert v{\Vert }_{\mathrm{\infty }}=\Vert A^{k}v{\Vert }_{\mathrm{\infty }}\ge \Vert A^k{\Vert }_{\mathrm{\infty }}\underset{i}{min}(v_i),\Rightarrow \Vert A^k{\Vert }_{\mathrm{\infty }}\le \Vert v\Vert /\underset{i}{min}(v_i)}$So ‖A^k‖_∞ is bounded for allk. This gives another proof that there are no eigenvalues which have greater absolute value than Perron–Frobenius one. It also contradicts the existence of the Jordan block for any eigenvalue which has absolute value equal to 1 (in particular for the Perron–Frobenius one), because existence of the Jordan block implies that ‖A^k‖_∞ is unbounded. For a two by two matrix:

hence ‖J^k‖_∞ = |k +λ| (for |λ| = 1), so it tends to infinity whenk does so. SinceJ^k =C⁻¹A^kC, thenA^k ≥J^k/ (C⁻¹C ), so it also tends to infinity. The resulting contradiction implies that there are no Jordan blocks for the corresponding eigenvalues.

Combining the two claims above reveals that the Perron–Frobenius eigenvaluer is simple root of the characteristic polynomial. In the case of nonprimitive matrices, there exist other eigenvalues which have the same absolute value asr. The same claim is true for them, but requires more work.

Given positive (or more generally irreducible non-negative matrix)A, the Perron–Frobenius eigenvector is the only (up to multiplication by constant) non-negative eigenvector forA.

Other eigenvectors must contain negative or complex components since eigenvectors for different eigenvalues are orthogonal in some sense, but two positive eigenvectors cannot be orthogonal, so they must correspond to the same eigenvalue, but the eigenspace for the Perron–Frobenius is one-dimensional.

Assuming there exists an eigenpair (λ,y) forA, such that vectory is positive, and given (r,x), wherex – is the left Perron–Frobenius eigenvector forA (i.e. eigenvector forA^T), thenrx^Ty = (x^TA)y =x^T (Ay) =λx^Ty, alsox^Ty > 0, so one has:r =λ. Since the eigenspace for the Perron–Frobenius eigenvaluer is one-dimensional, non-negative eigenvectory is a multiple of the Perron–Frobenius one.^[31]

Given a positive (or more generally irreducible non-negative matrix)A, one defines the functionf on the set of all non-negative non-zero vectorsx such thatf(x) is the minimum value of [Ax]_i / x_i taken over all thosei such thatx_i ≠ 0. Thenf is a real-valued function, whosemaximum is the Perron–Frobenius eigenvaluer.

For the proof we denote the maximum off by the valueR. The proof requires to show R = r. Inserting the Perron-Frobenius eigenvectorv intof, we obtainf(v) = r and concluder ≤ R. For the opposite inequality, we consider an arbitrary nonnegative vectorx and letξ=f(x). The definition off gives0 ≤ ξx ≤ Ax (componentwise). Now, we use the positive right eigenvectorw forA for the Perron-Frobenius eigenvaluer, then ξ w^T x = w^T ξx ≤ w^T (Ax) = (w^T A)x = r w^T x. Hencef(x) = ξ ≤ r, which impliesR ≤ r.^[32]

LetA be a positive (or more generally, primitive) matrix, and letr be its Perron–Frobenius eigenvalue.

1. There exists a limitA^k/r^k fork → ∞, denote it byP.
2. P is aprojection operator:P² =P, which commutes withA:AP =PA.
3. The image ofP is one-dimensional and spanned by the Perron–Frobenius eigenvectorv (respectively forP^T—by the Perron–Frobenius eigenvectorw forA^T).
4. P =vw^T, wherev,w are normalized such thatw^Tv = 1.
5. HenceP is a positive operator.

HenceP is aspectral projection for the Perron–Frobenius eigenvaluer, and is called the Perron projection. The above assertion is not true for general non-negative irreducible matrices.

Actually the claims above (except claim 5) are valid for any matrixM such that there exists an eigenvaluer which is strictly greater than the other eigenvalues in absolute value and is the simple root of the characteristicpolynomial. (These requirements hold for primitive matrices as above).

Given thatM is diagonalizable,M is conjugate to a diagonal matrix with eigenvaluesr₁, ... ,r_n on the diagonal (denoter₁ =r). The matrixM^k/r^k will be conjugate (1, (r₂/r)^k, ... , (r_n/r)^k), which tends to (1,0,0,...,0), fork → ∞, so the limit exists. The same method works for generalM (without assuming thatM is diagonalizable).

The projection and commutativity properties are elementary corollaries of the definition:MM^k/r^k =M^k/r^kM ;P² = limM^2k/r^2k =P. The third fact is also elementary:M(Pu) =M limM^k/r^ku = limrM^k+1/r^k+1u, so taking the limit yieldsM(Pu) =r(Pu), so image ofP lies in ther-eigenspace forM, which is one-dimensional by the assumptions.

Denoting byv,r-eigenvector forM (byw forM^T). Columns ofP are multiples ofv, because the image ofP is spanned by it. Respectively, rows ofw. SoP takes a form(a v w^T), for somea. Hence its trace equals to(a w^T v). Trace of projector equals the dimension of its image. It was proved before that it is not more than one-dimensional. From the definition one sees thatP acts identically on ther-eigenvector forM. So it is one-dimensional. So choosing (w^Tv) = 1, impliesP =vw^T.

For any non-negative matrixA its Perron–Frobenius eigenvaluer satisfies the inequality:

${\displaystyle r\le \underset{i}{max}\sum _j{a}_{ij}.}$

This is not specific to non-negative matrices: for any matrixA with an eigenvalue${\displaystyle \lambda }$ it is truethat${\displaystyle |\lambda |\le \underset{i}{max}\sum _j|a_{ij}|}$. This is an immediate corollary of theGershgorin circle theorem. However another proof is more direct:

Anymatrix induced norm satisfies the inequality${\displaystyle \Vert A\Vert \ge |\lambda |}$ for any eigenvalue${\displaystyle \lambda }$ because, if${\displaystyle x}$ is a corresponding eigenvector,${\displaystyle \Vert A\Vert \ge |Ax|/|x|=|\lambda x|/|x|=|\lambda |}$. Theinfinity norm of a matrix is the maximum of row sums:${\displaystyle {\Vert A\Vert }_{\mathrm{\infty }}=\underset{1\le i\le m}{max}\sum _{j=1}^n|a_{ij}|.}$ Hence the desired inequality is exactly${\displaystyle \Vert A{\Vert }_{\mathrm{\infty }}\ge |\lambda |}$ applied to the non-negative matrixA.

Another inequality is:

${\displaystyle \underset{i}{min}\sum _j{a}_{ij}\le r.}$

This fact is specific to non-negative matrices; for general matrices there is nothing similar. Given thatA is positive (not just non-negative), then there exists a positive eigenvectorw such thatAw =rw and the smallest component ofw (sayw_i) is 1. Thenr = (Aw)_i ≥ the sum of the numbers in rowi ofA. Thus the minimum row sum gives a lower bound forr and this observation can be extended to all non-negative matrices by continuity.

Another way to argue it is via theCollatz-Wielandt formula. One takes the vectorx = (1, 1, ..., 1) and immediately obtains the inequality.

The proof now proceeds usingspectral decomposition. The trick here is to split the Perron root from the other eigenvalues. The spectral projection associated with the Perron root is called the Perron projection and it enjoys the following property:

The Perron projection of an irreducible non-negative square matrix is a positive matrix.

Perron's findings and also (1)–(5) of the theorem are corollaries of this result. The key point is that a positive projection always has rank one. This means that ifA is an irreducible non-negative square matrix then the algebraic and geometric multiplicities of its Perron root are both one. Also ifP is its Perron projection thenAP =PA = ρ(A)P so every column ofP is a positive right eigenvector ofA and every row is a positive left eigenvector. Moreover, ifAx = λx thenPAx = λPx = ρ(A)Px which meansPx = 0 if λ ≠ ρ(A). Thus the only positive eigenvectors are those associated with ρ(A). IfA is a primitive matrix with ρ(A) = 1 then it can be decomposed asP ⊕ (1 − P)A so thatAⁿ =P + (1 − P)Aⁿ. Asn increases the second of these terms decays to zero leavingP as the limit ofAⁿ asn → ∞.

The power method is a convenient way to compute the Perron projection of a primitive matrix. Ifv andw are the positive row and column vectors that it generates then the Perron projection is justwv/vw. The spectral projections aren't neatly blocked as in the Jordan form. Here they are overlaid and each generally has complex entries extending to all four corners of the square matrix. Nevertheless, they retain their mutual orthogonality which is what facilitates the decomposition.

The analysis whenA is irreducible and non-negative is broadly similar. The Perron projection is still positive but there may now be other eigenvalues of modulus ρ(A) that negate use of the power method and prevent the powers of (1 − P)A decaying as in the primitive case whenever ρ(A) = 1. So we consider theperipheral projection, which is the spectral projection ofA corresponding to all the eigenvalues that have modulusρ(A). It may then be shown that the peripheral projection of an irreducible non-negative square matrix is a non-negative matrix with a positive diagonal.

Suppose in addition that ρ(A) = 1 andA hash eigenvalues on the unit circle. IfP is the peripheral projection then the matrixR =AP =PA is non-negative and irreducible,R^h =P, and the cyclic groupP,R,R², ....,R^h−1 represents the harmonics ofA. The spectral projection ofA at the eigenvalue λ on the unit circle is given by the formula${\displaystyle h^{-1}\sum _1^h{\lambda }^{-k}{R}^k}$. All of these projections (including the Perron projection) have the same positive diagonal, moreover choosing any one of them and then taking the modulus of every entry invariably yields the Perron projection. Some donkey work is still needed in order to establish the cyclic properties (6)–(8) but it's essentially just a matter of turning the handle. The spectral decomposition ofA is given byA = R ⊕ (1 − P)A so the difference betweenAⁿ andRⁿ isAⁿ − Rⁿ = (1 − P)Aⁿ representing the transients ofAⁿ which eventually decay to zero.P may be computed as the limit ofA^nh asn → ∞.

The matricesL =,P =,T =,M = provide simple examples of what can go wrong if the necessary conditions are not met. It is easily seen that the Perron and peripheral projections ofL are both equal toP, thus when the original matrix is reducible the projections may lose non-negativity and there is no chance of expressing them as limits of its powers. The matrixT is an example of a primitive matrix with zero diagonal. If the diagonal of an irreducible non-negative square matrix is non-zero then the matrix must be primitive but this example demonstrates that the converse is false.M is an example of a matrix with several missing spectral teeth. If ω = e^iπ/3 then ω⁶ = 1 and the eigenvalues ofM are {1,ω²,ω³=-1,ω⁴} with a dimension 2 eigenspace for +1 so ω and ω⁵ are both absent. More precisely, sinceM is block-diagonal cyclic, then the eigenvalues are {1,-1} for the first block, and {1,ω²,ω⁴} for the lower one^{[citation needed]}

A problem that causes confusion is a lack of standardisation in the definitions. For example, some authors use the termsstrictly positive andpositive to mean > 0 and ≥ 0 respectively. In this articlepositive means > 0 andnon-negative means ≥ 0. Another vexed area concernsdecomposability andreducibility:irreducible is an overloaded term. For avoidance of doubt a non-zero non-negative square matrixA such that 1 + A is primitive is sometimes said to beconnected. Then irreducible non-negative square matrices and connected matrices are synonymous.^[33]

The nonnegative eigenvector is often normalized so that the sum of its components is equal to unity; in this case, the eigenvector is the vector of aprobability distribution and is sometimes called astochastic eigenvector.

Perron–Frobenius eigenvalue anddominant eigenvalue are alternative names for the Perron root. Spectral projections are also known asspectral projectors andspectral idempotents. The period is sometimes referred to as theindex of imprimitivity or theorder of cyclicity.

• Min-max theorem – Variational characterization of eigenvalues of compact Hermitian operators on Hilbert spaces
• Z-matrix (mathematics) – Square matrix whose off-diagonal entries are nonpositive
• M-matrix – Matrix in mathematics
• P-matrix – Complex square matrix for which every principal minor is positive
• Routh–Hurwitz matrix – Matrix used to analyze the stability of a polynomial by its coefficients
• Metzler matrix (Quasipositive matrix)
• Positive operator – In mathematics, a linear operator acting on inner product space
• Krein–Rutman theorem – A generalization of the Perron–Frobenius theorem to Banach spaces

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• Abraham Berman,Robert J. Plemmons,Nonnegative Matrices in the Mathematical Sciences, 1994, SIAM.ISBN 0-89871-321-8.
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• Varga, Richard S. (2002),Matrix Iterative Analysis (2nd ed.), Springer-Verlag.

• Matrix theory
• Theorems in linear algebra
• Markov processes

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