[proofplan] We first prove the Perron eigenvector statement by perturbing $M$ to a strictly positive matrix and applying Brouwer's fixed point theorem on the standard simplex. Passing to the limit as the perturbation vanishes gives a non-negative eigenvector for $M$, and irreducibility upgrades it to a strictly positive eigenvector. A coordinatewise comparison argument proves that the corresponding eigenvalue is the spectral radius and that positive right and left eigenvectors are unique up to scale. In the aperiodic case, primitivity removes all non-Perron peripheral spectrum, and the finite-dimensional spectral decomposition gives convergence to the Perron projection $rl^\top$. [/proofplan]

[step:Fix the irreducibility and aperiodicity conventions]Throughout the proof, irreducibility is used in the matrix-power form: for every $i,j \in \{1,\dots,k\}$ there exists $n \in \mathbb{N}$ such that $(M^n)_{ij}>0$. Aperiodicity means that the common period \begin{align*} d:=\gcd\{n \in \mathbb{N}: (M^n)_{ii}>0\} \end{align*} is equal to $1$; for an irreducible non-negative matrix this integer is independent of the chosen index $i$.[/step]

[guided]We record the conventions used in the proof before beginning the construction. The matrix-power formulation of irreducibility says that for every ordered pair of indices $i,j \in \{1,\dots,k\}$ there is an exponent $n \in \mathbb{N}$ for which $(M^n)_{ij}>0$. The aperiodicity condition is expressed through the common period \begin{align*} d:=\gcd\{n \in \mathbb{N}: (M^n)_{ii}>0\}. \end{align*} For an irreducible non-negative matrix, this greatest common divisor is independent of the index $i$. Thus the aperiodic case is exactly the case $d=1$.[/guided]

[step:Produce positive eigenvectors for strictly positive matrices by normalising on the simplex]Let $\Delta \subset \mathbb{R}^k$ denote the standard simplex \begin{align*} \Delta := \{x \in [0,\infty)^k : x_1 + \cdots + x_k = 1\}. \end{align*} Let $A \in (0,\infty)^{k \times k}$ be a strictly positive matrix. Define the map $F_A:\Delta \to \Delta$ by \begin{align*} F_A(x) := \frac{Ax}{\sum_{i=1}^k (Ax)_i}. \end{align*} For every $x \in \Delta$, the vector $Ax$ belongs to $(0,\infty)^k$, so the denominator is positive and $F_A(x)$ is well-defined. The map $F_A$ is continuous, and $\Delta$ is a compact convex subset of $\mathbb{R}^k$. By the [Generalized Brouwer Fixed Point Theorem](/theorems/81), there exists $r_A \in \Delta$ such that $F_A(r_A)=r_A$. Define \begin{align*} \lambda_A := \sum_{i=1}^k (Ar_A)_i. \end{align*} Then $\lambda_A>0$ and $Ar_A=\lambda_A r_A$. Since $Ar_A \in (0,\infty)^k$, the equality $Ar_A=\lambda_A r_A$ implies $r_A \in (0,\infty)^k$. Applying the same argument to $A^\top$ gives a vector $l_A \in (0,\infty)^k$ and a number $\mu_A>0$ such that \begin{align*} A^\top l_A = \mu_A l_A. \end{align*} Equivalently, $l_A^\top A=\mu_A l_A^\top$.[/step]

[guided]The simplex is used because it gives a compact normalisation for non-negative vectors. We define \begin{align*} \Delta := \{x \in [0,\infty)^k : x_1 + \cdots + x_k = 1\}. \end{align*} If $A \in (0,\infty)^{k \times k}$ and $x \in \Delta$, then at least one coordinate of $x$ is positive and every entry of $A$ is positive. Therefore every coordinate of $Ax$ is positive. In particular \begin{align*} \sum_{i=1}^k (Ax)_i > 0. \end{align*} Thus the normalised map \begin{align*} F_A:\Delta \to \Delta, \qquad F_A(x) := \frac{Ax}{\sum_{i=1}^k (Ax)_i} \end{align*} is well-defined. It is continuous because it is obtained from the linear map $x \mapsto Ax$ by dividing by a positive continuous scalar function. The set $\Delta$ is compact and convex in the finite-dimensional Euclidean space $\mathbb{R}^k$. The [Generalized Brouwer Fixed Point Theorem](/theorems/81) therefore applies to $F_A$ and gives a point $r_A \in \Delta$ with $F_A(r_A)=r_A$. Define \begin{align*} \lambda_A := \sum_{i=1}^k (Ar_A)_i. \end{align*} The fixed point equation is exactly \begin{align*} \frac{Ar_A}{\lambda_A}=r_A, \end{align*} so \begin{align*} Ar_A=\lambda_A r_A. \end{align*} Because $Ar_A$ has strictly positive coordinates and $\lambda_A>0$, the vector $r_A$ also has strictly positive coordinates. The same construction applied to the transpose matrix $A^\top$ gives $l_A \in (0,\infty)^k$ and $\mu_A>0$ with \begin{align*} A^\top l_A=\mu_A l_A. \end{align*} Taking transposes of this equation gives the left eigenvector equation \begin{align*} l_A^\top A=\mu_A l_A^\top. \end{align*}[/guided]

[step:Pass from positive perturbations to an eigenvector of the irreducible nonnegative matrix]Let $J \in \mathbb{R}^{k \times k}$ be the matrix whose entries are all equal to $1$. For each $\varepsilon>0$, define \begin{align*} M_\varepsilon := M+\varepsilon J. \end{align*} Then $M_\varepsilon \in (0,\infty)^{k \times k}$. By the previous step, there exist $r_\varepsilon \in \Delta \cap (0,\infty)^k$ and $\lambda_\varepsilon>0$ such that \begin{align*} M_\varepsilon r_\varepsilon=\lambda_\varepsilon r_\varepsilon. \end{align*} The sequence $(r_\varepsilon)$ lies in the compact set $\Delta$ as $\varepsilon \downarrow 0$. Also, since $r_\varepsilon \in \Delta$ and each column of $J$ has sum $k$, \begin{align*} 0<\lambda_\varepsilon=\sum_{i=1}^k (M_\varepsilon r_\varepsilon)_i \leq \max_{j \in \{1,\dots,k\}} \sum_{i=1}^k M_{ij}+\varepsilon k. \end{align*} Indeed, the column-sum functional satisfies \begin{align*} \sum_{i=1}^k (M_\varepsilon r_\varepsilon)_i=\sum_{j=1}^k r_{\varepsilon,j}\sum_{i=1}^k (M_{ij}+\varepsilon J_{ij}), \end{align*} and this is a convex combination of the column sums of $M_\varepsilon$. Hence, along a sequence $\varepsilon_m \downarrow 0$, we may assume that $r_{\varepsilon_m}\to r$ in $\mathbb{R}^k$ and $\lambda_{\varepsilon_m}\to \lambda$ in $\mathbb{R}$, where $r \in \Delta$ and $\lambda \geq 0$. Passing to the limit in \begin{align*} M_{\varepsilon_m}r_{\varepsilon_m}=\lambda_{\varepsilon_m}r_{\varepsilon_m} \end{align*} gives \begin{align*} Mr=\lambda r. \end{align*} We prove that $r \in (0,\infty)^k$. Let \begin{align*} S:=\{j \in \{1,\dots,k\}: r_j>0\}. \end{align*} Since $r \in \Delta$, the set $S$ is non-empty. If $i \notin S$, then $r_i=0$, and the $i$th coordinate of $Mr=\lambda r$ gives \begin{align*} \sum_{j=1}^k M_{ij}r_j=0. \end{align*} Every summand is non-negative, so $M_{ij}=0$ for every $j \in S$. It follows by induction on $n$ that \begin{align*} (M^n)_{ij}=0 \end{align*} for every $n \in \mathbb{N}$, every $i \notin S$, and every $j \in S$. This contradicts irreducibility unless $S=\{1,\dots,k\}$. Therefore $r \in (0,\infty)^k$. Since $M$ is irreducible, each row of $M$ has at least one positive entry. With $r \in (0,\infty)^k$, this gives $Mr \in (0,\infty)^k$. From $Mr=\lambda r$ we obtain $\lambda>0$. Applying the same argument to $M^\top$, which is irreducible whenever $M$ is irreducible, gives $l \in (0,\infty)^k$ and $\mu>0$ such that \begin{align*} M^\top l=\mu l. \end{align*} Equivalently, \begin{align*} l^\top M=\mu l^\top. \end{align*}[/step]

[guided]The perturbation step lets us use the strictly positive case and then remove the perturbation. Let $J$ be the $k\times k$ matrix whose entries are all $1$, and for $\varepsilon>0$ set \begin{align*} M_\varepsilon:=M+\varepsilon J. \end{align*} Every entry of $M_\varepsilon$ is strictly positive. The previous step therefore gives $r_\varepsilon\in\Delta\cap(0,\infty)^k$ and $\lambda_\varepsilon>0$ with \begin{align*} M_\varepsilon r_\varepsilon=\lambda_\varepsilon r_\varepsilon. \end{align*} The vectors $r_\varepsilon$ lie in the compact simplex $\Delta$. The scalars $\lambda_\varepsilon$ are bounded above because \begin{align*} \lambda_\varepsilon=\sum_{i=1}^k(M_\varepsilon r_\varepsilon)_i =\sum_{j=1}^k r_{\varepsilon,j}\sum_{i=1}^k(M_{ij}+\varepsilon J_{ij}), \end{align*} which is a convex combination of the column sums of $M_\varepsilon$. Hence along a sequence $\varepsilon_m\downarrow 0$ we have $r_{\varepsilon_m}\to r\in\Delta$ and $\lambda_{\varepsilon_m}\to\lambda\geq 0$. Passing to the limit in \begin{align*} M_{\varepsilon_m}r_{\varepsilon_m}=\lambda_{\varepsilon_m}r_{\varepsilon_m} \end{align*} gives \begin{align*} Mr=\lambda r. \end{align*} It remains to show that $r$ has no zero coordinate. Let \begin{align*} S:=\{j:r_j>0\}. \end{align*} The set $S$ is non-empty because $r\in\Delta$. If $i\notin S$, then the $i$th coordinate of $Mr=\lambda r$ gives \begin{align*} \sum_{j=1}^kM_{ij}r_j=0. \end{align*} All terms are non-negative, so $M_{ij}=0$ for every $j\in S$. Inductively, $(M^n)_{ij}=0$ for every $n\in\mathbb N$, every $i\notin S$, and every $j\in S$, contradicting irreducibility unless $S=\{1,\dots,k\}$. Thus $r\in(0,\infty)^k$. Since each row of an irreducible non-negative matrix has a positive entry, $Mr\in(0,\infty)^k$, and $Mr=\lambda r$ forces $\lambda>0$. Applying the same perturbation argument to $M^\top$ gives $l\in(0,\infty)^k$ and $\mu>0$ with \begin{align*} M^\top l=\mu l, \end{align*} equivalently \begin{align*} l^\top M=\mu l^\top. \end{align*}[/guided]

[step:Compare eigenvalues against the positive eigenvector]Let $\alpha \in \mathbb{C}$ be an eigenvalue of $M$, and let $v \in \mathbb{C}^k \setminus \{0\}$ be an eigenvector with $Mv=\alpha v$. Define \begin{align*} c:=\max_{i \in \{1,\dots,k\}} \frac{|v_i|}{r_i}. \end{align*} Since $r \in (0,\infty)^k$ and $v\neq 0$, we have $c>0$. Choose $p \in \{1,\dots,k\}$ such that $|v_p|=c r_p$. Taking absolute values in the $p$th coordinate of $Mv=\alpha v$ and using the triangle inequality gives \begin{align*} |\alpha|c r_p=|\alpha v_p|=|(Mv)_p|\leq \sum_{j=1}^k M_{pj}|v_j|. \end{align*} By the definition of $c$, one has $|v_j|\leq c r_j$ for every $j$. Therefore \begin{align*} \sum_{j=1}^k M_{pj}|v_j|\leq c\sum_{j=1}^k M_{pj}r_j=c(Mr)_p=c\lambda r_p. \end{align*} Since $c r_p>0$, we conclude that $|\alpha|\leq \lambda$. Because $\lambda$ itself is an eigenvalue of $M$, this proves that the spectral radius of $M$ is $\lambda$. Applying the same comparison to the positive left eigenvector $l$, equivalently to the positive right eigenvector $l$ of $M^\top$, shows that the left eigenvalue $\mu$ found above must also equal the spectral radius. Hence $\mu=\lambda$, and \begin{align*} l^\top M=\lambda l^\top. \end{align*}[/step]

[guided]Let $\alpha$ be any complex eigenvalue of $M$, and choose a non-zero eigenvector $v\in\mathbb C^k$ with \begin{align*} Mv=\alpha v. \end{align*} Because every coordinate of $r$ is strictly positive, the comparison constant \begin{align*} c:=\max_{i\in\{1,\dots,k\}}\frac{|v_i|}{r_i} \end{align*} is finite and positive. Choose $p$ with $|v_p|=cr_p$. Looking at the $p$th coordinate and using the triangle inequality gives \begin{align*} |\alpha|cr_p=|\alpha v_p|=|(Mv)_p|\leq\sum_{j=1}^kM_{pj}|v_j|. \end{align*} Since $|v_j|\leq cr_j$ for every $j$, \begin{align*} \sum_{j=1}^kM_{pj}|v_j|\leq c\sum_{j=1}^kM_{pj}r_j=c(Mr)_p=c\lambda r_p. \end{align*} Dividing by $cr_p>0$ gives $|\alpha|\leq\lambda$. Since $\lambda$ is itself an eigenvalue, it is the spectral radius. The same reasoning applied to $M^\top$ shows that the positive left eigenvalue $\mu$ also equals the spectral radius. Hence $\mu=\lambda$, and the left eigenvector equation is \begin{align*} l^\top M=\lambda l^\top. \end{align*}[/guided]

[step:Prove uniqueness of positive right and left eigenvectors]Let $u \in (0,\infty)^k$ and $\nu>0$ satisfy $Mu=\nu u$. Applying the comparison argument from the previous step with the positive eigenvector $r$ gives $\nu\leq \lambda$, and applying it with the positive eigenvector $u$ gives $\lambda\leq \nu$. Hence $\nu=\lambda$. Define \begin{align*} t:=\max_{i \in \{1,\dots,k\}} \frac{u_i}{r_i}, \qquad w:=tr-u. \end{align*} Then $t>0$, $w \in [0,\infty)^k$, and at least one coordinate of $w$ is zero. Since $Mr=\lambda r$ and $Mu=\lambda u$, we have \begin{align*} Mw=\lambda w. \end{align*} Choose $p \in \{1,\dots,k\}$ with $w_p=0$. For every $n \in \mathbb{N}$, \begin{align*} M^n w=\lambda^n w. \end{align*} Taking the $p$th coordinate gives \begin{align*} \sum_{j=1}^k (M^n)_{pj}w_j=0. \end{align*} For each $j$, irreducibility gives an integer $n_j \in \mathbb{N}$ such that $(M^{n_j})_{pj}>0$. Since all terms in the displayed sum are non-negative, the equality with $n=n_j$ forces $w_j=0$. Thus $w=0$, so $u=tr$. Therefore every positive right eigenvector is a positive scalar multiple of $r$. Applying the same argument to $M^\top$ proves that every positive left eigenvector of $M$ corresponding to $\lambda$ is a positive scalar multiple of $l$.[/step]

[guided]Suppose $u\in(0,\infty)^k$ and $\nu>0$ satisfy \begin{align*} Mu=\nu u. \end{align*} The spectral-radius comparison just proved, first using $r$ and then using $u$, gives $\nu\leq\lambda$ and $\lambda\leq\nu$. Hence $\nu=\lambda$. To compare $u$ with $r$, set \begin{align*} t:=\max_{i\in\{1,\dots,k\}}\frac{u_i}{r_i} \end{align*} and \begin{align*} w:=tr-u. \end{align*} Then $w\in[0,\infty)^k$ and at least one coordinate of $w$ is zero. Since both $r$ and $u$ are $\lambda$-eigenvectors, \begin{align*} Mw=\lambda w. \end{align*} Choose $p$ with $w_p=0$. For every $n\in\mathbb N$, \begin{align*} M^nw=\lambda^n w. \end{align*} The $p$th coordinate is therefore \begin{align*} \sum_{j=1}^k(M^n)_{pj}w_j=0. \end{align*} For each $j$, irreducibility gives $n_j$ with $(M^{n_j})_{pj}>0$. Since every term is non-negative, the equality with $n=n_j$ forces $w_j=0$. Thus $w=0$ and $u=tr$. This proves uniqueness of the positive right eigenvector up to positive scalar. Applying the same argument to $M^\top$ proves the corresponding uniqueness for positive left eigenvectors.[/guided]

[step:Use aperiodicity to remove the non-Perron peripheral spectrum]Assume now that $M$ is aperiodic in the matrix-power sense stated above. We prove the required primitivity criterion from the directed graph of positive entries of $M$. Fix a vertex $i$. Since the gcd of the full set $\{n\in\mathbb N:(M^n)_{ii}>0\}$ is $1$, some finite subset already has gcd $1$; choose return lengths $a_1,\dots,a_s \in \mathbb N$ with $(M^{a_t})_{ii}>0$ for every $t$ and \begin{align*} \gcd(a_1,\dots,a_s)=1. \end{align*} The additive semigroup generated by $a_1,\dots,a_s$ contains all sufficiently large integers: indeed, modulo $a_1$, the residues of this semigroup form the subgroup generated by $a_1,\dots,a_s$, which is all of $\mathbb Z/a_1\mathbb Z$; choosing one representative $m_c$ in the semigroup for each residue class $c$ shows that every integer larger than $\max_c m_c$ is $m_c+q a_1$ for some $q\geq 0$. Hence there is $N_0\in\mathbb N$ such that \begin{align*} (M^n)_{ii}>0 \end{align*} for every $n\geq N_0$, by concatenating return loops at $i$. For arbitrary vertices $p,q$, irreducibility gives path lengths $u_p,v_q\in\mathbb N$ such that \begin{align*} (M^{u_p})_{pi}>0 \end{align*} and \begin{align*} (M^{v_q})_{iq}>0. \end{align*} If $n\geq N_0+u_p+v_q$, then the product of a path from $p$ to $i$, a return loop at $i$ of length $n-u_p-v_q$, and a path from $i$ to $q$ gives \begin{align*} (M^n)_{pq}>0. \end{align*} Taking $N$ larger than all finitely many thresholds $N_0+u_p+v_q$ gives a single exponent for which every entry is positive. Thus there exists $N \in \mathbb{N}$ such that \begin{align*} M^N \in (0,\infty)^{k \times k}. \end{align*} Let $\alpha \in \mathbb{C}$ be an eigenvalue of $M$ with $|\alpha|=\lambda$, and let $v \in \mathbb{C}^k \setminus \{0\}$ satisfy $Mv=\alpha v$. Then \begin{align*} M^N v=\alpha^N v. \end{align*} Define \begin{align*} c:=\max_{i \in \{1,\dots,k\}}\frac{|v_i|}{r_i}. \end{align*} Choose $p \in \{1,\dots,k\}$ with $|v_p|=cr_p$. Applying the comparison estimate to the strictly positive matrix $A:=M^N$ gives \begin{align*} \lambda^Ncr_p=|\alpha|^N|v_p|=|(Av)_p|\leq \sum_{j=1}^k A_{pj}|v_j|\leq c\sum_{j=1}^k A_{pj}r_j=c\lambda^Nr_p. \end{align*} Thus equality holds in both inequalities. Since $A_{pj}>0$ for every $j$, equality in the [second inequality](/theorems/2136) forces $|v_j|=cr_j$ for every $j$. Equality in the triangle inequality forces the non-zero complex numbers $A_{pj}v_j$ to have a common argument, so all $v_j/r_j$ have the same argument. Hence $v=\zeta c r$ for some $\zeta \in \mathbb{C}$ with $|\zeta|=1$ and some $c>0$. Substituting into $Mv=\alpha v$ and using $Mr=\lambda r$ gives \begin{align*} \zeta c\lambda r=\alpha \zeta c r. \end{align*} Since $\zeta c r\neq 0$, this implies $\alpha=\lambda$. Thus $\lambda$ is the only eigenvalue of $M$ on the circle $\{z \in \mathbb{C}: |z|=\lambda\}$. The eigenspace of $M$ for $\lambda$ is one-dimensional by the uniqueness of the positive right eigenvector and the preceding peripheral-spectrum argument. Moreover, $\lambda$ has no non-trivial Jordan chain. Indeed, if a non-zero generalized chain existed, then because the eigenspace is $\operatorname{span}\{r\}$, there would be $u \in \mathbb{C}^k$ and $a \in \mathbb{C}\setminus\{0\}$ such that \begin{align*} (M-\lambda I)u=ar. \end{align*} Replacing $u$ by $a^{-1}u$ gives a vector satisfying \begin{align*} (M-\lambda I)u=r. \end{align*} Multiplying on the left by $l^\top$ would then give \begin{align*} 0=l^\top(M-\lambda I)u=l^\top r, \end{align*} contradicting $l^\top r>0$. Hence $\lambda$ is algebraically simple.[/step]

[guided]Aperiodicity is used exactly to turn irreducibility into strict positivity after taking a power. Here aperiodicity means that the period $\gcd\{n \in \mathbb{N}: (M^n)_{ii}>0\}$ is $1$, independent of $i$ because $M$ is irreducible. We verify the needed graph-theoretic criterion directly. Fix a vertex $i$. Since the gcd of all return lengths at $i$ is $1$, a finite subset of return lengths already has gcd $1$. Thus there are return lengths $a_1,\dots,a_s$ with \begin{align*} (M^{a_t})_{ii}>0 \end{align*} for each $t$, and with \begin{align*} \gcd(a_1,\dots,a_s)=1. \end{align*} The additive semigroup generated by these lengths contains every sufficiently large integer. To see this, work modulo $a_1$: the generated residues form all of $\mathbb Z/a_1\mathbb Z$, so choose one generated representative $m_c$ for each residue class $c$. If $n$ is large and has residue $c$, then $n=m_c+qa_1$ for some $q\geq 0$, so $n$ is also generated by $a_1,\dots,a_s$. Concatenating the corresponding return loops gives \begin{align*} (M^n)_{ii}>0 \end{align*} for all sufficiently large $n$. Now take arbitrary vertices $p$ and $q$. Irreducibility gives a path from $p$ to $i$ and a path from $i$ to $q$, so for some $u_p,v_q\in\mathbb N$, \begin{align*} (M^{u_p})_{pi}>0 \end{align*} and \begin{align*} (M^{v_q})_{iq}>0. \end{align*} For all large $n$, the middle length $n-u_p-v_q$ is an allowed return length at $i$, and therefore concatenating the three paths gives \begin{align*} (M^n)_{pq}>0. \end{align*} There are only finitely many pairs $(p,q)$, so one exponent $N$ works for all of them. Thus $M^N$ has every entry strictly positive. Set $A:=M^N$. Since $Mr=\lambda r$, we have $Ar=\lambda^N r$. If $Mv=\alpha v$ and $|\alpha|=\lambda$, then $Av=\alpha^N v$. Define \begin{align*} c:=\max_{i \in \{1,\dots,k\}}\frac{|v_i|}{r_i}. \end{align*} Because $r_i>0$ for every $i$ and $v\neq 0$, this number satisfies $c>0$. Choose $p$ with $|v_p|=cr_p$. The comparison estimate at the $p$th coordinate gives \begin{align*} \lambda^Ncr_p=|\alpha|^N|v_p|=|(Av)_p|\leq \sum_{j=1}^k A_{pj}|v_j|\leq c\sum_{j=1}^k A_{pj}r_j=c\lambda^Nr_p. \end{align*} The left and right endpoints are equal, so both inequalities are equalities. The [second inequality](/theorems/2899) was obtained from $|v_j|\leq cr_j$. Since every coefficient $A_{pj}$ is strictly positive, equality in the weighted sum forces $|v_j|=cr_j$ for every $j$. The [first inequality](/theorems/2897) was the triangle inequality for the complex sum $\sum_j A_{pj}v_j$. Equality in the triangle inequality forces all non-zero summands $A_{pj}v_j$ to have the same argument. Since $A_{pj}>0$, this means all ratios $v_j/r_j$ have the same argument. Therefore $v=\zeta c r$ for some complex number $\zeta$ with $|\zeta|=1$. Substituting this form into $Mv=\alpha v$ gives \begin{align*} \zeta c\lambda r=\alpha \zeta c r. \end{align*} The vector $\zeta c r$ is non-zero, so $\alpha=\lambda$. This proves that no eigenvalue other than $\lambda$ lies on the spectral circle. Finally, the eigenspace for $\lambda$ is one-dimensional. Indeed, any $\lambda$-eigenvector has modulus dominated by the comparison argument above, and the equality case forces it to be a scalar multiple of $r$. There is also no non-trivial Jordan chain at $\lambda$. If such a chain existed, then for some $u\in\mathbb C^k$ one would have \begin{align*} (M-\lambda I)u=r. \end{align*} Multiplying by the positive left eigenvector gives \begin{align*} 0=l^\top(M-\lambda I)u=l^\top r, \end{align*} which contradicts $l^\top r>0$. Thus $\lambda$ is algebraically simple.[/guided]

[step:Decompose along the Perron projection and take the limit]Normalize $l$ by multiplying it by a positive scalar so that \begin{align*} l^\top r=1. \end{align*} Define the rank-one [linear map](/page/Linear%20Map) $P:\mathbb{R}^k \to \mathbb{R}^k$ by \begin{align*} P x := r(l^\top x). \end{align*} Its matrix is $rl^\top$. Since $r,l \in (0,\infty)^k$, every entry of $rl^\top$ is positive. Also $P^2=P$, because $l^\top r=1$. Let \begin{align*} H:=\{x \in \mathbb{R}^k : l^\top x=0\}. \end{align*} Then $\mathbb{R}^k=\operatorname{span}\{r\}\oplus H$. The subspace $H$ is invariant under $M$, since for $x \in H$, \begin{align*} l^\top(Mx)=(l^\top M)x=\lambda l^\top x=0. \end{align*} On $\operatorname{span}\{r\}$, the matrix $M$ acts as multiplication by $\lambda$. Define $M|_H:H\to H$ to be the restriction of $M$ to the invariant subspace $H$. Every complex eigenvalue of $M|_H$ has modulus strictly smaller than $\lambda$, by the previous step. Indeed, no eigenvalue can have modulus larger than $\lambda$, no eigenvalue other than $\lambda$ can have modulus $\lambda$, and $\lambda$ itself cannot occur in $H$: if $x\in H$ and $Mx=\lambda x$, then the one-dimensionality of the $\lambda$-eigenspace gives $x=cr$, and $0=l^\top x=cl^\top r=c$ forces $x=0$. We complexify the invariant real subspace by setting $H_{\mathbb{C}}:=H\otimes_{\mathbb{R}}\mathbb{C}$ and extend $M|_H$ complex-linearly to $H_{\mathbb{C}}$. By the [Jordan Normal Form](/theorems/864) applied to this complexified operator, each Jordan block of $\lambda^{-1}M|_{H_{\mathbb{C}}}$ has eigenvalue of modulus strictly less than $1$. Powers of such a block converge to zero because the polynomial growth from the nilpotent part is dominated by the exponential decay of the eigenvalue modulus. Restricting the resulting convergence back to the real subspace $H$ gives \begin{align*} \lambda^{-n}(M|_H)^n \to 0 \end{align*} as $n \to \infty$. Therefore, for every $x \in \mathbb{R}^k$, writing \begin{align*} x=P x+(x-Px) \end{align*} with $Px \in \operatorname{span}\{r\}$ and $x-Px \in H$, we obtain \begin{align*} \lambda^{-n}M^n x=Px+\lambda^{-n}M^n(x-Px)\to Px. \end{align*} Thus $\lambda^{-n}M^n$ converges in $\mathbb{R}^{k \times k}$ to the matrix of $P$, namely $rl^\top$. This proves the asserted convergence to a positive rank-one matrix.[/step]

[guided]Normalize the left eigenvector so that \begin{align*} l^\top r=1. \end{align*} Define the projection \begin{align*} P:\mathbb R^k\to\mathbb R^k \end{align*} by \begin{align*} Px:=r(l^\top x). \end{align*} The matrix of $P$ is $rl^\top$, every entry of $rl^\top$ is positive, and $P^2=P$ because $l^\top r=1$. Let \begin{align*} H:=\{x\in\mathbb R^k:l^\top x=0\}. \end{align*} Every vector decomposes uniquely as \begin{align*} x=Px+(x-Px), \end{align*} with $Px\in\operatorname{span}\{r\}$ and $x-Px\in H$. The hyperplane $H$ is invariant because if $x\in H$, then \begin{align*} l^\top(Mx)=(l^\top M)x=\lambda l^\top x=0. \end{align*} On $\operatorname{span}\{r\}$, the matrix $M$ acts by multiplication by $\lambda$. On $H$, all complex eigenvalues of $M|_H$ have modulus strictly smaller than $\lambda$: the previous step excluded all other eigenvalues on the spectral circle, and $\lambda$ itself cannot occur on $H$ because a $\lambda$-eigenvector would be a multiple of $r$, while $l^\top r=1$. After complexifying $H$, Jordan normal form applies to $\lambda^{-1}M|_H$. Each Jordan block has eigenvalue of modulus less than $1$, so its powers converge to zero. Hence \begin{align*} \lambda^{-n}(M|_H)^n\to 0. \end{align*} Therefore \begin{align*} \lambda^{-n}M^n x=Px+\lambda^{-n}M^n(x-Px)\to Px \end{align*} for every $x\in\mathbb R^k$. This is convergence of matrices to the matrix of $P$, namely \begin{align*} rl^\top. \end{align*}[/guided]