Convergence of a sequence of eigenvectors (nonnegative matrix)

Question

Let $A$ be a $n\times n$ matrix with coefficients in $ [0,1] $. Let $ B $ be the matrix filled up only with the value $ \frac{1}{2} $: $$B = \begin{pmatrix} \frac{1}{2} & \dots & \frac{1}{2} \\ \vdots & \ddots & \vdots \\ \frac{1}{2} & \dots & \frac{1}{2} \end{pmatrix}\,.$$ For all $ t \in ]0,1] $ let $ A(t) = tB + (1-t)A $. The matrix $ A(t) $ is primitive for any fixed $ t $. Hence, from the Perron-Frobenius theorem, we have that $ \rho(t) $ (i.e. the spectral radius of the matrix $ A(t) $) is a simple eigenvalue, with the relative eigenvector that can be taken positive (i.e. every component is strictly positive). Let me call it $ x(t) $, choosing it such that $ \|x(t)\|_1 = \rho(t) $ (i.e. the sum of all the components is equal to the spectral radius of the matrix). In this way, I obtain the following properties: \begin{matrix} A(t)x(t) = \rho(t)x(t) \\ \|x(t)\|_1 = \rho(t)\\ x(t) > 0 \end{matrix} My question is: does there exist $ \lim_{t \to 0^+}x(t) $?

I have read that spectral radius is continuous with respect to any matrix norm, and then I would have: $ \lim_{t \to 0^+}A(t) = A \Rightarrow \lim_{t \to 0^+}\rho(t) = \rho(0) $. Is it correct?

Anyway, passing to the sequences with $ t = \frac{1}{n} $, I did not succeed in proving that the generated sequence $ x_n = x(\frac{1}{n}) $ is a Cauchy sequence (this fact would imply that $ x_n $ is convergent, because of the sequentially compactness of $ [0,1]^n $).

I think that the better way in order to prove the existence of the limit above is to prove the monotonicity of the components of $ x_n $ using the monotonicity of the coefficients $ a_{ij}(t) $ of $ A(t) $. It is only an idea but I do not know if it works.

I really thank you in advance.

Answer 1

For $4$ days, I have been stuck essentially at the place where Surb arrived. I detail the basic reasoning

Note that, when $A(t)$ is continuous, the spectral radius $\rho(t)$ of $A(t)$, is always continuous on $[0,1]$.

When $t>0$: $A(t)$ is an analytic function of $t$ and a positive matrix; moreover $\rho(t)>0$ is a simple eigenvalue s.t. $A(t)x(t)=\rho(t)x(t)$ where $x(t)$ is a positive vector satisfying $||x(t)||=1$ for some norm. Then, according to

https://www.math.upenn.edu/~kazdan/504/eigenv.pdf

$\rho(t)$ and $x(t)$ are analytic functions on $(0,1]$.

$\textbf{Proposition 1.}$ If $A$ is symmetric, or more generally, $A$ is normal and $(A-A^T)B=B(A-A^T)$, then $x(t)$ converges.

$\textbf{Proof.}$ Since $A(t)$ is normal over $[0,1]$, $\rho(t)$ and $x(t)$ are analytic over $[0,1]$. cf. Theorem(A) in

https://arxiv.org/pdf/1111.4475v2.pdf

$\square$

$\textbf{Proposition 2.}$ The cluster points of $x(t)$ are some non-negative vectors in $\ker(A-\rho(A)I)$.

$\textbf{Proof}.$ Consider a sequence $(t_p)$ that tends to $0$ s.t. $x(t_p)\rightarrow x\;(\geq 0)$; then $Ax=\rho(A)x$. $\square$

$\textbf{Remark.}$ It "remains " to show that all these subsequences tend to the same limit.

$\textbf{Corollary.}$ If $dim(\ker(A-\rho(A)I))=1$, then $x(t)$ converges.

EDIT.

$\textbf {Proposition 3.}$ $\lim x(t)$ (if it exists) is not a continuous function of $A$.

$\textbf{Proof.}$ Let $A=\begin{pmatrix}1&a&0\\0&4/5&1/10\\0&2/5&4/5\end{pmatrix}$ where $\rho(A)=1$. We use $||.||_{\infty}$.

If $a=0$, then $dim(\ker(A-I))=2$ and $x(t)\rightarrow [2/3,1/2,1]^T$.

If $a\not= 0$, then $dim(\ker(A-I))=1$ and $x(t)\rightarrow [1,0,0]^T$. $\square$

EDIT 2. About the sasquires' post and the following reference that user1551 gave (indirectly)

[1] A note on perturbations of stochastic matrices, Huppert, Willems, Journal of Algebra (2000). -in free access-

https://reader.elsevier.com/reader/sd/pii/S0021869300985442?token=A1062A5C61A49B95037719BF2E60DBE3462EA3D00AD1980BC8CFC9C479C5C1250AC05536F459134264505903FCFFA812

One has $2$ questions. i) Does $x(t)$ converge ? ii) If yes, towards what ?

i) For almost stochastic matrices, the authors of [1] claim yes in Theorem 3-2 (a). Yet, I am not convinced by the end of the proof of (a); could someone read this proof and say what he/she thinks ?

ii) sasquires claims, in his post, that the following is true (I think that too)

$\textbf{Conjecture.}$ Let $A\in M_n$ be non-negative where $\rho(A)>0$ is semi simple and (do we need that ?) is the unique eigenvalue with modulus $\rho(A)$. IF $x(t)$ CONVERGES, then the limit can be explicitly calculated as follows

Let $(v_i)_{i\leq k},(u_i)_{i\leq k}$ be bases of $\ker(A-\rho(A) I)$ and $\ker(A^T-\rho(A)I)$ that satisfy $u_i^Tv_i=1$ and, when $i\not= j$, $u_i^Tv_j=0$.

-Of course, one must prove that such bases exist-

$R$ is the $k\times k$ matrix $R_{i,j}=u_i^T(B-A)v_j$; $c$ is the eigenvector associated to $\max( spectrum(R))$; finally -up to a factor- $\lim x(t)=\sum_i c_iv_i$.

sasquires, it's your idea; then, it's your job to write a flawless proof.

$\textbf{Remark.}$ If $x(t)$ converges and the above conjecture is true, then

$\textbf{Lemma.}$ Let $a,b>0$ and let $y(t)$ be the eigenvector associated to $(aA,bB)$. Then $\lim y(t)=\lim x(t)$.

$\textbf{Proof.}$ We may assume that $b=1$. Then the new matrix $R$ is $S_{i,j}=u_i^T(B-aA)v_j=u_i^TBv_j-au_i^TAv_j$ and $S=[u_i^TBv_j]-a\rho(A)I$. Finally we may replace $R$ with the matrix $[u_i^TBv_j]$. $\square$

In particular, we may replace $B$ with the matrix $[1/n]$, that is, a stochastic matrix and $A$ with a matrix satisfying $\rho(A)=1$.

Answer 2

Edit: This answer has been mostly rewritten, since the previous versions contain a lot of text, most of which is no longer valuable (given that this answer is more comprehensive).

I read a paper that reminded me of this problem and decided to revisit it recently.  This problem is not nearly as difficult as I thought it was at the time of my last answer.  (I had previously considered an argument along the lines of the following, but I rejected it before thinking it through properly.)

This answer should be rigorous (and in danger of being pedantic), so I hope it will satisfy the OP's request.  But please let me know if you find any errors in the reasoning.

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Introduction: Instead of considering the problem as stated by the OP, I will consider a similar and more general problem.  Let $$ X(t) = A + t B $$ where $A$ is any square nonnegative matrix and $B$ is any positive matrix of the same dimensions.

The OP's original question can be restored by substituting $t \to \frac{t}{1-t}$, $B = \tfrac{1}{2} 1 1^T$, and $A(t) = (1 - t)X(t)$.  (Obviously this would pose a problem if we cared what happens at $t=1$, but the question is only concerned with what happens in a neighborhood of $t=0$.)

Definitions: By the Frobenius-Perron theorem, the spectral radius of $A$, $\rho(A)$, is associated with one or more real nonnegative eigenvectors.  Let $R$ denote the right eigenspace associated to $\rho(A)$ and $L$ denote the associated left eigenspace.

Proposition 1: $\rho(X(t))$ is continuous as a function of $t$ on $[0, 1]$.

Proof: This is an immediate consequence of the fact that any eigenvalue of a continuously varying matrix can be expressed as a continuous function.  See this answer, for example.

Definitions: Let the vectors $v_i$ be a basis for $R$ and $u_i$ be a basis for $L$.  By the Frobenius-Perron theorem, all $u_i$ and $v_i$ must be nonnegative.  (Note that $\dim R = \dim L$ since every matrix is similar to its transpose.)

Proposition 2: The $u_i$ can be chosen such that $u_i^T v_j = 0$ when $i \ne j$ and $u_i^T v_i = 1$.  We will assume that this has been done below.

Proof: This is a well-known theorem.  See this answer, for example.

Definitions: Let $V$ be a matrix containing $v_i$ as columns and $U$ be a matrix containing $u_i$ as columns.  Let $P = V U^T$ and $Q = I - P$.  Note that $P$ and $Q$ have the same dimensions as $A$ and $B$.

Proposition 3: $P$ is a projection operator onto $R$ and $P r = r$ for any $r \in R$.  

Proof: This is also well-known, but here is a quick proof. $$ P^2 = (V U^T) (V U^T) = V (U^T V) U^T = V I U^T = V U^T = P$$ The main step $U^T V = I$ is a consequence of Proposition 2 and the normalization chosen there.  For any $r \in R$, $r = \sum_j c_j v_j$, so $(U^T r)_i = \sum_j c_j u_i^T v_j = c_i$ and $P r = \sum_i c_i v_i = r$.

Definitions: Since $X(t)$ is a positive matrix for $t>0$, then for all $t>0$ there is a unique, positive eigenvector associated with $\rho(X(t))$.  Denote this $x(t)$.

Note: All limits considered below are from the right (i.e., replace $t \to 0$ with $t \to 0^+$).

Proposition 4: $\lim_{t \to 0} Q x(t) = 0$.

Proof: This is the same as @loupblanc's Proposition 2 on this page.  The proof is as follows.  First, note that any limit point of $x(t)$ must be in $R$.  Consider any sequence $t_i \to 0$ such that $x(t_i)$ converges to some vector $x_*$.  Considering the limit in the eigenvalue equation, we find that $X(0) x_* = \rho(X(0)) x_*$, or $A x_* = \rho(A) x_*$, so $x_* \in R$.  This implies $P x_* = x_*$ by Proposition 3, so $Q x_* = 0$.

Definition: Let $Y = P B$.

Proposition 5: $Y$ is either a positive matrix or it has some zero rows and contains a positive submatrix.

Proof: $P$ is nonnegative (being the sum of products of nonnegative elements) and $B$ is positive.  $(PB)_{ij} = \sum_k P_{ik} B_{kj} > 0$ if there is any $k$ where $P_{ik} > 0$, so the only way for an element to be zero is if $P$ has an entire row that is zero.  Since the rank of $P$ is equal to $\dim R \ge 1$, then not all the rows of $P$ can be zero.  Excluding the indices where the rows of $P$ are zero, then the remaining submatrix must be positive.

Comment: Given Proposition 5, we can without loss of generality assume that $Y$ is a positive matrix.  If not, we can write $Y = \begin{pmatrix} Y_1 & Y_2 \\ 0 & 0 \end{pmatrix}$ where $Y_1$ and $Y_2$ are positive submatrices.  We can restrict the entire following discussion to the basis elements with positive entries, effectively replacing $Y$ with $Y_1$. Note that, in this case, the convergence of the bottom entries of $x(t)$ (on the rows where $P$ is zero) to zero is guaranteed by the fact that $\lim_{t \to 0} Qx(t) = 0$ and $Q = I - P$.

Definition: Let $w^T$ and $y$ be the left and right Frobenius-Perron eigenvectors of $Y$, respectively.  Note that since $Y$ is positive, both of these are positive.

Proposition 6. $\lim_{t \to 0} \frac{\rho(X(t)) - \rho(A)}{t} = \rho(Y)$.

Proof: $$ \rho(X(t)) P x(t) = PX(t) x(t) = P \left( A + tB \right) x(t) = P \rho(A) + tY x(t) $$ Rearranging gives $$ Y x(t) = \frac{\rho(X(t)) - \rho(A)}{t} P x(t) $$ Multiplying through by $w^T$ and dividing by $w^T P x(t)$ gives $$ \frac{\rho(X(t)) - \rho(A)}{t} = \rho(Y) \frac{w^T x(t)}{w^T P x(t)} = \rho(Y) \frac{w^T (P + Q) x(t)}{w^T P x(t)} = \rho(Y) \left( 1 + \frac{w^T Q x(t)}{w^T P x(t)} \right) $$

As $t \to 0$, $Qx(t) \to 0$.  Moreover, $w^T P x(t)$ cannot go to $0$ because $P x(t)$ is nonnegative and always has some nonzero elements and $w$ is positive.  Taking the limit on both sides proves the result.

Proposition 7: If $M(t)$ is a matrix varying with parameter $t$ that has a limit $L$ at $t=0$, and $m(t)$ is a vector which has a bounded norm for $t > 0$, and $\lim_{t \to 0} L m(t)$ exists, then $\lim_{t \to 0} M(t) m(t) = \lim_{t \to 0} L m(t)$ (i.e., the limit can be substituted for $M$).

Proof: There is probably an easier proof, but this is easy enough to prove directly.  Using any induced matrix norm, $$ 0 \le \| M(t) m(t) - L m(t) \| \le \| M(t) - L \| \| m(t) \| \to 0 $$ as $t \to 0$, where the last step uses the fact that $m(t)$ is bounded.  The result follows from the squeeze theorem.

Main theorem: $\lim_{t \to 0} x(t)$ exists and is equal to $y$.

Proof: From the proof to Proposition 6, we have $$ \left[ Y - P \left( \frac{\rho(X(t))-\rho(A)}{t} \right) \right] x(t) = 0 $$ Taking a limit and applying Proposition 7, we get that $$ \lim_{t \to 0} \left[ \left( Y - \rho(Y) P \right) x(t) \right] = 0 $$ We can subtract $\lim_{t \to 0} \rho(Y) Q x(t)$ from both sides and apply the theorem on limits of sums and $P+Q=I$ to get rid of $P$, i.e., $$ \lim_{t \to 0} \left[ \left( Y - \rho(Y) \right) x(t) \right] = 0 $$

This means that any limit points of $x(t)$ must lie in the eigenspace of $Y$ associated to $\rho(Y)$.  Since $Y$ is a positive matrix, then this eigenspace is one-dimensional.  Since $x(t)$ and $y$ are both normalized positive vectors belonging to the same one-dimensional eigenspace, they must be equal.