Let $I$ be the identity matrix and $P$ be an irreducible $n$ by $n$ row stochastic matrix. Let $d$ be a stochastic (column) vector and $e$ be an all one (column) vector. Let $t > 0$ be a real number. Define $$A(t) = \rho(P)I - P + t de^\top,$$ where $\rho(A)$ is the spectral radius of $P$. Since $P$ is stochastic, we know $\rho(P) = 1$. This $A(t)$ is then a rank one perturbation of the matrix $I - P$. Can we prove that $A(t)$ is positive stable for any $t > 0$, i.e., the real part of eigenvalues of $A(t)$ is strictly positive? The best result I can find is https://arxiv.org/abs/2007.11453. It provides counterexamples if $P$ is merely irreducible and nonnegative and their $d$ and $e$ are (almost) two arbitrary nonnegative vectors. I'm wondering by adding that the row sum of $P$ is 1 and more restrictions on $d$ and $e$, can we prove it? I tried to play with their counterexample but wasn't able to get a new counterexample under the stronger assumptions. If $d$ was the right eigenvector of $P$ or $e$ was the left eigenvector of $P$, things would be obvious according the references therein.
1 Answers
To get an idea of what is going on, we can use perturbation theory and properties of nonnegative matrices.
First of all, note that $I_{n}-P$ has a single eigenvalue at zero and all the others have positive real part. This is a consequence of the Perron-Frobenius theorem and the fact that $P$ is irreducible.
We now look at what happens when $t>0$ is small, perturbation theory says that all simple eigenvalues of $I-P$ obeys
$$ \lambda(I_n-P+tde^T)=\lambda(I_n-P)+tu^Tde^Tv+o(t) $$ where $u,v$ are the normalized (i.e. $u^Tv=1$) right and left eigenvectors of the considered eigenvalue.
For the zero eigenvalue, we have that $u>0$ is the stationary distribution and $v=e$. This yields $u^Tde^Tv=u^Tdn>0$. This means that the zero eigenvalue always bifurcates in towards the right half plane. By the continuity property of eigenvalues, we can conclude that there exists a $t_0>0$ such that the eigenvalues of $I_n-P+tde^T$ have positive real part for all $0<t\le t_0$. So, all is fine in that case.
Now we can look at what happens when $t=\infty$. For instance, we can divide everything by $t$ and use the change of variable $\epsilon=1/t$ to get the expression
$$ de^T+\epsilon(I_n-P) $$ on which we can do the same analysis. It can be shown that the zero eigenvalue of $de^T$ is semisimple and that its left and right eigenvectors $U^T,V$ are normalized; i.e. $U^TV=I_{n-1}$. This yields the bifurcation expression
$$ \lambda(de^T+\epsilon(I_n-P))=\lambda(de^T)+\epsilon \lambda_i(U^T(I_n-P)V)=\epsilon \lambda_i(U^T(I_n-P)V) $$ where $\lambda_i$ denotes the $i$-th eigenvalue. Simple calculations yield $$ U^T(I_n-P)V=I_{n-1}-U^TPV $$ and therefore that there exists a $t_1>0$ such that the eigenvalues of $I-P+tde^T$ have positive real part for all $t>t_1$ if all the eigenvalues of $I-U^TPV$ have positive real part.
In fact, if we define $D(s):=\det(sI_n-I_n+P)$ and $N(s):=\det(sI_{n-1}-I_{n-1}+U^TPV)$, we have that $Z_t(s):=\det(sI_n-I_n-P-tde^T)=D(s)-tN(s)$.
This shows that as $t\to+\infty$, the roots of $Z_t(s)$ tend to those of $N(s)$ which have positive real part if and only if the eigenvalues of $I_{n-1}-U^TPV$ have positive real parts as well. Under this condition, the root going to infinity as $t\to+\infty$ does so along the real axis in the positive direction.
Now, what can we say about $I_{n-1}-U^TPV$? First, we can say it is nonsingular. Indeed, observe that $[V\ \ e]$ and $[U\ \ d]$ are both nonsingular, then we get that $$ [U\ \ d]^T(I_n-P)[V\ \ e]=\begin{bmatrix}U^T(I-P)V & 0\\ \star & 0\end{bmatrix}. $$ Since $I_n-P$ has only one eigenvalue at zero, then this is also the case of $[U\ \ d]^T(I_n-P)[V\ \ e]$, which means that $U^T(I-P)V$ must be nonsingular.
I am currently looking into proving that the eigenvalues have positive real part.
Proving that this the case for all $t>0$ requires more work on my side but it should be possible to obtain conditions on for which this is the case.
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