Is it possible to determine the largest number $\tau$ such that the spectral radius $\rho(A\pm \tau ee^T) < 1$

Question

Let $A \in M_n(\mathbb R)$ with no particular structure assumed and the spectral radius $\rho(A) < 1$. Let us denote the all $1$ vector by $e = (1, \dots, 1)^T$. I would like to determine a number $\tau \in \mathbb R_+$ such that $\rho(A \pm \tau e e^T) < 1$. Since we know $\{X \in \mathbb R: \rho(X) < 1\}$ is open, we could always have some positive number $\tau$. I would be happy if we can find a number $\tau = \tau(A)$ dependent on $A$ such that is sufficient to guarantee $\rho(A\pm \tau ee^T)<1$.

Answer 1

In this post, we assume that $A$ is symmetric (otherwise I do not see what to show).

Let $spectrum(A)=\{\lambda_1\geq \cdots\geq\lambda_n\}$. We may assume that $\rho(A)=\lambda_1\geq 0$.

$\textbf{Proposition 1}.$ If $\tau< \dfrac{1-\rho(A)}{n}$, then $\rho(A\pm \tau ee^T)<1$.

$\textbf{Proof}$. Let $U=ee^T$ (symmetric $\geq 0$). Note that $\rho(U)=n$ and $\rho(A+\tau U)\leq \lambda_1+\tau n<1$.$\square$

We can do better when $\rho(A)$ is closed to $1$, that we assume in the sequel.

$\textbf{proposition 2}$. Let $\theta =\sup\{|u_1+\cdots+u_n|; Au=\rho(A) u,||u||=1\}$.

then $\rho(A\pm \tau ee^T)<1$ when $\tau<\tau_0$ where $\tau_0\approx \dfrac{1-\rho(A)}{\theta^2}$.

$\textbf{Proof}$. Let $A(\tau)=A+\tau U$; it's an analytic function. Then the eigenvalues and a basis of (unit length) eigenvectors of $A(\tau)$ are globally analytically parametrizable (even if the eigenvalues (for example $\rho(A)$) present some mutiplicities.

Let $u(\tau)$ be a unitary vector function ($u^Tu'=0$) s.t. $A(\tau)u(\tau)=\lambda(\tau)u(\tau)$. Then $A'u+Au'=\lambda'u+\lambda u'$ and consequently $\lambda'=u^TA'u=u^TUu=(u_1+\cdots+u_n)^2$.

Finally $\lambda(\tau)\approx \rho(A)+\tau \theta^2$. Note that $\theta^2\leq n$.