$A=\begin{pmatrix} B+T && E\\E^T&& I\end{pmatrix}$
Here $E=\begin{pmatrix} 1& 1& 1& \ldots &1\\ 0& 0& 0& \ldots &0\\ \ldots &\ldots &\ldots&\ldots&\ldots\\ \ldots &\ldots &\ldots&\ldots&\ldots\\ 0& 0& 0& \ldots &0\end{pmatrix}$ and $T=\begin{pmatrix} n& 0& 0& \ldots &0\\ 0& 0& 0& \ldots &0\\ \ldots &\ldots &\ldots&\ldots&\ldots\\ \ldots &\ldots &\ldots&\ldots&\ldots\\ 0& 0& 0& \ldots &0\end{pmatrix}$
Here the spectral radius of $B$ is given as $\rho(B)$. Here $A$ is a $2n\times 2n$ matrix and each block is a $n\times n$ matrix
How to find a lower bound on $\rho(A)?$
My try
I found that $\rho(A)\ge \rho(B)+t$ where $t$ is the smallest eigenvalue of $\begin{pmatrix} T& E\\ E^T& I\end{pmatrix}$
I found that the smallest eigenvalue of $\begin{pmatrix} T& E\\ E^T& I\end{pmatrix}$ is $0$.
So I got $\rho(A)>\rho(B)$.
I was wondering if a more sharp lower bound is possible. Is it possible to find a more sharp lower bound of $\rho(A)$ in terms of $\rho(B)$?
I will be grateful if some help could be provided.
There is no sharper lower bound. Note for instance that in the case of $n=2$, if we take $$ B = \pmatrix{0&0\\0&\alpha} $$ for any $\alpha \geq 2$, then we find that $\rho(A) = \rho(B) = \alpha$.