If $A \in M_{n\times n}$ is a transition matrix such that $A_{ii} \geq \frac{4}{5}$, what is the smallest disk in $C$ containing all eigenvalues of $A$?
I understand the concept of the Gerschgorin's Disks for each eigenvalue of this matrix. I can also intuit what's necessary here: the furthest an eigenvalue can be from $\frac{4}{5}$ is in the disk centered there with radius $n-\frac{4}{5}$, and the furthest an eigenvalue can be from $1$ (since all diagonals must be between $\frac{4}{5}$ and $1$ for the given transition matrix) is in the disk centered there with radius $n-1$.
Thus, I need to construct the smallest such circle that contains those two eigenvalues. However, I'm having a bit of trouble figuring out exactly how to prove/construct this.
By transition matrix I assume you mean that the row sums are equal to $1$ and all entries are nonnegative. Thus, $\frac{4}{5}\leq A_{i,i}\leq 1$ for each $i$, and the Gershgorin disks are $$D_i=\{z\in\mathbb C:|z-A_{i,i}|\leq r_i\}$$ where $0\leq r_i\leq \frac{1}{5}$ since $r_i$ is the sum of the absolute values of the off-diagonal entries in the $i$-th row.