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From the Gerschgorin Theorem we know that each eigenvalue located in the circles which radius are:$$r_{i}= \sum_{j=1,j\ne i}^n | a_{ij}|$$(which means add each entries in a row except for the diagonal)$$$$But we also know that the $A^T$and A has the same eigenvalue. So we can conclude from that for each $r_{i}$,we have $$r_{j}= \sum_{i=1,i\ne j}^n | a_{ij}|$$(simply add the abs of each entries in each colum except the diagonal). So from that we can know the minimum radius of the circle of the ith eigenvalus is the minimum of $r_{i}$ and $r_{j}$. But why the textbook just told us that the minimum is $r_{i}$?

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Personal opinion: because there's no need to state everything. Your reasoning is a simple corollary. –  Git Gud Jan 13 '13 at 14:26
"The $i$th eigenvalue" doesn't really have a meaning: the matrix has $n$ eigenvalues, but they don't come with a natural ordering, nor is a particular eigenvalue naturally associated with a particular diagonal entry (in general). The theorem says that the $n$ eigenvalues are contained in the union of the $n$ "row disks". Your reasoning does show that the $n$ eigenvalues are contained in the union of the $n$ "column disks". But the eigenvalue contained in the 3rd row disk might be in the 7th column disk, for example. –  Greg Martin Feb 3 at 19:30
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