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If $B$ is a symmetric matrix then how do its eigenvalues compare to the eigenvalues of $B - vv^T$? ( where $v$ is a vector of the same dimension as $B$) I guess that the eigenvalues of $B - vv^T$ ...

Given two positive integers $n$ and $a \leq \frac{n}{2}$ consider a $n \times n$ matrix $A$ such that, all the diagonal entries are either $a$ or $a+1$ all the non-zero off-diagonal entries are $\pm ...