Let $A=(a_{ij})$ and $B=(b_{ij})$ be two complex Hermitian matrices of order $n$ each such that for all $(i,j)$-entry and $i\le j$, $b_{ij}=a_{ij}$ , except for two positions (say)- the $(i_1,j_1),(i_2,j_2)$-entries, where $b_{i_1j_1}=\overline{a}_{i_1j_1}$ and $b_{i_2j_2}=\overline{a}_{i_2j_2}$. So $B$ is generated from $A$ by simply interchanging two of its entries around its diagonal. Then, is there any relationship between the spectral radii of $A$ and $B$?
comparision between spectral radii of two Hermitian matrices
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matrices
eigenvalues-eigenvectors
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0The way you wrote the relation, either $B$ is not Hermitian, or $B=A$. – 2017-01-19
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0As $B$ is Hermitian, hence $b_{i_1j_1}=\overline{a}_{i_1j_1}$ implies $b_{j_1i_1}=\overline{a}_{j_1i_1}=a_{i_1j_1}$ and similarly. – 2017-01-19
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1Fair enough, but that's not what the question says. You are replacing four entries, not two. – 2017-01-19
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0Well, the sum of the eigenvalues is obviously the same, which gives you some bound. But I think there is something more sophisticated. Btw, I do not understand entirely your definition of the matrix. Can you please check whether the matrix is not circulant? – 2017-01-19
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0Yes, four entries are replaced. Thanks @Martin Argerami. – 2017-01-20
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0@Jane Maths. The matrix need not be circular. Only two upper half entries are swapped with their corresponding lower half entries. – 2017-01-20