Let $A$ and $B$ be two Hermitian matrices of order $n$ such that $\mbox{rank}(B)=2$. Then how is the spectral radius of $A$, denoted by $\rho(A)$, related to that of $A+B$?
That is - Is there any relationship exist between $\rho(A)$ and $\rho(A+B)$?
Effect on the spectral radius of a Hermitian matrix by the addition of a rank 2 Hermitian matrix
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linear-algebra
matrices
eigenvalues-eigenvectors
1 Answers
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For Hermitian matrices, we have $\rho(A+B) \leq \rho(A) + \rho(B)$. In general, we have $\|A+B\|\leq \|A\| + \|B\|$ where $\|\cdot\|$ denotes the spectral norm. When $A$ is Hermitian, $\|A\| = \rho(A)$.
If you want to incorporate the rank of $B$, then there is more we could say about the particular eigenvalues using Weyl's inequalities.