Let $A$ be a matrix. If $A$ is "almost" equal to $A^*$, it follows from an argument of continuity that the eigenvalues of $A$ are "almost" real. Same argument can be made for $A$ "almost" $-A^*$, in which case the eigenvalues are "almost" purely imaginary. Question: Is there a way of getting a quantitative estimate on the relative or absolute value of the imaginary and real parts of the eigenvalues?
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Suppose $A=B+C$ where $B$ is self adjoint. Then you can diagonalize $B$, and then by applying Gersgorin's criterion, you get the following: if $\{\lambda_i\}$ are the eigenvalues of $B$ and $\hat \lambda$ is an eigenvalue of $A$, then there is an $i$ so that $|\hat \lambda-\lambda_i|\leq \|C\|_2$. (Here, $\|C\|_2$ is the Frobenius norm of $C$.)
answered Sep 2, 2013 at 12:04
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$\begingroup$ I can't follow, assuming B is self adjoint I write A=PDP^+C so A-PDP^=C, the only way I see at the moment to arrive to your conclusion is by assuming that A is at least triangularizable under the same unitary transformation P, can you clarify? $\endgroup$ Commented Sep 2, 2013 at 12:24
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$\begingroup$ @Craven use the matrix that diagonalizes B to conjugate A to a matrix of the form D+C' where C' is conjugate to C and hence has the same spectral data. Then apply Gershforin. If this point remains unclear then perhaps you could take further questions to math.stackexchange.com $\endgroup$ Commented Sep 2, 2013 at 17:22
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$\begingroup$ Gee, should have seen that right away. Thx for the clarification $\endgroup$ Commented Sep 2, 2013 at 17:28