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Let $A$ be an $N$ by $N$ Hermitian matrix with elements $a_{ij}$. What will be the bound on the elements $b_{ij}$ where $B=A^{-1}$? If $A$ is a diagonal matrix, solution is trivial. Also for tri-diagonal matrix, bounds exists.

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For tridiagonal matrices, bounds on inverse elements exist. Want to know about Hermitian matrices. –  user8856 Mar 29 '11 at 19:41
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if you are more specific, you can (possibly) get more help... –  yoyo Mar 29 '11 at 19:48

1 Answer 1

If H is a Hermitian matrix whose eigenvalues all have absolute value $\ge r$, then the matrix elements of $H^{-1}$ are bounded by $1/r$.

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Can you provide a reference? –  svenkatr Mar 29 '11 at 20:51
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Diagonalize $H = U^* D U$ where $U$ is unitary, $D$ is diagonal and * is Hermitian conjugate. Moreover, $D = R |D|$ where $R$ is diagonal with diagonal elements $\pm 1$ (and thus unitary). Then for any unit column vectors $u$ and $v$, $u^* H^{-1} v = (R U u)^* |D|^{-1} (U v)$ where $RUu$ and $Uv$ are unit vectors. Now use Cauchy-Schwarz for the inner product $(x,y) \to x^* |D|^{-1} y$, and $x^* |D|^{-1} x \le (1/r) x^* x$. –  Robert Israel Mar 30 '11 at 4:37
    
Thanks for your answer. I didn't expect such a succinct proof sketch. –  svenkatr Mar 30 '11 at 19:43

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