Matrix elements of an inverse Hermitian matrix
The comments on my question helped me make the necessary logical jump to solve this problem.
First, it is important to realize that if $\mathbf{A} = \mathbf{U}\mathbf{D}\mathbf{U}^{\dagger}$, then $f(\mathbf{A}) = \mathbf{U}f(\mathbf{D})\mathbf{U}^{\dagger}$ (where $\mathbf{D}$ is a diagonal matrix and $f(\mathbf{D})$ is a diagonal matrix with elements $f(D_{ii})$). The jump I needed to make was to realize that $(\omega\mathbf{1}-\mathbf{A})^{-1}$ is a function of $\mathbf{A}$, not just of $(\omega\mathbf{1} - \mathbf{A})$, suggesting
$$\mathbf{G}(\omega) = (\omega\mathbf{1}-\mathbf{A})^{-1} = \mathbf{U}(\omega\mathbf{1}-\mathbf{a})^{-1}\mathbf{U}^{\dagger}.$$
$(\omega\mathbf{1}-\mathbf{a})^{-1}$ is a diagonal matrix and the eigenvalues of $(\omega\mathbf{1}-\mathbf{A})^{-1}$ are therefore $(\omega-a_{\alpha})^{-1}$ an
$$[\mathbf{G}(\omega)]_{ij} = \sum_{\alpha} U_{i\alpha}(\omega-a_{\alpha})^{-1}U_{j\alpha}^{*} = \sum_{\alpha} \frac{U_{i\alpha}U_{j\alpha}^{*}}{\omega-a_{\alpha}},$$
as required by the original problem.
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James Womack
Updated on November 23, 2020Comments
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James Womack almost 3 years
I am trying to tackle the following exercise in a quantum chemistry textbook:
Show that: If $\mathbf{G}(\omega) = (\omega \mathbf{1}-\mathbf{A})^{-1}$, and $\mathbf{A}$ is Hermitian (i.e. $\mathbf{U}^{\dagger}\mathbf{A}\mathbf{U} =\mathbf{a}$ where $(\mathbf{a})_{ij} = a_{i}\delta_{ij}$ ) that $[\mathbf{G}(\omega)]_{ij} = \sum_{\alpha = 1}^{N} \frac{U_{i\alpha}U^{*}_{j\alpha}}{\omega-a_{\alpha}}$ (where $a_{\alpha}$ are the eigenvalues of $\mathbf{A}$).
I think that if $\mathbf{A}$ is Hermitian, then $(\omega\mathbf{1}-\mathbf{A})$ is also Hermitian, since the property $M_{ij} = M^{*}_{ji}$ is conserved when $\mathbf{A}$ is multiplied by $-1$ and when it's diagonal matrix elements are modified (assuming $\omega$ is real).
How should I approach this problem?
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J. M. ain't a mathematician almost 12 yearsThe idea is that if you have the eigendecomposition of $\mathbf A$, it is a simple matter to evaluate $f(\mathbf A)$: if $\mathbf A=\mathbf U\mathbf D\mathbf U^\dagger$, then $f(\mathbf A)=\mathbf U f(\mathbf D)\mathbf U^\dagger$, and it is a simple matter to evaluate the matrix function of a diagonal matrix...
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James Womack almost 12 yearsAh thanks. I knew this, but I think what I missed that not only is $(\omega\mathbf{1}-\mathbf{A})^{1}$ a function of $(\omega\mathbf{1}-\mathbf{A})$, it is also a function of $\mathbf{A}$. I'll work though the problem and write up the solution as an answer if I get there.
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