Matrix elements of an inverse Hermitian matrix

Question

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?

Answer 1

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.