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Let the resolvent matrix of $\mathbf{X}$, a symmetric matrix with real entries, be defined as \begin{align} R_{\mathbf{X}}(\lambda):=\bigl(\mathbf{X}-\lambda\mathbf{I}\bigr)^{-1}, \qquad \lambda \in \mathbb{C}\backslash\mathbb{R}. \end{align} with $\mathbf{I}$ the identity of matrix (same dimensions as $\mathbf{X}$).

I must show that \begin{align} R_{\mathbf{X}}(\lambda)= - \frac{1}{\lambda} \mathbf{I} - \frac{1}{\lambda}\mathbf{X}R_{\mathbf{X}}(\lambda). \end{align}

My naive approach is:

By definition \begin{align} \bigl(\mathbf{X}-\lambda\mathbf{I}\bigr) R_{\mathbf{X}}(\lambda)= \mathbf{I} \end{align} Thus, direct inversion yields \begin{align} \mathbf{X}R_{\mathbf{X}}(\lambda)-\lambda R_{\mathbf{X}}(\lambda) = \mathbf{I}\\ R_{\mathbf{X}}(\lambda) = -\frac{1}{\lambda}\mathbf{I}+\frac{1}{\lambda}\mathbf{X}R_{\mathbf{X}}(\lambda) \end{align}

which seems to contradictory the stated result... what did I missed?

jgyou
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    What you've done looks right. Make up an example of a matrix $X$, and see whether the thing you "must show" is actually true. – Gerry Myerson Nov 27 '13 at 04:26
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    I should have though of that! For a simple 2x2 matrix, one can already see that my result is correct, and not the other way around. – jgyou Nov 28 '13 at 02:02

1 Answers1

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When there is only one operator present, you can treat them as if they are numbers. So what you were asked to prove was $$ \frac1{x-\lambda} = - \frac1\lambda - \frac x{x-\lambda} .$$ Now it is really obvious that the statement you were asked to prove is wrong.

Stephen Montgomery-Smith
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