Resolvent recurrence relation 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?

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.
