linear algebra - Why did I write this equivalency involving an eigenvalue decomposition?

Going through some notes from the past quarter, I stumbled across the following equation:
\begin{equation}
(\tilde{\lambda}I-A)^{-1}=Q(\tilde{\lambda}I-\Lambda)^{-1}Q^*
\end{equation}

I'm sure it made sense to me at the time, but now I can't get back to the intermediate steps other than the usual eigenvalue decomposition $A=Q\Lambda Q^*$.

$A$ is a normal matrix, $\tilde{\lambda}$ is an eigenvalue of $A+\delta A$, $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ on its diagonal, and $Q$ is unitary.

Answer

If we take the inverse of both sides, we see that your equation is equivalent to
$$
(\lambda I - A) = Q(\lambda I - \Lambda)Q^*
$$
(where I have removed the $\sim$ from your $\lambda$). To see that this is true, it suffices to multiply out the right side of the equation:
$$

Q(\lambda I - \Lambda)Q^* = Q(\lambda I)Q^* - Q\Lambda Q^* = \lambda I - A
$$
which is what we wanted.