TITLE: Spectrum of a matrix after applying an element-wise function (e.g. elementwise log)
QUESTION [6 upvotes]: Let $M \in \mathbb{R_{\geq0}}^{n \times n}$ be symmetric matrix with a given eigenvalue decomposition $M = U \Lambda U^T$. Suppose now we apply an element-wise $\log$ function to the non-zero entries of $M$ to obtain $M'$.
Can anything be said about the spectrum of $M'$ in terms of the spectrum of $M$?
More specifically can we say anything about the eigenvalues $\lambda_i(M')$ in terms of the $\lambda_i(M)$. Perhaps we can bound them?
Some ideas I've considered:

Approximate the log with a series and use eigenvalue inequalities about the Hadamard product of matrices + eigenvalue inequalities about the sum of matrices 
Reason about the distribution of elements of $M$ and $M'$ and the consequent distribution of their spectrums
Express $M' = M + \Delta M$ and use eigenvalue perturbation to obtain the spectrum of $M'$. However, I believe $\Delta M$ is not sufficiently small for this to be valid.

REPLY [4 votes]: I do not think you can bind the eigenvalues. Consider the following matrix for $0.2 \leq c\leq 0.8$:
$$A = \begin{pmatrix}c & \sqrt{c-c^2-0.16} \\ \sqrt{c-c^2-0.16} & 1-c\end{pmatrix}.$$
From the trace and the determinant it is easy to see that the eigenvalues are 0.2 and 0.8. The trace and determinant of $\log(A)$ (elementwise) are $\log(c(1-c))$ and $\log(c)\log(1-c) - 0.25\log(c(1-c)-0.16)^2$. When $c$ gets very close to 0.2, the sum of the eigenvalues is approximately $\log(0.16)$, while their product can be made arbitrarily small (as in $-\infty$). That means that the eigenvalues are unbounded.