Let $Z$ be a positive semidefinite matrix with nonnegative entries, and define $X=\log(1+Z)$, where the $\log$ is taken entrywise, i.e., $X\_{ij}=\log(1+Z\_{ij})$. Are there some simple sufficient conditions that guarantee that $X$ is positive semidefinite? 


**Note 1**
Intuitively, this should work when $Z$ is small enough, because then $X \simeq Z \succeq 0$.


**Note 2** A simple counter-example is given by $Z=\left[ \begin{array}{cc}
3 & 5\\
5 & 9
\end{array}
\right]$.


**Note 3**
It is easy to verify that the converse relation allways holds, that is,
$$X\succeq 0 \Longrightarrow Z:=exp(X)-1\succeq 0,$$
where the above operations are elementwise: $Z\_{ij} = e^{X\_{ij}} -1$. Indeed, we have the relation $Z=\sum\_{k=1}^\infty X^{\circ k}/k!$, where $X^{\circ k}$ is the $k$th power of $X$ with respect to the Hadamard (elementwise) product, and it is well-known that the Hadamard product of 2 positive semidefinite matrices is positive semidefinite.


**Note 4** Some background on this question: I ask myself when a random variable $Y$ with mean vector $\mu\in\mathbb{R}^n$ and variance-covariance matrix $\Sigma \succeq 0 \in \mathbb{R}^{n\times n}$ can be fitted to a multivariate log-normal distribution using the method of moments. Applying the formulas, we find that the elements of variance-covariance matrix $V$ of $\log Y$ must satisfy
$$V\_{ij}= \log(\frac{\Sigma\_{ij}}{\mu\_i \mu\_j} +1),$$
but in general this matrix is not guaranteed to be positive semidefinite.



when does elementwise-log preserve positive-semidefiniteness?
It's not true that it works for $Z$ small enough. Consider the $2 \times 2$ case
$$ Z = \pmatrix{t & 2t\cr 2t & 4t\cr} $$
which is positive semidefinite for $t \ge 0$.
Then $$\det(X) = \log(1+t)\log(1+4t) - \log(1+2t)^2 $$
which appears to be negative for all $t > 0$, and certainly is negative for small $t > 0$: its Maclaurin series is
$ \det(X) = -2 t^3 + O(t^4)$.


What is true is that if $Z$ is positive **definite**, $\log(1+tZ)$ will be positive definite for sufficiently small $t > 0$. This can be obtained using the
Maclaurin series for $\log(1+z)$.


EDIT:


Let $g(z) = z - \log(1+z)$, and 
write $X = Z - G$, where $G\_{ij} = g(Z\_{ij})$.
Note that $g$ is increasing on $[0,1]$. If all $|Z\_{ij}| \le b$, then 
$|W\_{ij}| \le g(b)$. If your matrices are $N \times N$, 
for any vector $x$ we have by Cauchy-Schwarz $x^T G x \le N g(b) \|x\|^2$.
Thus to have $X \succeq 0$ it suffices for $Z$ to be positive definite with least eigenvalue $\lambda$ and elementwise bound $b$ where
$$ \lambda \ge N g(b) $$


