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.


1 Answer 1


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 $\log(1+z)$.


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 $$ \lambda \ge N g(b) $$

  • 1
    $\begingroup$ Thank you, this is a very enlightening answer. But now the question is: How do I recognize if a positive definite matrix $Z$ is small enough (so that $X\succeq 0$)? Maybe we can find a criterion relying on the eigenvalues of $Z$ ? $\endgroup$
    – guigux
    Aug 19, 2015 at 9:29
  • $\begingroup$ Very nice, his is exactly the kind of criterion I was looking for ! $\endgroup$
    – guigux
    Aug 20, 2015 at 12:33

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