Showing that a certain complex matrix is positive definite

Question

Consider the $n$-variable complex function $f(z)=f(z_1,\dots,z_n)=\log (1+|z|^2)=\log (1+|z_1|^2+\cdots+|z_n|^2)$. I am trying to show that the Hessian (the $n\times n$ matrix whose $(i,j)$-entry is $\frac{\partial^2 f}{\partial z_i\partial \bar{z}_j}$) is complex positive definite. By calculation I've shown that the Hessian is given by $\frac{1}{(1+|z|^2)^2} (a_{ij})$ where $a_{ij}=-z_j\bar{z}_i$ if $i\neq j$ and $a_{ii}=1+\sum_{j\neq i}|z_j|^2$. Clearly it suffices to show that the matrix $(a_{ij})$ is positive definite, but it seems that the calculation showing this directly is quite complicated. Any hints please?

Answer 1

The common denominator does not affect the result, so it suffices to show that the matrix $A = (a_{ij})$ is positive definite. $A$ can be written as a rank-one modification of a diagonal matrix: $$ A = (1 + |z|^2) I - \overline z z^T $$ where $I$ is the $n$-dimensional identity matrix and $\overline z$ denotes the component-wise complex conjugate of the column vector $z$. Then $$ w^* A w = (1 + |z|^2) |w|^2 - \overline w^T z z^T w = (1 + |z|^2) |w|^2 - |z^T w|^2 \ge |w|^2 > 0 $$ for all non-zero vectors $w$, using the Cauchy-Schwarz inequality.