I´m triying to show that an $n\times n$ matrix $M$ such that $M_{ij}=e^{\gamma_{ij}}$ where $\gamma_{ji}=\gamma_{ij}^{\ast}$ are complex numbers is a positive semi-definite matrix.
I made a proof of the claim but assuming $M=e^{B}$ where $B$ is some hermitian matrix. In this case I can diagonalize $B$ and the rest is easy.
But I am not sure at all that $M=e^{B}$ as above. I also know that if $M$ is invertible then $log(M)$ exists but $B=log(M)$ is not necesarily hermitian, so I don't know if I can diagonalize $B$ or not and consecuently I can't do the same as above.
I apreciate any help.
What do you mean by $M_{ij}$? If you mean the element of $M$ in row $i$ and column $j$, your conjecture is incorrect.A real number is a complex number. Let $$\gamma_{11}=1,\gamma_{12}=2,\gamma_{21}=2,\gamma_{22}=2$$ The matrix is $$\begin{bmatrix}e&e^2\\e^2&e^2\end{bmatrix}$$ which is not positive semi-definite. If by $M_{ij}$, you mean the co-factor, the proof is a bit harder.