Let A, B, C, D be positive semi-definite matrices and $\alpha$ a real number between zero and one.
It is fulfilled:
$Tr(A \log(B)) \geq Tr(A \log(C))$,
where $Tr$ is a trace. Does from this follow:
a) $Tr(A \log(B+D)) \geq Tr(A \log(C+D))$
b)$Tr(A \log(\alpha B)) \geq Tr(A \log(\alpha C))$
c) $Tr(A \log(B+A)) \geq Tr(A \log(C+A))$
?
I would be also grateful for naming me some literature which can me help with these problems. "Topics in matrix analysis" wasn't helpful (or I have missed something).
Ok, I have found an answer for part b): Yes.
The proof:
We use the following property of the matrix logarithm:
$\log(GF)=\log(G)+\log(F)$,
where $G$ and $F$ commute. Accordingly, where $F$ will be in our case $\alpha 1$ (where 1 is identity matrix, so it commutes with everything)
$Tr(A\log(B)) \geq Tr(A\log(C)) \Rightarrow$
$Tr(A\log(B)) + Tr(A\log(\alpha 1)) \geq Tr(A\log(C)) + Tr(A\log(\alpha 1)) \Rightarrow$
$Tr(A\log(\alpha B)) \geq Tr(A\log(\alpha C)).$
According a) and c): No, as the matrices
$ a=\left( \begin{array}{cc} 0.584421 & 0.295001\, +0.310138 i \\ 0.295001\, -0.310138 i & 0.415579 \\ \end{array} \right)$,
$b=\left( \begin{array}{cc} 0.385804 & 0.286079\, -0.213992 i \\ 0.286079\, +0.213992 i & 0.614196 \\ \end{array} \right),$
$c=\left( \begin{array}{cc} 0.688838 & 0.37651\, -0.227299 i \\ 0.37651\, +0.227299 i & 0.311162 \\ \end{array} \right)$
does not satisfy c).