Logarithm of sum of orthogonal matrices Assume that X, Y are non-negative matrices. Because of the monotonicity of the logarithm it is then fulfilled:
$\text{Tr}(X\log(X+Y))\geq \text{Tr}(X\log(X))$, (1)
where $\text{Tr}$ denotes trace.
However, when $X$ and $Y$ are orthogonal ($\text{Tr}(XY)=0$), the both sides of (1) are equal:
$\text{Tr}(X\log(X+Y))= \text{Tr}(X\log(X))$.
Why is it so?
Is the following also satisfied:
$\text{Tr}(X\log(X+Y))\geq \text{Tr}(X\log(Y))$, (2)
or, general
$\text{Tr}(A\log(X+Y))\geq \text{Tr}(A\log(Y))$, (3)
where A non-negative
?
 A: As you have answered (2) and (3) I will only answer (1).
Note that if $X$ is positive definite then $\mathrm{Tr}[X Y] = 0 \implies Y = 0$ and we are done. Otherwise let $X$ be positive semidefinite and consider the spectral decomposition of $X$, i.e.
$$
X = \sum_i \lambda_i |i \rangle\langle i |.
$$
In this basis $X$ takes the block matrix form
$$
X = \begin{pmatrix} X_0 & 0 \\ 0 & 0 \end{pmatrix}.
$$
where $X_0$ is a positive definite matrix. A priori we can also write Y in this basis as
$$
Y = \begin{pmatrix} Y_0 & Y_1 \\ Y_1^* & Y_2 \end{pmatrix},
$$
where $Y_0,Y_2$ are positive semidefinite. Now note that the restriction $\mathrm{Tr}[X Y] = 0$ implies that
$$
Y = \begin{pmatrix} 0 & 0 \\ 0 & Y_2 \end{pmatrix}.
$$
To see this note that $0=\mathrm{Tr}[XY] = \mathrm{Tr}[X_0 Y_0] \implies Y_0=0$ as $X_0$ is positive definite and $Y_0$ is positive semidefinite. Now for $Y$ to be PSD we must have $x^* Y x \geq 0$. By suitably choosing vectors $x$ we can also show that if $Y_1 \neq 0$ then $Y \not \geq 0$ hence $Y_1 = 0$ also.
Thus
$$
\begin{aligned}
\mathrm{Tr}[X \log(X + Y)] &= \mathrm{Tr}\left[ \begin{pmatrix} X_0 & 0 \\ 0 & 0 \end{pmatrix} \log \begin{pmatrix} X_0 & 0 \\ 0 & Y_1 \end{pmatrix} \right] \\ 
&= \mathrm{Tr}\left[ \begin{pmatrix} X_0 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \log X_0 & 0 \\ 0 & \log Y_1 \end{pmatrix} \right] \\
&= \mathrm{Tr}[X_0 \log X_0] \\
&= \mathrm{Tr}[X \log X].
\end{aligned}
$$
A: I have found an answer for (2) and (3) and yes, they are true.
First note that
$X+Y \geq Y$,
as X is positive semidefinite. From monotonicity of the matrix logarithm
we have
$\log(X+Y) \geq \log(Y)$,
and, as A is positive semidefinite,
$\sqrt{A}\log(X+Y)\sqrt{A} \geq \sqrt{A}\log(Y)\sqrt{A}$
and using the cyclic property of the trace
$Tr(A\log(X+Y)) \geq Tr(A\log(Y))$.
