Trace of a matrix exponential with tensor products, and Von Neumann entropy

Question

$\def\T{\operatorname{Tr}}$ $\def\1{\mathbb{1}}$ Let $H=H_1\otimes H_2\otimes H_3$ be a finite dimensional Hilbert space, and let $\rho_{123}$ be a self-adjoint matrix with $\rho_{123}\geq 0$ (positive eigenvalues) and $\T{\rho_{123}}=1$, so a density matrix.

Furthermore let $i,j,k\in\{1,2,3\}$ different, and define $\rho_i=\T_{jk}(\rho_{123})$ and $\rho_{jk}=\T_{i}(\rho_{123})$, so for example if $\rho=A_1\otimes B_2\otimes \1_3$ then $\rho_{1}=A_1$ and $\rho_{13}=A_1\otimes\1_3$.

I was reading this paper - https://aip.scitation.org/doi/pdf/10.1063/1.1497701?class=pdf (IV, equation 30) - and there is one property that it takes for granted, but I don't find it at all obvious. Even though it doesn't explicitly say it, I think it's implied there's an equality here:

$$ \T\big(e^{\log{\rho_{12}}\otimes\1_3+\1_1\otimes\log{\rho_{23}}} (\log{\rho_{12}}\otimes\1_3+\1_1\otimes\log{\rho_{23}})\big) = \T_{12}\big({\rho_{12} \log{\rho_{12}}}\big) + \T_{23}\big({\rho_{23} \log{\rho_{23}}}\big) $$

I am trying to prove the following generalization of the above property $(A_{12} = \log(\rho_{12}))$

$$ \T_3( e^{A_{12}\otimes \1_3 + \1_1 \otimes B_{12}}) = e^{A_{12}}. $$

If this is true then the above proposition follows since

$$ \T(A_{123}\cdot(B_{12}\otimes \1_3))= \T_{12}(\T_3(A_{123})\cdot B_{12}). $$

I have a feeling like the generalization I am trying to prove is false.

Any different proofs of $$ \T\big(e^{r_{12}+r_{23}}(r_{12}+r_{23})\big) = \T(e^{r_{12}}r_{12}) + \T(e^{r_{23}}r_{23}) $$ are welcome, and any help is appreciated. Thanks.

EDIT:

If the following is true, it solves all my problems $$ S(e^{A_{12}\otimes \1_3 + \1_1 \otimes B_{23}}) = S(e^{A_{12}} \otimes e^{B_{23}}) $$ where $S$ is the Von Neumann entropy, since $$ S(e^{A}\otimes e^{B}) = S(e^{A}) + S(e^B) $$ and this leads to the desired result.

Note that LHS matrix operates on $H$, while the RHS one has a greater dimension. I don't know how to apply the known formula $$ e^{A\otimes \1_2 + \1_1\otimes B} = e^{A} \otimes e^{B} $$ in this case since this equation requires $\1_1$ and $A$ to operate on the same space, and this is not true in my case.

Answer 1

Having looked at the original, here's what I think is going on:

The paper instructs:

A second, more powerful subadditivity inequality... $$ S(\rho_{123}) \leq S(\rho_{12}) + S(\rho_{23}) $$ under the constraint $\operatorname{tr}(\rho_{123}) = 1$. To prove this, choose $A = \rho_{123}$ and $B = e^{\log\rho_{12} + \log \rho_{23}}$ in Klein's inequality to obtain $$ -S(\rho_{123})+ S(\rho_{12}) + S(\rho_{23}) \geq 1 - \operatorname{tr} e^{\log\rho_{12} + \log \rho_{23}} $$

For reference,

Klein's inequality: $$ \operatorname{tr}A(\log A - \log B) \geq \operatorname{tr}(A - B) $$

With that, we apply Klein's inequality as follows: $$ \operatorname{tr}A(\log A - \log B) \geq \operatorname{tr}(A - B) \implies\\ \operatorname{tr}\rho_{123}(\log \rho_{123} - \log [e^{\log\rho_{12} + \log \rho_{23}}]) \geq \operatorname{tr}(\rho_{123} - e^{\log\rho_{12} + \log \rho_{23}}) \implies\\ \operatorname{tr}\rho_{123}(\log \rho_{123}) - \operatorname{tr}\rho_{123}(\log\rho_{12} + \log \rho_{23}) \geq \operatorname{tr}(\rho_{123})- \operatorname{tr}(e^{\log\rho_{12} + \log \rho_{23}}) \implies\\ -S(\rho_{123}) - \operatorname{tr}\rho_{123}(\log\rho_{12} + \log \rho_{23}) \geq 1 - \operatorname{tr}(e^{\log\rho_{12} + \log \rho_{23}}) \implies\\ -S(\rho_{123}) - \operatorname{tr}\rho_{123}\log\rho_{12} - \operatorname{tr}\rho_{123}\log \rho_{23} \geq 1 - \operatorname{tr}(e^{\log\rho_{12} + \log \rho_{23}}) $$

With that in mind, it seems to me that the equality being implied is $$ \operatorname{tr}\rho_{123}\log\rho_{12} = \operatorname{tr}\rho_{12}\log\rho_{12} $$ Or, more explicitly, $$ \operatorname{tr}[\rho_{123} \log[\operatorname{tr}_3(\rho_{123}) \otimes 1_3]] = \operatorname{tr}[\operatorname{tr}_3(\rho_{123}) \log[\operatorname{tr}_3(\rho_{123})]] $$ and similarly for $\rho_{23}$. It seems to me that this equality indeed holds; let me know if you would like a proof of it.