I'm trying to understand the steps in a certain proof, where I think the following claim is used: $S( |i \rangle \langle i| \otimes \rho) = S(\rho)$, where $S()$ is the entropy, $|i \rangle$ is a state from some orthonormal basis, and $\rho$ is a density operator. Here's my attempt to see why this holds:
$$
\begin{align}
S( |i \rangle \langle i| \otimes \rho) &= -tr\left[\left(|i \rangle \langle i| \otimes \rho\right) \times \log \left(|i \rangle \langle i| \otimes \rho\right)\right]\\
&= -tr[(|i \rangle \langle i| \times \log |i \rangle \langle i|) \otimes \rho \log \rho]\\
&= -tr[(|i \rangle \langle i| \times \log |i \rangle \langle i|)] \times tr[\rho \log \rho]\\
&= 0 \times S(\rho)
\end{align}$$
so I am clearly making a mistake somewhere, that $0$ should not be there. The first equality comes from the definition of $S()$, the second because $(A \otimes B)(C \otimes D) = AC \otimes BD$, the third because $tr(A \otimes B) = tr(A) \times tr(B)$. Are any of these incorrect?
Best Answer
$$\log(A\otimes B)=\log A \otimes I + I\otimes \log B$$ You can show this by diagonalizing $A$ and $B$ and using the idea that the matrix log of a diagonal matrix is the log of the elements. In components,
$$\log(A\otimes B)_{ii',jj'}=\log(A_i B_{i'})I_{ij}I_{i'j'}=(\log A_i I_{ij})I_{i'j'}+I_{ij}(\log B_{i'}I_{i'j'})$$
Then if you use your correct rules for tensor product manipulation in your question you will prove the statement.