In the proof of the joint entropy theorem, why are $p_i\lambda_i^j$ the eigenvalues?

Question

From section 11.3.2 of Nielsen & Chuang:

(4) let $\lambda_i^j$ and $\left|e_i^j\right>$ be the eigenvalues and corresponding eigenvectors of $\rho_i$. Observe that $p_i\lambda_i^j$ and $\left|e_i^j\right>$ are the eigenvalues and eigenvectors of $\sum_ip_i\rho_i$ and thus \begin{align}S\left(\sum_ip_i\rho_i\right) &= -\sum_{ij}p_i\lambda_i^j\log p_i\lambda_i^j \\ &= - \sum_ip_i\log p_i - \sum_ip_i\sum_j\lambda_i^j\log\lambda_i^j \\ &= H\left(p_i\right) + \sum_ip_iS\left(\rho_i\right)\end{align}

So this lemma is used to prove the joint entropy of $S(\rho_{AB})$ with $\rho_{AB}$ equal to $$\rho_{AB} = \sum_i p_i (\rho_i)_A \otimes (|e_i\rangle\langle e_i|)_B.$$ In the attached picture (the lemma), they do it first for only the A-system $\rho_A = \sum_i p_i (\rho_i)_A$, and then they say that this result directly leads to the same result for the above described AB-system (exactly the same?).

First of all, I don't understand how they go from the first to the second line. I understand the fact that $p_i \lambda^j_i $ are the eigenvalues, but I really don't see how.

The second question being, how can you use this result to infer the same result for $S(\rho_{AB})$? I mean you have your extra system in $H_B$ "hanging" on your A-system and I don't see how this just magically disappears or eventually comes down to a factor of 1 when explicitly calculating $S(\rho_AB)$.

I feel pretty silly because this is already the proof aka explanation in the book Nielsen & Chuang and I don't even get it. So maybe explain it really simple and down to Earth for me please.

Answer 1

I don't understand how they go from the first to the second line

So, you're starting from $-\sum_{ij}p_i\lambda^j_i\log(p_i\lambda^j_i)$. Remember that $\log(ab)=\log(a)+\log(b)$, so this is the same as $$ -\sum_{ij}p_i\lambda^j_i\log(p_i)-\sum_{ij}p_i\lambda^j_i\log(\lambda^j_i) $$ For the first term, do the sum over $j$: $\sum_j\lambda^j_i=\text{Tr}(\rho_i)=1$. This gives you $$ -\sum_{i}p_i\log(p_i)-\sum_{i}p_i\sum_j\lambda^j_i\log(\lambda^j_i), $$ which was the second line you were after.

how can you use this result to infer the same result for $S(\rho_{AB})$

Observe that the eigenvectors of $\rho_{AB}$ are $|e^j_i\rangle|e_i\rangle$ with eigenvalue $p_i\lambda^j_i$. Hence, $$ S(\rho_{AB})=-\sum_{ij}p_i\lambda^j_i\log(p_i\lambda^j_i), $$ and you're back to that first line again...

PS for anybody else coming in and trying to make sense of the question, it is important to state that a condition of the theorem that is being proven is that the $\rho_i$ have support on orthogonal subspaces. This way, if a given $\rho_i$ has an eigenvector $|e^j_i\rangle$ with a non-zero eigenvalue $\lambda^j_i$, then $|e^j_i\rangle$ is also an eigenvector of all other $\rho_k$ for $k\neq i$, but $\lambda^j_k=0$. That is the key reason why we can write that the eigenvalues of $\sum_ip_i\rho_i$ are $p_i\lambda^j_i$.

Answer 2

That lemma has condition that all $p_i$ ($1\le i \le n$) have support on orthogonal subspaces. This means that there is a decomposition $H = \oplus_i H_i$ of the space $H$ such that $p_i(H_j) = 0$ if $i\neq j$ and $p_i(v_i)\neq 0$ if $v_i \in H_i$. Note that there can be component $H_0$ such that $p_i(H_0)=0$ for all $i \ge 1$.
Now there is a little oversimplification in their explanation. In fact we pick only nonzero eigenvalues $\lambda_i^j$, so vectors $|e_i^j\rangle$ lay in $H_i$. Now it is easy to see that $\sum_ip_i\rho_i |e_k^j\rangle = p_k\lambda_k^j|e_k^j\rangle$. Moreover, the union of all vectors $|e_i^j\rangle$ span the support of $\sum_ip_i\rho_i$. So this union is a complete set of orthogonal eigenvectors (excluding those from $H_0$).

As for your second question, we apply lemma to the states $\rho_i^\prime = (\rho_i)_A \otimes (|e_i\rangle\langle e_i|)_B$ on the system $AB$. That lemma is general, it is not "just for the A-system". You can check that all $\rho_i^\prime$ indeed have orthogonal support.