Lemma 41.4.9.label Let $A$ be a $C^{*}$-algebra and $P \in M_{n}(A)$ be positive, then there exists $\bracsn{a_{k, j}}_{j, k = 1}^{n}$ such that:
- (1)
$P = \sum_{k = 1}^{n} \sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j})$.
- (2)
For each $1 \le k \le n$,
\[\sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j}) = \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^{*}\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}\ge 0\]
Proof, [Lemma 3.13, Po02]. Since $P \ge 0$, there exists $\bracsn{a_{j, k}}_{j, k = 1}^{n}$ such that
\begin{align*}P&= \braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}}^{*}\braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}}\\&= \sum_{i, j, k, l = 1}^{n} (E_{ij}\otimes a_{i, j})^{*}(E_{kl}\otimes a_{k, l}) = \sum_{i, j, k, l = 1}^{n} (E_{ji}E_{kl}) \otimes (a_{i, j}^{*} a_{k, l})\end{align*}
For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl}= E_{il}$ if $i = k$, and $E_{ji}E_{kl}= 0$ otherwise. Thus after relabeling,
\[P = \sum_{j, k, l = 1}^{n}E_{jl}\otimes (a_{k, j}^{*}a_{k, l}) = \sum_{k = 1}^{n} \sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j})\]
Finally, for each $1 \le k \le n$, undoing the product shows that
\begin{align*}\sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j})&= \sum_{i, j = 1}^{n} (E_{ik}E_{kj}) \otimes (a_{k, i}^{*}a_{k, j}) \\&= \sum_{i, j = 1}^{n} (E_{ki}\otimes a_{k, i})^{*} (E_{kj}\otimes a_{k, j}) \\&= \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^{*}\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}\ge 0\end{align*}
$\square$
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