Lemma 41.8.1.label Let $H$ be a complex Hilbert space, $A = \sum_{i, j = 1}^{n} E_{ij}\otimes a_{ij}\in M_{n}(B(H))$ be positive, and $\Lambda = \sum_{i, j = 1}^{n} \lambda_{ij}E_{ij}\in M_{n}(B(H))$ be positive, then $S_{A}(\Lambda) = \sum_{i, j = 1}^{n} \lambda_{ij}E_{ij}\otimes a_{ij}$ is also positive.
Proof. Using the Spectral Theorem, assume without loss of generality that there exists $\bracsn{\lambda_j}_{1}^{n} \subset [0, \infty)$ such that $\Lambda = \text{diag}(\lambda_{1}, \cdots, \lambda_{n})$. In which case,
\[S_{A}(\lambda) = \sum_{i = 1}^{n} \lambda_{i} E_{ii}\otimes a_{ii}\]
is a sum of positive operators, and hence positive.$\square$
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