Proof of Theorem 2.7, p. 14, observations (i)–(ii) — diagonal blocks are scalar, the $S_0$ rows and columns vanish

Let $\mathcal A \subseteq \mathbb C^{n\times n}$ be a matrix ∗-algebra, and let $\mathcal B$ be an inclusionwise maximal commutative ∗-subalgebra of $\mathcal A$ that is in diagonal form $\mathcal B = \{\lambda_1 I_1 + \dots + \lambda_k I_k : \lambda_i \in \mathbb C\}$ for a partition $[n] = S_0 \cup S_1 \cup \dots \cup S_k$ with $S_1, \dots, S_k$ nonempty, $I_i$ the zero-one diagonal matrix of $S_i$. Then for every $X \in \mathcal A$:

1. (i) for each $i = 1, \dots, k$ there is $c \in \mathbb C$ with $I_i X I_i = c\, I_i$, and $I_0 X I_0 = 0$;
2. (ii) $I_0 X = 0$ and $X I_0 = 0$.

In the paper's block notation $X_{ij}$ (the restriction of $I_iXI_j$ to rows $S_i$ and columns $S_j$): $X_{ii}$ is a multiple of the identity, $X_{00} = 0$, and $X_{0i}$, $X_{i0}$ are zero for all $i = 0, \dots, k$. Consequently $I_1 + \dots + I_k$ acts as the identity on $\mathcal A$.

Formalization Note The block $X_{ij}$ is represented by the full $n\times n$ product $I_iXI_j$, which carries the same entries padded with zeros. The paper's "we may assume $\mathcal B$ is in diagonal form" (after a unitary change of basis supplied by Theorem 2.2) is taken as the hypothesis IsDiagForm B S. Since $I_0 + I_1 + \dots + I_k = I$, item (ii) for all $i$ is exactly $I_0X = 0 = XI_0$. Partition indices are 0-based, with none for $S_0$.