Proof. (4, $\Leftarrow$): If $X = UV$, then
\begin{align*}\begin{bmatrix}1_{A}&U&UV \\ 0&0&0 \\ 0&0&0\end{bmatrix}^{*}\begin{bmatrix}1_{A}&U&UV \\ 0&0&0 \\ 0&0&0\end{bmatrix}&= \begin{bmatrix}1_{A}&0&0 \\ U^{*}&0&0 \\ V^{*}U^{*}&0&0\end{bmatrix}\begin{bmatrix}1_{A}&U&UV \\ 0&0&0 \\ 0&0&0\end{bmatrix} \\&= \begin{bmatrix}1_{A}&U&UV \\ U^{*}&1_{A}&U^{*}UV \\ V^{*}U^{*}&VV^{*}U^{*}&_{A}\end{bmatrix}\\&= \begin{bmatrix}1_{A}&U&UV \\ U^{*}&1_{A}&U^{*}UV \\ V^{*}U^{*}&V^{*}U^{*}U&A_{A}\end{bmatrix}\\&= \begin{bmatrix}1_{A}&U&UV \\ U^{*}&1_{A}&V \\ V^{*}U^{*}&V^{*}&A_{A}\end{bmatrix}\end{align*}
so the given matrix is positive.
(4, $\Rightarrow$): Suppose that the given matrix is positive. By Lemma 41.5.9, there exists $\bracsn{a_{k, j}}_{j, k = 1}^{3} \subset A$ such that
\[\begin{bmatrix}1_{A}&U&X \\ U^{*}&1_{A}&V \\ X^{*}&V^{*}&1_{A}\end{bmatrix} = \sum_{k = 1}^{3}\begin{bmatrix}a_{k, 1}^{*}a_{k,1}&a_{k, 1}^{*}a_{k,2}&a_{k, 1}^{*}a_{k,3}\\ a_{k, 2}^{*}a_{k,1}&a_{k, 2}^{*}a_{k,2}&a_{k, 2}^{*}a_{k,3}\\ a_{k, 3}^{*}a_{k,1}&a_{k, 3}^{*}a_{k,2}&a_{k, 3}^{*}a_{k,3}\\\end{bmatrix}\]
Using an elementary column operation,
\[\begin{bmatrix}1_{A}&U - U&X \\ U^{*}&1_{A} - U^{*}U&V \\ X^{*}&V^{*} - X^{*}U&1_{A}\end{bmatrix} = \sum_{k = 1}^{3}\begin{bmatrix}a_{k, 1}^{*}a_{k,1}&a_{k, 1}^{*}(a_{k,2}- a_{k, 1}U)&a_{k, 1}^{*}a_{k,3}\\ a_{k, 2}^{*}a_{k,1}&a_{k, 2}^{*}(a_{k,2}- a_{k, 1}U)&a_{k, 2}^{*}a_{k,3}\\ a_{k, 3}^{*}a_{k,1}&a_{k, 3}^{*}(a_{k,2}- a_{k, 1}U)&a_{k, 3}^{*}a_{k,3}\\\end{bmatrix}\]
and an elementary row operation,
\[\begin{bmatrix}1_{A}&0&X \\ 0&0&V - U^{*}X \\ X^{*}&V^{*} - X^{*}U&1_{A}\end{bmatrix}\]
is equal to
\[\sum_{k = 1}^{3}\begin{bmatrix}a_{k, 1}^{*}a_{k,1}&a_{k, 1}^{*}(a_{k,2}- a_{k, 1}U)&a_{k, 1}^{*}a_{k,3}\\ (a_{k, 2}- a_{k, 1}U)^{*}a_{k,1}&(a_{k, 2}- a_{k, 1}U)^{*}(a_{k,2}- a_{k, 1}U)&(a_{k, 2}- a_{k, 1}U)^{*}a_{k,3}\\ a_{k, 3}^{*}a_{k,1}&a_{k, 3}^{*}(a_{k,2}- a_{k, 1}U)&a_{k, 3}^{*}a_{k,3}\\\end{bmatrix}\]
In particular,
\[\sum_{k = 1}^{3} (a_{k, 2}- a_{k, 1}U)^{*}(a_{k,2}- a_{k, 1}U) = 0\]
so $a_{k, 2}= a_{k, 1}U$ for $1 \le k \le 3$. Thus
\[\begin{bmatrix}1_{A}&0&X \\ 0&0&V - U^{*}X \\ X^{*}&V^{*} - X^{*}U&1_{A}\end{bmatrix} = \sum_{k = 1}^{3}\begin{bmatrix}a_{k, 1}^{*}a_{k,1}&0&a_{k, 1}^{*}a_{k,3}\\ 0&0&0 \\ a_{k, 3}^{*}a_{k,1}&0&a_{k, 3}^{*}a_{k,3}\\\end{bmatrix}\]
and $V = U^{*}X$, so $X = UV$.
(5): Let $U, V \in U(A)$, then by (4),
\[\begin{bmatrix}1_{A}&U&UV \\ U^{*}&1_{A}&V \\ V^{*}U^{*}&V^{*}&A_{A}\end{bmatrix} \ge 0\]
If the identity is completely positive, then
\[\begin{bmatrix}1_{A}&U&VU \\ U^{*}&1_{A}&V \\ U^{*}V^{*}&V^{*}&A_{A}\end{bmatrix} = \begin{bmatrix}1_{A}&U&U \circ V \\ U^{*}&1_{A}&V \\ V^{*} \circ U^{*}&V^{*}&A_{A}\end{bmatrix} \ge 0\]
in $M_{3}(A^{\text{op}})$. By (4), $VU = U \circ V$. Therefore $A$ is commutative.$\square$
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