\section{Completely Positive and Completely Bounded Maps}
\label{section:completely-positive-bounded}

\begin{proposition}
\label{proposition:matrix-cstar}
    Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, 
    \[
    M_n(A) \times M_n(A) \to M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab
    \]
    and the involution

    \[
    *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^*
    \]

    then there exists a unique norm on $M_n(A)$ such that $M_n(A)$ is a $C^*$-algebra.
\end{proposition}
\begin{proof}
    After applying a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $A$ is unital. 
    
    By the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, there exists a faithful representation $(H, \pi)$ of $A$. Under the identification that $H^n = \complex^n \otimes H$, $\pi$ induces a faithful unital *-representation
    \[
    \pi_n: M_n(A) \to B(H^n) \quad \pi_n(M \otimes a)(\lambda \otimes \xi) = (M\lambda) \otimes (\pi(a)\xi)
    \]

    whose image is a uniformly closed subalgebra of $B(H^n)$. As such, $B(H^n)$ induces a norm on $M_n(A)$, making it a $C^*$-algebra. 

    By \autoref{corollary:c-star-unique-norm}, this norm is unique.
\end{proof}

\begin{definition}[Extended Map]
\label{definition:matrix-extended-map}
    Let $A$ be a unital $C^*$-algebra, $B$ be a $C^*$-algebra, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$, then the mapping
    \[
    \phi_n = \text{Id}_{M_n(\complex)} \otimes \phi: M_n(S) \to M_n(B)
    \]

    is the \textbf{extended map}/\textbf{amplification} of $\phi$ to $M_n(S)$. 
\end{definition}


\begin{proposition}[Tomiyama]
\label{proposition:tomiyama-matrix-norm-bound}
    Let $A, B$ be unital $C^*$-algebras and $S \subset A$ be an operator system, then:
    \begin{enumerate}
        \item For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(A)$, 
        \[
        \norm{a} \le \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} \le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)} 
        \]

        where all inequalities are sharp. 
        \item For any bounded linear map $\phi: S \to B$, $\norm{\phi_n}_{L(M_n(S); M_n(B))} \le n\norm{\phi}_{L(S; B)}$. 
    \end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Exercise 3.10]{Paulsen}}}. ]
    (1): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $A$ is a $C^*$-subalgebra of $B(H)$. 

    For each $\xi = (\xi_1, \cdots, \xi_n), \eta = (\eta_1, \cdots, \eta_n) \in H^n$,
    \begin{align*}
        \dpn{a\xi, \eta}{H^n} &= \sum_{i, j = 1}^n \dpn{(E_{ij} \otimes a_{ij})\xi, \eta}{H^n} = \sum_{i, j = 1}^n \dpn{a_{ij}\xi_j, \eta_i}{H} \\
        |\dpn{a\xi, \eta}{H^n}| &\le \sum_{i, j = 1}^n \norm{a_{ij}}_A\norm{\xi_j}_H\norm{\eta_i}_H \\
        &= \sum_{i, j = 1}^n\dpn{\norm{a_{ij}}_AE_{ij}(\norm{\xi_1}_H, \cdots, \norm{\xi_n}_H), (\norm{\eta_1}_H, \cdots, \norm{\eta_n}_H)}{\complex^n} \\ 
        &\le \norm{\xi}_{H^n} \cdot \norm{\eta}_{H^n} \cdot \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} 
    \end{align*}

    The sharp case is given by the identity.

    For each $\xi = (\xi_1, \cdots, \xi_n) \in \complex^n$, by the \hyperref[Cauchy-Schwarz Inequality]{proposition:cauchy-schwarz}, 
    \begin{align*}
        \norm{\sum_{i, j = 1}^n \norm{a_{ij}}_AE_{ij}\xi}_{\complex^n}^2 &= \sum_{i = 1}^n \abs{\sum_{j = 1}^n \norm{a_{ij}}_{A}\xi_j}^2 \\
        &\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}\braks{\sum_{j = 1}^n |\xi_j|^2} \\
        \norm{\sum_{i, j = 1}^n \norm{a_{ij}}_AE_{ij}\xi}_{\complex^n} &\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}^{1/2} \cdot \norm{\xi}_{\complex^n}
    \end{align*}

    The sharp case is given by operators of the form $E_{ij} \otimes b$ for $b \in A$. 
    
    Finally, 
    \[
    \sum_{i, j = 1}^n \norm{a_{ij}}_A^2 = \sum_{i, j = 1}^n \norm{E_{ii} \cdot a \cdot E_{jj}}_{M_n(A)}^2 \le n^2\norm{a}_{M_n(A)}^2
    \]

    and
    \[
    \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)}
    \]

    The sharp case is given by operators of the form $\sum_{i, j = 1}^n E_{ij} \otimes b$ for $b \in A$. 
    
    (2): For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(S)$, 
    \begin{align*}
        \normn{\phi(a)}_{M_n(B)}^2 &\le \sum_{i, j = 1}^n \norm{\phi(a_{ij})}_{B}^2 \le \norm{\phi}_{L(S; B)}^2\sum_{i, j = 1}^n \norm{a_{ij}}_A^2 \\
        &\le n^2\norm{\phi}_{L(S; B)}^2\norm{a}_{M_n(A)}^2
    \end{align*}
\end{proof}

\begin{proposition}
\label{proposition:positive-column-test}
    Let $A$ be a unital $C^*$-algebra and $b = \sum_{i, j = 1}^n E_{ij} \otimes b_{ij} \in M_n(A)$, then the following are equivalent:
    \begin{enumerate}
        \item $b$ is positive in $M_n(A)$. 
        \item For each $\seqf{a_j} \subset A$, $\sum_{i, j = 1}^n a_ib_{ij}a_j^*$ is positive in $A$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1) $\Rightarrow$ (2): For any $\seqf{a_j} \subset A$, let $a = \sum_{i = 1}^n E_{1i} \otimes a_i$, then
    \begin{align*}
        0 \le aba^* &= \braks{\sum_{k = 1}^n E_{k1} \otimes a_k^*}\braks{\sum_{j, k = 1}^n E_{jk} \otimes b_{jk}}\braks{\sum_{i = 1}^n E_{i1} \otimes a_i} \\
        &= \sum_{j, k = 1}^n (E_{1j} \cdot E_{jk} \cdot E_{k1}) \otimes (a_jb_{jk}a_k^*) = E_{11} \otimes \sum_{i, j = 1}^n a_ib_{ij}a_j^*
    \end{align*}
    
    so $\sum_{i, j = 1}^n a_ib_{ij}a_j^* \ge 0$ in $A$. 

    (2) $\Rightarrow$ (1): For each state $\phi \in S(A)$, let $(H_\phi, \pi_\phi, \xi_\phi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\phi$. Let $H = [l^2(S(A)); H_\phi]$,
    \[
    \pi : A \to B(H) \quad \pi(a)(\xi)_\phi = \pi_\phi(a)(\xi_\phi)
    \]

    be the universal representation of $A$, and identify $A$ as a $C^*$-subalgebra of $B(H)$ through $\pi$. 
    
    For any $\seqf{a_j} \subset A$ and $\phi \in S(A)$, 
    \begin{align*}
    0 \le \sum_{i, j = 1}^n \dpn{a_i^*b_{ij}a_j , \phi}{A} &= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij}a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} \\
    &= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi}
    \end{align*}

    Identify $H_\phi$ as a subspace of $H$ and let $\xi = (\pi_\phi(a_1)\xi_\phi, \cdots, \pi_\phi(a_n)\xi_\phi) \in H^n$, then 
    \[
    0 \le \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} = \dpn{b\xi, \xi}{H^n}
    \]

    As $(H_\phi, \pi_\phi)$ is a cyclic representation, the above implies that $\dpn{b\eta, \eta}{H^n} \ge 0$ for all $\eta \in H_\phi^n$. Since this holds for all $\phi \in S(A)$, $b \ge 0$ in $M_n(A)$. 
\end{proof}


\begin{definition}[Completely Positive Map]
\label{definition:completely-positive-map}
    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$, then $\phi$ is \textbf{$n$-positive} if $\phi_n: M_n(S) \to M_n(B)$ is positive. The map $\phi$ is \textbf{completely positive} if $\phi$ is $n$-positive for all $n \in \natp$. 
\end{definition}

\begin{definition}[Completely Bounded Map]
\label{definition:completely-bounded-map}
    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a linear map, then
    \[
    \norm{\phi}_{\text{cb}} = \sup_{n \in \natp}\norm{\phi_n}_{L(M_n(S); M_n(B))}
    \]

    is the \textbf{completely bounded norm} of $\phi$. 
    
    The map $\phi$ is \textbf{completely bounded} if $\norm{\phi}_{\text{cb}} < \infty$, and \textbf{completely contractive} if $\norm{\phi}_{\text{cb}} \le 1$. 
\end{definition}

\begin{lemma}
\label{lemma:completely-mental-gymnastics}
    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$. 
    \begin{enumerate}
        \item If $\phi$ is $n$-positive, then $\phi$ is $k$-positive for all $1 \le k \le n$. 
        \item For each $1 \le k \le n$, $\norm{\phi_k}_{L(M_k(S); M_k(B))} \le \norm{\phi_n}_{L(M_n(S); M_n(B))}$. 
    \end{enumerate}
\end{lemma}
\begin{proof}
    For any $n \in \natp$ and $1 \le k < n$, there is a natural *-homomorphism
    \[
    e: M_k(\complex) \to M_n(\complex) \quad M \mapsto \begin{bmatrix}
        M & 0 \\
        0 & 0
    \end{bmatrix}
    \]

    such that the following diagram commutes:
    \[
    \xymatrix{
    M_k(S) \ar@{->}[d]_{e \otimes \text{Id}_S} \ar@{->}[r]^{\phi_k} & M_k(B) \ar@{->}[d]^{e \otimes \text{Id}_B} \\
    M_n(S) \ar@{->}[r]_{\phi_n} & M_n(B)
    }
    \]

    (1): Since $e \otimes \text{Id}_S$ is the restriction of a *-homomorphism, it is positive. On the other hand, for any $x \in M_k(B)$, $x \ge 0$ if and only if $(e \otimes \text{Id}_B)(x) \ge 0$. Given that $\phi_n$ is positive, $\phi_k$ must also be positive. 

    (2): The maps $e \otimes \text{Id}_S$ and $e \otimes \text{Id}_B$ are both isometric. As such, the norm of $\phi_k$ is bounded above by the norm of $\phi_n$. 
\end{proof}


\begin{proposition}
\label{proposition:completely-positive-norm}
    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then
    \[
    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
    \]
\end{proposition}
\begin{proof}
    After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. 
    
    It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then
    \[
    \begin{bmatrix}
            1_{M_n(A)} &T \\
            T^* &1_{M_n(A)}
    \end{bmatrix} \ge 0
    \]

    in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, 
    \[
    \phi_{2n}\begin{bmatrix}
            1_{M_n(A)} &T \\
            T^* &1_{M_n(A)}
    \end{bmatrix} = \begin{bmatrix}
            \phi_n(1_{M_n(A)}) &\phi_n(T) \\
            \phi_n(T)^* & \phi_n(1_{M_n(A)})
    \end{bmatrix} \ge 0
    \]

    First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then
    
    \begin{align*}
        \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\
        1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\
        \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} 
    \end{align*}
    
    By the \hyperref[Schur complement condition]{proposition:schur-complement-positive}, 
    \begin{align*}
    \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\
    &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\
    \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 
    \end{align*}

    and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$.
    
    Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, 
    \[
    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
    \]
\end{proof}


\begin{proposition}
\label{proposition:unital-2-positive-contractive}
    Let $A, B$ be unital $C^*$-algebras, $S \subset A$ be an operator system, $\phi: S \to B$ be a unital $2$-positive map, then $\norm{\phi}_{L(S; B)} \le 1$. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 3.2]{Paulsen}}}. ]
    Let $x \in S$ with $\norm{x}_A \le 1$, then
    \[
    \phi_2\begin{bmatrix}
            1 &x \\
            x^* &1
        \end{bmatrix} = \begin{bmatrix}
            1 &\phi(x) \\
            \phi(x)^* &1
        \end{bmatrix} \ge 0
    \]

    By \autoref{lemma:norm-matrix-test}, $\norm{\phi(x)}_B \le 1$. 
\end{proof}

\begin{proposition}[Kadison-Schwarz Inequality]
\label{proposition:cauchy-schwarz-2-positive}
    Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a $2$-positive map, then for each $x \in A$,
    \[
    \phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B \phi(x^*x)
    \]
\end{proposition}
\begin{proof}
    Since
    \[
    \begin{bmatrix}
            1_A &x \\
            x^* &x^*x
        \end{bmatrix} = \begin{bmatrix}
            1_A &x \\
            0 &0
        \end{bmatrix}^*\begin{bmatrix}
            1_A &x \\
            0 &0
        \end{bmatrix} \ge 0
    \]

    and $\phi$ is $2$-positive, 
    \[
    \begin{bmatrix}
            1_B\norm{\phi(1_A)}_B &\phi(x) \\
            \phi(x)^* &\phi(x^*x)
        \end{bmatrix} \ge \begin{bmatrix}
            \phi(1_A) &\phi(x) \\
            \phi(x)^* &\phi(x^*x)
        \end{bmatrix} = \phi_2\begin{bmatrix}
            1_A &x \\
            x^* &x^*x
        \end{bmatrix} \ge 0
    \]

    By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B\phi(x^*x)$. 
\end{proof}



\begin{example}
    \label{example:transpose-not-completely-positive}
    Let $T: M_2(\complex) \to M_2(\complex)$ be the transpose map, then $T$ is positive, but not $2$-positive. 
\end{example}
\begin{proof}
    Let
    \[
    C = \begin{bmatrix}
            1&0&0&1 \\
            0&0&0&0 \\
            0&0&0&0 \\ 
            1&0&0&1
    \end{bmatrix} = (1, 0, 0, 1) \otimes (1, 0, 0, 1) \ge 0
    \]

    then
    \[
    T_2(C) = \begin{bmatrix}
            1&0&0&0 \\
            0&0&1&0 \\
            0&1&0&0 \\
            0&0&0&1
    \end{bmatrix}
    \]

    is invertible, with negative determinant, so $T_2(C)$ is not positive. 
\end{proof}


\begin{example}[Choi]
    \label{example:choi-roots-of-unity}

    Let $A$ be a $C^*$-algebra, $\lambda \in \complex$ with $|\lambda| = 1$, and $U_\lambda = \text{diag}(\lambda^1, \cdots, \lambda^n)$, then:
    \begin{enumerate}
        \item For each $X = \sum_{i, j = 1}^n E_{ij} \otimes x_{ij} \in M_n(A)$, 
        \[
        U_\lambda^* X U_{\lambda} = \sum_{i, j = 1}^n E_{ij} \otimes (\lambda^{j - i}x_{ij})
        \]
        \item The mapping $\phi: M_n(A) \to M_n(A)$ defined by $\phi(X) = n\text{diag}(X) - X$ is completely positive.
        \item The mapping $\psi: M_n(A) \to M_n(A)$ defined by $\psi(X) = (n-1)\text{diag}(X) - X$ is not completely positive. 
    \end{enumerate}
\end{example}
\begin{proof}[Proof, {{\cite[Exercise 3.6]{Paulsen}}}. ]
    Throughout this proof, if $A$ is not unital, view $A$ as an operator system in its unitisation, and $M_n(A)$ as an operator system in $M_n$ of its unitisation. 

    (1): Let $X = \sum_{i, j = 1}^n E_{ij} \otimes x_{ij} \in M_n(A)$, then
    \begin{align*}
        U_\lambda^* X U_{\lambda} &= \sum_{i, j, k, l = 1}^n (E_{ii} \otimes \lambda^{-i})(E_{jk} \otimes x_{jk})(E_{ll} \otimes \lambda^l) \\
        &= \sum_{j, k = 1}^n (E_{jj} \cdot E_{jk} \cdot E_{kk}) \otimes (\lambda^{-j} \cdot x_{jk} \cdot \lambda^k) \\
        &=  \sum_{j, k = 1}^n E_{jk} \otimes (\lambda^{k-j} x_{jk})
    \end{align*}
    
    (2): Let $\omega = e^{2\pi i /n}$, then the $n$-th roots of unity are $\bracsn{\omega^k}_{k = 1}^{n}$. By (1), for any $X = \sum_{i, j = 1}^n E_{ij} \otimes x_{ij}$, 
    \[
        \sum_{k = 1}^{n} U_{\omega^k}^*XU_{\omega^k} = \sum_{i, j, k = 1}^n E_{ij} \otimes \omega^{k(j - i)}x_{ij}
    \]

    where
    \[
    \sum_{k = 1}^n \omega^{k(j-i)} = \begin{cases}
        0 &j \ne i \\
        n &j = i
    \end{cases}
    \]

    so
    \begin{align*}
    \sum_{k = 1}^n U_{\omega^k}^*XU_{\omega^k} &= \sum_{i = 1}^n E_{ii} \otimes (nx_{ii}) = n\text{diag}(X) \\
    \phi(X) = \sum_{k = 1}^{n-1} U_{\omega^k}^*XU_{\omega^k} &= n\text{diag}(X) - U_1^*XU_1 = n\text{diag}(X) - X
    \end{align*}

    By \autoref{proposition:multiply-completely-bounded}, $\phi$ is completely positive.
    
    (3): Let $x \in A$ with $x \ge 0$ and $x \ne 0$, and $X = \sum_{i, j = 1}^n E_{ij} \otimes x$, then
    \[
    X = \sum_{i, j = 1}^n E_{ij} \otimes x = \braks{\sum_{i = 1}^n E_{1i} \otimes x^{1/2}}^*\braks{\sum_{i = 1}^n E_{1i} \otimes x^{1/2}} \ge 0
    \]

    However, if $V = \sum_{i = 1}^n E_{i1} \otimes 1_A$, then
    \begin{align*}
        V^*\psi(X)V &= (n-1)V^*\text{diag}(X)V -  V^*XV \\
        V^*\text{diag}(X)V &= \sum_{i, j, k = 1}^n (E_{1i} \otimes 1_A)(E_{jj} \otimes x)(E_{k1} \otimes 1_A) \\
        &= \sum_{i = 1}^n (E_{1i} \otimes 1_A)(E_{ii} \otimes x)(E_{i1} \otimes 1_A) = n(E_{11} \otimes x) \\
        V^*XV &= \sum_{i, j, k, l = 1}^n  (E_{1i} \otimes 1_A)(E_{jk} \otimes x)(E_{l1} \otimes 1_A) \\
        &= \sum_{i, j = 1}^n (E_{1i} \otimes 1_A)(E_{ij} \otimes x)(E_{j1} \otimes 1_A) = n^2(E_{11} \otimes x) \\
        V^*\psi(X)V &= (n-1)n(E_{11} \otimes x) - n^2(E_{11} \otimes x) = -n(E_{11} \otimes x)
    \end{align*}
    
    As $x \ne 0$, $-n(E_{11} \otimes x)$ is not positive. Therefore $\psi(X)$ is not positive, and $\psi$ is not a positive map. 
    
\end{proof}

\begin{proposition}
\label{proposition:cstar-reverse-positive}
    Let $A$ be a $C^*$-algebra, $A^{\text{op}}$ denote $A$ with the same norm and *-operation, but with the product
    \[
    A^{\text{op}} \times A^{\text{op}} \to A^{\text{op}} \quad (x, y) \mapsto x \circ y = y \cdot x
    \]

    then:
    \begin{enumerate}
        \item $A^{\text{op}}$ is a $C^*$-algebra, $\text{Id}: A \to A^{\text{op}}$ is positive.
        \item The transpose map $M_n(\complex) \to M_n(\complex)^{\text{op}}$ is a *-isomorphism. 
        \item $\text{Id}: M_2(\complex) \to M_2(\complex)^{\text{op}}$ is not 2-positive. 
        \item (\textbf{Walter}) For each $U, V \in U(A)$ and $X \in A$, 
        \[
        \begin{bmatrix}
            1_A & U & X \\
            U^* & 1_A & V \\
            X^* & V^* & 1_A
        \end{bmatrix} \ge 0
        \]

        if and only if $X = UV$. 
        \item $\text{Id}: A \to A^{\text{op}}$ is completely positive if and only if $A$ is commutative. 
    \end{enumerate}
\end{proposition}
\begin{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^* & V^*U^*U & 1_A
        \end{bmatrix}\\
        &= \begin{bmatrix}
            1_A & U & UV \\
            U^* & 1_A & U^*UV \\
            V^*U^* & V^*U^*U & 1_A
        \end{bmatrix}
    \end{align*}
    
    so the given matrix is positive.

    (4, $\Rightarrow$): Suppose that the given matrix is positive. By \autoref{lemma:cstar-positive-matrix-decompose-sum}, 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), 
    \[
    \begin{bmatrix}
        1_A & U & UV \\
        U^* & 1_A & V \\
        V^*U^* & V^* & 1_A
    \end{bmatrix} \ge 0
    \]

    If the identity is completely positive, then
    \[
    \begin{bmatrix}
        1_A & U & V \circ U \\
        U^* & 1_A & V \\
        U^* \circ V^* & V^* & 1_A
    \end{bmatrix} = \begin{bmatrix}
        1_A & U & UV \\
        U^* & 1_A & V \\
        V^*U^* & V^* & 1_A
    \end{bmatrix} \ge 0
    \]

    in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V = V \circ U = UV$. Therefore $A$ is commutative.
\end{proof}