Theorem 8.3 in Paulsen's 'Completely bounded maps and operator algebras". I'm trying to understand the proof of theorem 8.3 in Paulsen's book "Completely bounded maps and operator algebras". Here is (part of) the proof:

Why can we conclude that
$$\Phi\begin{pmatrix} p & 0 \\ 0 & 0\end{pmatrix}= \begin{pmatrix} * & 0 \\ 0 & 0\end{pmatrix}?$$
I guess it might be sufficient to show that
$$\begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} \le \begin{pmatrix} a & b \\ b^* & c\end{pmatrix} \le \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\implies b = c = 0$$
where the middle matrix is a positive matrix. However, I can't show this and it should be straightforward.
 A: I'll call the three matrices in question $0$, $A$ and $E_{11}$. For $\xi\in \mathcal H$ the inequality
$$
0\leq \left\langle A\binom{0}{\xi},\binom{0}{\xi}\right\rangle\leq\left\langle E_{11}\binom{0}{\xi},\binom{0}{\xi}\right\rangle
$$
implies $0\leq\langle c\xi,\xi\rangle\leq0$, hence $c=0$.
Moreover,
$$
0\leq\left\langle A\binom{\xi}{\eta},\binom{\xi}{\eta}\right\rangle=\langle a\xi,\xi\rangle+2\mathrm{Re}\langle \xi,b\eta\rangle.
$$
If $b\neq 0$, then there exist $\xi,\eta\in H$ such that $\langle\xi,b\eta\rangle\neq 0$. Multiplying $\eta$ by a suitable scalar makes this inner product negative and arbitrarily small, while $\langle a\xi,\xi\rangle$ stays unchanged. Thus we can make $\langle a\xi,\xi\rangle+2\mathrm{Re}\langle \xi,b\eta\rangle$ negative, a contradiction.
A: This depends on two basic facts about a positive elements:

*

*its diagonal elements are positive;


*if a diagonal element is zero, then its whole row and column are zero.
If you know that elements in a C$^*$-algebra are of the form $X^*X$, you have that if $\begin{bmatrix} a&b\\ b^*&c\end{bmatrix}\geq0$ then
$$
\begin{bmatrix} a&b\\ b^*&c\end{bmatrix}
=\begin{bmatrix} x&y\\ y^*&z\end{bmatrix}^*\begin{bmatrix} x&y\\ y^*&z\end{bmatrix}
=\begin{bmatrix} x^*x++yy^*&x^*y+yz\\ y^*x+z^*y^*&y^*y+z^*z\end{bmatrix}.
$$
From this you get that diagonal elements are positive. And if a diagonal element is zero, say $y^*y+z^*z=0$, then $y=z=0$.
