Questions about Stinespring's theorem for completely positive maps I have a question about Stinespring's Theorem:  
Let $A$ be a $C^*$-algebra, $H$ be a complex Hilbert space and $L(H)$ the set of bounded linear operators on H. Let $\Phi: A\to L(H)$ be a completely positive map. Then there exists a complex Hilbert space $K$, a $*$-homomorphism $\pi:A\to L(K)$ and a bounded linear map $V:H\to K$ such that
$$\Phi(a)=V^*\pi(a)V,$$ for all $a\in A$.
Furthermore, we have $\|\Phi\|=\|V\|^2$. 
If A is unital, $\phi$ is unital and if $\phi(e_A)=Id_{L(H)}$, V is isometric. In this case you can consider H as a sub-hilbertspace of K and it is $\phi(a)=P_H\pi(a)_{|H}$ for all $a\in A$. Here is $P_H$ the projection of K onto H.
My questions are: 
Is $V^*V\le Id_{L(H)}$ true in general?
It is equivalent to prove that $\langle Id-V^*Vx,x\rangle \ge 0$ for all $x\in H$ and this means $\|x\|\ge \|Vx\|$ for all $x\in H$. But I don't think this is true in general for the V above in Stinespring's Theorem. 
And why is "In this case ... it is $\phi(a)=P_H\pi(a)_{|H}$ for all $a\in A$" true? I only know if $V$ is an isometry, it is $V^*V=Id_{L(H)}$. 
I appreciate your help. Regards
 A: Of course it is not true in general. As you mentioned, $\|V\|^2=\|\Phi\|$, so if $V^*V\leq\text{Id}_{L(H)}$, this implies that $\|\Phi\|\leq1$. 
The equality $\Phi(a)=P_H\,\pi(a)|_H$ is not really an equality; there is a unitary hanging around, but it is not worth writing. What happens is this: if $\Phi$ is unital, then $V$ is an isometry. So $H_0=VH$ is a subspace of $K$, isomorphic to $H$; and if we restrict the codomain of $V$ to $H_0$, then $V$ is bijective. Also, $VV^*\in L(K)$ is a projection, because $VV^*VV^*=VV^*$. For $\xi\in H_0$, we have $\eta=V^*\xi\in H$, and $VV^*\xi=\xi$. For  $\xi\in H_0^\perp$ and $\nu\in H_0$, 
$$
\langle VV^*\xi,\nu\rangle=\langle\xi,VV^*\nu\rangle=\langle\xi,\nu\rangle=0.
$$
So $VV^*$ maps $H_0^\perp$ to zero, which shows that $VV^*=P_{H_0}$. Now, for $\xi\in H_0$ and $\eta=V^*\xi\in H$,
$$
P_{H_0}\pi(a)\xi=VV^*\pi(a)V\eta=V\Phi(a)\eta=V\Phi(a)V^*\xi.
$$
So, on $H_0$, $P_{H_0}\pi(a)=V\Phi(a)V^*$ where $V$ is a unitary $H\to H_0$. In other words, up to a unitary that implements the isomorphism between $H$ and $H_0$, 
$$
P_{H_0}\pi(a)|_{H_0}=\Phi(a).
$$
