# Connection between Stinespring's factorization theorem and the spectral theorem for bounded operators

I know at least 2 versions of a Spectral theorem for operators, one of them is the following

Theorem: Let H be a separable complex Hilbert space, $A\in L(H)$ self-adjoint ($L(H)$ are the bounded linear operators on a Hilbert space H). Then there exists a $\sigma$-finite measure space $(\Omega ,\Sigma ,\mu)$, a bounded measurable function $f:\Omega \to\mathbb{R}$ and a unitary operator $V:H\to L^2(\Omega)$, such that:
$(VAV^{-1})\varphi =f\cdot \varphi \;\; \mu$-almost everywhere.

Now, I'will read up on dilations, Stinespring and the theory of completely bounded maps between $C^*$-algebras.

I want to know if there is a connection between the theorem above and the Stinespring factorization theorem. But although the theorems look similar, the theorem above don't seem to be a special case of

the theorem:

Let $A$ be a unital $C^*$-algebra, $H$ be a Hilbert space and $L(H)$ as above. For every completely positive map $\Phi: A\to L(H)$ there exists a Hilbert space $K$ and a unital $*$-homomorphism $\pi:A\to L(K)$ such that $$\Phi(a)=V\pi(a)V^*,$$ where $V:K\to H$ is a bounded operator. Furthermore, we have $\|\Phi(1)\|=\|V\|^2$.

Is it correct?

• Note that you missed "selfadjoint" for your operator $A$. – Martin Argerami May 28 '15 at 17:24
• No, I wrote that A is self-adjoint. But in a new line^^ I will change it. Thanks – banach-c May 28 '15 at 18:52

## 1 Answer

I don't think there is any natural connection. Note that the $V$ in your Spectral Theorem is a unitary, while in Stinespring it can be any operator. Also, in the Spectral Theorem your map goes from a Hilbert space to a Hilbert space, while in Stinespring it maps a C$^*$-algebra into $L(H)$.

• Ok, I thought the same. Thanks. It's a pitty – banach-c May 28 '15 at 18:50