I'm trying to find some motivation for the study of operator spaces and on https://noncommutativeanalysis.wordpress.com/2012/10/11/william-arveson/ I came across the following interesting theorem:
Theorem. Let $A$ and $B$ be two irreducible compact operators on $H$. Then $A$ and $B$ are unitarily equivalent if and only if the two dimensional operator spaces $\text{span} \{ I, A\}$ and $\text{span} \{ I, B\}$ are completely isometric.
What is the proof of this theorem?
As stated, the theorem is false. What is missing is that the complete isometry should map $A$ to $B$ and be unital. For an easy example where the theorem fails as stated, take $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ \ B=\begin{bmatrix} 1&1\\0&1\end{bmatrix} $$ (they generate the same unital operator space, but they are obviously not unitarily equivalent).
Even in that case, the nontrivial implication ($\Leftarrow$) is definitely not straightforward. Here is a very rough sketch of an argument.
A unital completely isometry between the operator spaces $\text{span}\,\{I,A\}$ and $\text{span}\,\{I,B\}$ can be extended uniquely to a ucp map $\phi$ between the operator systems $\text{span}\,\{I,A, A^*\}$ and $\text{span}\,\{I,B,B^*\}$ with ucp inverse (i.e. a complete order isomorphism).
Since $C^*(I,A)=I+K(H)$ (because $A$ is irreducible), by Arveson's Boundary Theorem the identity map $C^*(I,A)\to C^*(I,A)$ is a boundary representation, and similarly for $B$.
Arveson's Choquet theorem then guarantees that $\phi$ extends to a C$^*$-isomorphism between $C^*(I,A)$ and $C^*(I,B)$. As these are irreducible and contain compact operators, one can show that the isomorphism is implemented by unitary conjugation.
There is a rather complete proof in this paper by Doug Farenick (I'm not sure if this preprint is as updated as the published version in LAA).