'Shrinking' the Hilbert Space a $\mathrm{C}^*$-algebra is represented on This is a follow on question to this.
Suppose that there is a subspace $U\subset H$ a Hilbert space such that for all $f\in A\subset B(H)$ a $\mathrm{C}^*$-algebra, $f$ restricted to $U$ is scalar.
Would we get an isomorphic $\mathrm{C}^*$-algebra if we choose a unit vector $e_u\in U$ and collapse $U$ to one-dimensional space spanned by $e_u$?
 A: The answer to your question is "yes". Rather than give a full answer, I will just explain why the situation you are concerned about in your comment above  (in which $v \mapsto u_1 + 2u_2 + v$) cannot possibly occur. I suspect that you will be able to put the rest of the story together yourself.
Suppose that $S \subset B(H)$ is some collection of operators. We say that a closed subspace $U \subset B(H)$ is $\mathbf{S}$-invariant if $Tu \in U$ whenever $T\in S$ and $u \in U$. It is simple to check, using the definition of the adjoint of an operator, that $U$ is $S$-invariant if and only if $U^\bot$ is $S^*$-invariant. Thus, if $S$ is closed under adjoint, then $U$ is $S$-invariant if and only if $U^\bot$ is $S$-invariant. In particular, since a C*-algebra $A$ is closed under adjoint, the orthogonal complement of any $A$-invariant subspace is also an invariant subspace.
In your situation, you have a subspace $U$ on which $A$ acts by scalar multiplications. Such a subspace is obviously $A$-invariant and so, by the above discussion, its orthogonal complement is also preserved by $A$.
