Let $V$ be a TRO i.e. closed subspace of $B(H,K)$ such that $xy^*z \in V$ for all $x,y,z \in V$. Let $C(V)$ and $D(V)$ denotes the $C^{\ast}$-algebra generated by $VV^{\ast}$ and $V^*V$ respectively. We define $A(V)$, the linking $C^*$-algebra of $V$ as follows:
$$A(V) = \begin{bmatrix} C(V) & V\\ V^* & D(V) \end{bmatrix}$$
Let $V$ be a TRO such that $A(V)= \mathbb{C}$, what can we say about $V$?
So, I think when we write $$ A(V) = \left[ \begin{matrix} C(V) & V \\ V^* & D(V) \end{matrix} \right] $$ we implicitly mean taking the linear span. Thus $A(V) = \mathbb C$ means that $A(V)$ is spanned by a single matrix, which is necessarily a projection, as $A(V)$ is a $C^*$-algebra. So $V$ must certainly be one-dimensional as well, say spanned by $v\in B(H,K)$. However, if $C(V)$ or $D(V)$ is non-zero, then because we take linear spans, $A(V)$ will still be more than one-dimensional. As $\|v^*v\| = \|v\|^2$, we must have that $C(V)$ and $D(V)$ are non-zero. So I believe this is impossible.
If $V$ were spanned by a single partial isometry then $A(V)$ will be four dimensional.
