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:


$$A(V) = \begin{bmatrix}
 C(V) & V\\
 V^\* & D(V)
\end{bmatrix}$$



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> Let $V$ be a TRO such that $A(V)= \mathbb{C}$, what can we say about $V$?
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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.




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If $V$ were spanned by a single [partial isometry](https://en.wikipedia.org/wiki/Partial_isometry) then $A(V)$ will be four dimensional.


