First, let us consider the algebra of bounded operators on a Hilbet space $\mathcal H$, denoted as $\mathcal B(\mathcal H)$. For an operator $B \in \mathcal B(\mathcal H)$ let $[B]$ be its range projection (i.e., the projection operator that projects onto the range of $B$). For a projection operator $P$, let $P^\perp = I - P$.
Let $R(B)$ be the range of $B$, $\text{ker}(B)$ be the kernel of $B$ and $B^*$ the adjoint of $B$. We know that $\text{ker}(B^*) = R(B)^\perp$.
- Since $x \in R(B)^\perp$ iff $[B]^\perp x = x$ iff $[B] x = 0$, it seems that the above statement can be equivalently written as $$ \text{ker}(B^*) = \text{ker}([B]). $$ Is this correct?
- The above statement is equivalent to $$ (*) \quad B^*A =0 \iff [B] A = 0, \quad \forall A \in \mathcal B (\mathcal H). $$ Is this correct?
- Now consider a general $C^*$-algebra $\mathcal C$. The question is whether $(*)$ holds (perhaps under extra conditions) in $\mathcal C$? In other words, can we give a proof of $(*)$ without using the inner product structure of $\mathcal H$? I can show one direction: Assume that $[B] A = 0$. Then, $ B^* [B] A = 0$, hence $([B] B)^* A = 0$ hence $B^* A = 0$ since $[B] B = B$. How about the other direction?
EDIT. Per QuantumSpace's comment, let's change the setting of 3 to the following: Assume that $\mathcal C \subset \mathcal B(\mathcal H)$ is a von Neumann algebra.