Consider the projection lattice of $\mathcal B(\mathcal H)$, the algebra of bounded operators on a Hilbert space $\mathcal H$. In particular, for two (orthogonal) projections $P, Q \in \mathcal B(\mathcal H)$ their join (supremum) $P \vee Q$ is defined as the projection onto the closed linear span of the union of the ranges of $P$ and $Q$.

For a general operator $A \in \mathcal B(\mathcal H)$, let $[A]$ denote the projection onto the closed linear span of the range of $A$ (the so-called range projection of $A$). Do we have the following $$ P \vee Q = P + [P^\perp Q] $$ where $P^\perp = I - P$?

If not, can we have a correct additive formula of the form $P \vee Q = P + R$ where $P$ and $R$ are orthogonal projections?

My thoughts so far: I seem to be able to show that if $(P \vee Q) x = x$ then, $(P + [P^\perp Q]) x = x$ for $x \in \mathcal H$. But I can't show that if $(P \vee Q) x = 0$ then $(P + [P^\perp Q]) x = 0$.

Consider the projection lattice of $\mathcal B(\mathcal H)$, the algebra of bounded operators on a Hilbert space $\mathcal H$. In particular, for two (orthogonal) projections $P, Q \in \mathcal B(\mathcal H)$ their join (supremum) $P \vee Q$ is defined as the projection onto the closed linear span of the union of the ranges of $P$ and $Q$.

For a general operator $A \in \mathcal B(\mathcal H)$, let $[A]$ denote the projection onto the closed linear span of the range of $A$ (the so-called range projection of $A$). Do we have the following $$ P \vee Q = P + [P^\perp Q] $$ where $P^\perp = I - P$?

If not, can we have a correct additive formula of the form $P \vee Q = P + R$ where $P$ and $R$ are orthogonal projections?

My thoughts so far: I seem to be able to show that if $(P \vee Q) x = x$ then, $(P + [P^\perp Q]) x = x$ for $x \in \mathcal H$. But I can't show that if $(P \vee Q) x = 0$ then $(P + [P^\perp Q]) x = 0$.

EDIT: I think I have a complete proof. I will add it as an answer. I appreciate if you let me know if you notice any bug. Any other proof is also appreciated.