An additive formula for the join of two projection operators

Question

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.

Answer 1

Since every $x \in \mathcal H$ can be deocmposed into a component in $\text{ran}(P \vee Q)$ and a component in $\text{ran}(P \vee Q)^\perp$, it is enough to show that the operator equality holds only over these two subspaces.

We first make a couple of observations: We have $\text{ran}(P^\perp Q) \subset \text{ran}(P \vee Q)$, that is, $[P^\perp Q] \le P \vee Q$. To see this, let $x \in \text{ran}(P^\perp Q)$. Then, $$x = P^\perp Q z = (I-P) Qz = Qz - PQ z \in \text{ran}(P \vee Q)$$ since it is a linear combination of a component in $\text{ran}(Q)$ and component in $\text{ran}(P)$.

Now assume that $x \in \mathcal H$ satisfies $(P \vee Q) x = 0$. Multiplying both sides by $P$ and noting that $P \le P \vee Q$, we have $P x = 0$. Similarly, multiplying both sides by $[P^\perp Q]$ and using $[P^\perp Q] \le P \vee Q$, we get $[P^\perp Q] x = 0$. Thus, $(P + [P^\perp Q])x = 0$ as desired.

Next, assume that $x \in \mathcal H$ is such that $(P \vee Q) x = x$. We have $x = x_1 + x_2$ where $x_1 \in \text{ran}(P)$ and $x_2 \in \text{ran}(Q)$. Let $x_3 = x_1 + P x_2 \in \text{ran}(P)$. Then, $x = x_3 + P^\perp x_2$. We can write $x_2 = Q z$ for some $z \in \mathcal H$. Let $R = [P^\perp Q]$ and note that $R$ and $P$ are orthogonal projection, i.e., $PR = 0$. We have \begin{align*} (P + R) x &= (P+R)(x_3 + P^\perp Q z) \\ &= Px_3 + R P^\perp Q z = x_3 + P^\perp Q z = x, \end{align*} which is the desired result. The last equality uses $[A]A = A$ which holds for any $A \in B(\mathcal H)$. The proof is complete.