Let $ S $ and $ T $ be two linear subspaces of $ \Bbb{R}^{2} $. Then is the sum of the projections $ P_{S} $ and $ P_{T} $ (i.e., $ P_{S} + P_{T} $) a projection?
I don’t think it is since the projection rule doesn’t hold ($ P_{M}^{2} $ does not equal $ P_{M} $), but I was hoping someone could solidify this. It would be very much appreciated!
The sum is a projection iff $$ p_1 \circ p_2 = - p_2 \circ p_1 $$ because \begin{align} (p_1 + p_2)\circ (p_1 + p_2) - (p_1 + p_2) &= \color{red}{p_1 \circ p_1} + p_1 \circ p_2 + p_2 \circ p_1 + \color{blue}{p_2 \circ p_2} - \color{red}{p_1} - \color{blue}{p_2} \\&= p_1 \circ p_2 + p_2 \circ p_1 \end{align}
Another condition is that the vector spaces need to be perpendicular. Here is the full proof.
Let $V_1, V_2$ be closed subspaces of the Hilbert space $H$ and let $p_{V_1}$ and $p_{V_2}$ denote the orthogonal projections on $V_1$ and $V_2$, respectively. Show that $p_{V_1} + p_{V_2}$ is a projection onto some closed subspace if and only if $V_1 \perp V_2$.
First, assume that $p_{V_1} + p_{V_2}$ is a projection. Thus, for any $x$, $p_{V_1}, p_{V_2}$, and $p_{V_1} + p_{V_2}$ are idempotents. Therefore,
\begin{equation} \begin{split} p_{V_1}(x) + p_{V_2}(x) & = (p_{V_1} + p_{V_2})(x)\\ & = (p_{V_1} + p_{V_2})^2(x)\\ & = (p_{V_1} + p_{V_2}) \Big((p_{V_1} + p_{V_2})(x) \Big)\\ & = p_{V_1}\Big((p_{V_1} + p_{V_2})(x)\Big) + p_{V_2}\Big((p_{V_1} + p_{V_2})(x)\Big)\\ & = p_{V_1}p_{V_1}(x)+p_{V_1}p_{V_2}(x)+p_{V_2}p_{V_1}(x)+p_{V_2}p_{V_2}(x)\\ & = p_{V_1}(x)+p_{V_1}p_{V_2}(x)+p_{V_2}p_{V_1}(x)+p_{V_2}(x)\\ \end{split} \end{equation}
which implies that $$p_{V_1}p_{V_2}=-p_{V_2}p_{V_1}$$ For simplicity, denote $p_{V_1}=p$ and $p_{V_2}=q$. Thus, for all $x$ we have that $$pq=-qp$$ Using the fact that $p$ and $q$ are idempotent, by multiplying both sides of the equation by $p$ on the left and $q$ on the right we get $$pq=-pqpq$$ Also notice that if we take $pq=-qp$ and multiply both sides from the left and from the right by $p$, we get $$pqp=-pqp$$ But this means that for any $x$, $$pqp(x)=pqp(\frac{1}{2}x)+pqp(\frac{1}{2}x)=0$$ Thus, multiplying both sides by $p$ on the left yields $$pqpq(x)=0$$ for all $x$. Thus, from the first relation we obtained, we get $$pq(x)=-pqpq(x)=0$$ Therefore, $\forall x \in V_2$, we get $$pq(x)=p(x)=0$$ Thus, since $x$ was an arbitrary element of $V_2$, we get that $V_2 \perp V_1$.
Conversely, assume that $V_1 \perp V_2$. Since the sum of closed orthogonal subspaces of a Hilbert space is a closed subspace, we know that $V_1+V_2$ is a closed subspace of $H$. Let $V:=V_1+V_2$. Since $V$ is closed, $p_V$ is defined. Thus, to prove that $p_{V_1} + p_{V_2}$ is a projection, it is enough to show that $$p_{V_1} + p_{V_2}=p_V$$ Let $x \in H$. Then $x=p_V(x)+p_{V^{\perp}}(x)$ Since $p_V(x) \in V$, $p_V(x)=x_1+x_2$, where $x_1 \in V_1$ and $x_2 \in V_2$. This gives us \begin{equation} \begin{split} (p_{V_1} + p_{V_2})(x) & =p_{V_1}(x) + p_{V_2}(x)\\ & =p_{V_1}\Big(p_V(x)+p_{V^{\perp}}(x)\Big) + p_{V_2}\Big(p_V(x)+p_{V^{\perp}}(x)\Big)\\ & =p_{V_1}\Big(x_1+x_2+p_{V^{\perp}}(x)\Big) + p_{V_2}\Big(x_1+x_2+p_{V^{\perp}}(x)\Big)\\ & =x_1+p_{V^1}p_{V^{\perp}}(x)+x_2+p_{V^2}p_{V^{\perp}}(x)\\ &=x_1+x_2\\ &=p_V(x) \end{split} \end{equation}
where we used the fact that $p_{V^1}p_{V^{\perp}}(x)=0$, since any element orthogonal to $V$ is orthogonal to $V_1$, and $p_{V^2}p_{V^{\perp}}(x)=0$, since any element orthogonal to $V$ is orthogonal to $V_2$. Thus, $p_{V_1} + p_{V_2}$ is a projection, as desired.
A linear operator $P$ is a projector iff $P^2=P$. We have:
$$ (P_s+P_t)^2=P_s^2+P_t^2+P_sP_t+P_tP_s= P_s+P_t+P_sP_t+P_tP_s $$ So the the sum is a projector iff $P_sP_t+P_tP_s=0$
Let us work on a generic Hilbert space $H$. If $P_S$ and $P_T$ are projectors, then $P_S + P_T$ is a projector iff $S \perp T$. Or, in other words, iff the ranges are perpendicular.
Recall that $P$ is a projector iff $P = P^\dagger$ (self-adjointness) and $P^2 = P$ (idempotency). We use this theorem for $P_S + P_T$.
Because self-adjoining is a linear operation, and $P_S$ and $P_T$ are projectors, we immediately have: $$(P_S + P_T)^\dagger = P_S^\dagger + P_T^\dagger = P_S + P_T\,.$$ So we need to focus on the idempotency only.
Let's go in one way. Suppose $P_S + P_T$ is a projector. Then, idempotency requires: $$(P_S + P_T)^2 = P_S + P_T\,.$$ and then: $$P_S^2 + P_SP_T + P_TP_S + P_T^2 = P_S + P_T\,.$$ Using the idempotency of $P_S$ and $P_T$: $$P_SP_T + P_TP_S = 0\,.$$ Now, suppose that $\exists x \neq 0 \in S \cap T$. Since $x$ belong to both ranges, we have $P_Sx = x$ and $P_Tx = x$. In particular, $P_SP_Tx = P_TP_Sx = x$. But then the condition found above implies that $x = 0$. So the intersection of the two ranges contains only the null vector and $P_SP_T x = 0$ for all $x\in H$. Then, computing the scalar product $\forall x,y \in H$, and using the self-adjointness of the projectors: $$0 = (P_SP_T x,y) = (P_Tx,P_Sy)\,,$$ which amounts to say that $S\perp T$ because $x$ and $y$ are generic and $P_Tx \in T$ and $P_Sy \in S$.
Going the other way around in order to prove the iff, let's suppose that the two ranges are perpendicular, $S \perp T$. Then, computing the scalar product $\forall x,y \in H$, and using the self-adjointness of the projectors: $$0 = (P_Tx,P_Sy) = (P_SP_T x,y)\,,$$ which tells us that $P_SP_T = P_TP_S = 0$, since the above equality holds $\forall x,y \in H$. Then: $$(P_S + P_T)^2 = P_S + P_T\,,$$ and so $P_S + P_T$ is idempotent, thereby also being a projector.
