On Sum and Product of Two Projections

Question

The following is Exercise 4 page 40 in Functional Analysis book of Conway :

Let $\operatorname{P}$ and $\operatorname{Q}$ be projections. Show: (a) $\operatorname{Ρ+Q}$ is a projection if and only if $\operatorname{ranP} \perp \operatorname{ranQ}$. If $\operatorname{P+Q}$ is a projection, then $\operatorname{ran(P+Q)} = \operatorname{ranP} + \operatorname{ranQ}$ and $\operatorname{ker(P+Q)} = \operatorname{kerΡ} \cap \operatorname{kerQ}$. (b) $\operatorname{PQ}$ is a projection if and only if $\operatorname{PQ} = \operatorname{QP}$. If $\operatorname{PQ}$ is a projection, then $\operatorname{ran(PQ)} = \operatorname{ranP} \cap \operatorname{ranQ}$ and $\operatorname{ker(ΡQ)} = \operatorname{kerΡ} + \operatorname{kerQ}$.

My attempt for (a) : Let $ran Ρ \perp ran Q$. If $x \in ran Ρ$ then $x \in (ran Q)^{\perp} = \operatorname{ker}(Q)$ where the equality is by def (See Definition. 3.1. of the book). Thus (QP+PQ)(x)=0. If $x \in \operatorname{ker}(Ρ)$ then by symmetry of argument (QP+PQ)(x)=0. Since $H = \operatorname{ker}(Ρ) + ran P$ so QP+PQ = 0 and thus $(P+Q)^2=P+Q$. To complete the "if" part of (a) I have to show that $\operatorname{ker}(Ρ+Q) = (ran (Ρ+Q))^{\perp}$ which I couldn't do. For the "only if" part, see here. The answer in that link uses the claim that proving being Hermitian of an operator is enough for being Projection which there is no such discussion in the book! for the second part of (a), if $x \in \operatorname{ker}(Ρ) \cap \operatorname{ker}(Q)$ then obviously $x \in \operatorname{ker}(P+Q)$; I don't know how to prove the converse.

My attempt for (b) : If PQ=QP then PQ is a projection if I can prove $ker (PQ) = (ran PQ)^{\perp}$ which is not so clear.

Projection is defined in the book as follows :

3.1. Definition. An idempotent on H is a bounded linear operator Ε on H such that E^2 = E. A projection is an idempotent Ρ such that $ker Ρ = (ran P)^{\perp}$.

For any other way of proving the statements in the exercise I need to prove them as well i.e. only the definition given is the basic valid one!

Answer 1

Here's a way to complete the proof for the "if" direction.

First, we show that $\operatorname{ran}(P + Q) = \operatorname{ran}(P) + \operatorname{ran}(Q)$.

It is clear that $\operatorname{ran}(P + Q) \subset \operatorname{ran}(P) + \operatorname{ran}(Q)$ (why?). For any $x \in \operatorname{ran}(P)$, we note that $Px = x$ and $Qx = 0$. Thus, $(P + Q)x = Px + Qx = x$. Thus, $x \in \operatorname{ran}(P + Q)$. So, $\operatorname{ran}(P) \subset \operatorname{ran}(P + Q)$. By a symmetrical argument, $\operatorname{ran}(Q) \subset \operatorname{ran}(P + Q)$.

Because $\operatorname{ran}(P) \subset \operatorname{ran}(P + Q)$ and $\operatorname{ran}(Q) \subset \operatorname{ran}(P + Q)$, conclude that $\operatorname{ran}(P) + \operatorname{ran}(Q) \subset \operatorname{ran}(P + Q)$. The conclusion follows.

Now, note that $(\operatorname{ran}(P) + \operatorname{ran}(Q))^\perp = \ker(P) \cap \ker(Q)$ (why?). To show that $\operatorname{ran}(P + Q) \perp \ker(P + Q)$, it suffices to show that $$ \ker(P + Q) = \ker(P) \cap \ker(Q). $$ It is clear that $\ker(P) \cap \ker(Q) \subset \ker(P + Q)$ (why?). On the other hand, suppose that $(P + Q)x = 0$. Because $\operatorname{ran}(P) \cap \operatorname{ran}(Q) = \{0\}$, $Px + Qx = 0 \implies Px = Qx = 0$. That is, we only have $(P + Q)x = 0$ if $x \in \ker(P) \cap \ker(Q)$, which is to say that $\ker(P + Q) \subset \ker(P) \cap \ker(Q)$.

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Here is one way to prove the "only if" direction. We make use of the following claim, which I will leave for you to prove.

Claim: If $P$ is a projection, then $\|Px\| \leq \|x\|$ for all $x$.

Now, suppose that $\operatorname{ran}(P)$ and $\operatorname{ran}(Q)$ are not perpendicular. Deduce that there exists an element $x \in \operatorname{ran}(P)$ such that $\operatorname{Re} \langle x, Qx \rangle > 0$. Consequently, we find that $$ \|(P + Q)x\|^2 = \langle Px + Qx, Px + Qx \rangle \\ = \langle x + Qx,x + Qx \rangle \\ = \|x\|^2 + 2 \operatorname{Re}\langle Qx,x \rangle + \|Qx\|^2 > \|x\|^2. $$ By the above claim, this means that $P+Q$ cannot be a projection.

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For the "if" direction of part b), I recommend that you show that $PQ = QP$ implies that $\operatorname{ran}(PQ) = \operatorname{ran}(P) \cap \operatorname{ran}(Q)$ and $\ker(PQ) = \ker(P) + \ker(Q)$ in a manner similar to my proof of the "if" direction for part a).

For the "only if" direction, (prove if necessary and) use the following fact.

Claim: If $P$ is a projection, then $P$ is self-adjoint (i.e. $P = P^*$).

Now, suppose that $PQ$ is a projection. It follows that $$ PQ = (PQ)^* = Q^*P^* = QP. $$ Thus, it follows that $PQ = QP$.

Answer 2

Let us start proving two easy lemmas:

Lemma 1 If $P$ is idempotent, then $P$ is a projection if and only if, for all $x, y \in H$, $\langle x- Px, Py \rangle=0$.

Proof: It is immediate. Just note that $\operatorname{ker} P =\{x -Px : x \in H\}$ and that $H= \operatorname{ker} P \oplus \operatorname{ran} P$. $\square$

Lemma 2 : If $P$ and $Q$ are idempotent, then $Ρ+Q$ is idempotent if and only if $PQ = -QP$.

Proof: It is straight forward. Just note that $Ρ+Q$ is idempotent if and only if $$P+Q = (P+Q)(P+Q)= PP + PQ + QP +QQ = P + PQ + QP + Q$$ if and only if $PQ+QP =0$ if and only if $PQ = -QP$. $\square$

Item a: Let $P$ and $Q$ be projections.

Let us prove that: $Ρ+Q$ is a projection if and only if $\operatorname{ran}P \perp \operatorname{ran}Q$.

$(\Leftarrow)$ Suppose $\operatorname{ran}P \perp \operatorname{ran}Q$. Since $\operatorname{ker}P = (\operatorname{ran}P)^\perp$, we have that $\operatorname{ran}Q \subseteq \operatorname{ker}P$. In a similar way have, $\operatorname{ran}P \subseteq \operatorname{ker}Q$. It follows that $PQ = QP =0$. So $PQ = -QP=0$. So, by lemma $2$, $Ρ+Q$ is idempotent.

Now, for all $x, y \in H$, we have $$\langle x- (P+Q)x, (P+Q)y \rangle=\langle x- Px, Py \rangle - \langle Qx, Py \rangle + \langle x- Qx, Qy \rangle - \langle Px, Qy \rangle =0 $$ (because $P$ and $Q$ be projections and $\operatorname{ran}P \perp \operatorname{ran}Q$). So, by lemma $1$, $Ρ+Q$ is a projection.

$(\Rightarrow)$ Suppose $\operatorname{ran}P \not \perp \operatorname{ran}Q$. Then, there is $y\in H$ such that $Qy \notin \operatorname{ker}P = (\operatorname{ran}P)^\perp$. It means $PQy \ne 0$.

Let us prove that $P+Q$ is not idempotent. To get a contradiction, suppose $P+Q$ is idempotent. Then, by lemma $2$, $PQ = -QP$. So, in particular, $$ PQ(y-Py) = PQy -PQPy = PQy +PPQy = PQy +PQy= 2PQy \ne 0$$ However $$ -QP(y-Py) = -QPy + QPPy = -QPy + QPy =0$$ So $PQ \ne -QP$. Contradiction. So $P+Q$ is not idempotent. So $P+Q$ is not a projection. $\square$

Now, let us prove that if $P+Q$ is a projection, then $\operatorname{ran}(P+Q) = \operatorname{ran}P + \operatorname{ran}Q$ and $\operatorname{ker}(P+Q) = \operatorname{ker}Ρ \cap \operatorname{ker}Q$.

Suppose $P+Q$ is a projection. It is immediate that: $$\operatorname{ran}(P+Q) \subseteq \operatorname{ran}P + \operatorname{ran}Q$$ and $$\operatorname{ker}(P+Q) \supseteq \operatorname{ker} P \cap \operatorname{ker}Q$$

Now, since $P+Q$ is a projection, we have already proved that $\operatorname{ran}P \perp \operatorname{ran}Q$. Moreover, we also know that $\operatorname{ker} P =(\operatorname{ran}P)^\perp$ and $\operatorname{ker} Q =(\operatorname{ran}Q)^\perp$. So we have that $PQ = QP=0$.

So, given any $x, y \in H$, we have $$ (P+Q) (Px+Qy) = PPx +PQy +QPx +QQy = Px +PQy + QPx + Qy = Px + Qy$$ So $ Px+Qy \in \operatorname{ran}(P+Q)$. So we have proved that $$\operatorname{ran}(P+Q) = \operatorname{ran}P + \operatorname{ran}Q$$

Now, given any $x \in \operatorname{ker}(P+Q)$, we have $0=(P+Q)x= Px +Qx$. So $Px = -Qx$. But since $P+Q$ is a projection, we have already proved that $\operatorname{ran}P \perp \operatorname{ran}Q$, so we have $Px = -Qx=0$. It means $x \in \operatorname{ker} P \cap \operatorname{ker}Q$. So we have $$\operatorname{ker}(P+Q) = \operatorname{ker} P \cap \operatorname{ker}Q$$

Item b: Let us start with a lemma.

Lemma 3: If $P$ is a projection, then $P^*=P$.

Proof: For all $x, y \in H$, we have, by lemma $1$, $\langle x -Px, Py \rangle=0$.

So $\langle P^*x -P^*Px, y \rangle=0$. So $P^*x -P^*Px =0$. So $P^* = P^*P$.

On the other hand, from $\langle x -Px, Py \rangle=0$, we also get $\langle x, Py \rangle - \langle Px, Py \rangle =0$. So $\langle x, Py \rangle - \langle x, P^*Py \rangle =0$. So $\langle x, Py - P^*Py \rangle =0$. So $Py - P^*Py =0$. So $ P = P^*P$. So we can conclude that $P^* = P$. $\square$

Now, let $P$ and $Q$ be projections. Let us prove that

$PQ$ is a projection if and only if $PQ = QP$.

$(\Leftarrow)$ Suppose $PQ = QP$. Then it is immediate that $PQPQ = PQQP = PQP =PPQ= PQ$. So $PQ$ is idempotent.

Now, for all $x,y \in H$, \begin{align*} \langle x- PQx, PQy \rangle & = \langle x- PQx, PQQy \rangle = & \text{because $Q=QQ$} \\ &= \langle x- PQx, QPQy \rangle = & \text{because $PQ=QP$}\\ & = \langle Q^*(x- PQx), PQy \rangle = \\ & = \langle Q(x- PQx), PQy \rangle = & \text{because $Q^*=Q$}\\ & = \langle Qx- QPQx), PQy \rangle = \\ & = \langle Qx- PQQx), PQy \rangle = & \text{because $PQ=QP$}\\ & = \langle Qx- PQx), PQy \rangle = 0 & \text{because $P$ is a projection} \end{align*}

So $PQ$ is a projection.

$(\Rightarrow)$ Suppose $PQ$ is a projection. Then, since $P$ and $Q$ are also projections, we have by lemma $3$.
$PQ = (PQ)^* = Q^*P^*= QP$. $\square$

The rest of item b can be proved in a similar way to item a. If you need I can post a complete answer to item b.