If $P$, $Q$ and $Ρ - Q$ are projections, then $ran(P - Q) = (ran Ρ) \cap (ran Q)^{\perp}$

Question

I have problem with solving the two equality :

If $P$, $Q$ and $Ρ - Q$ are projections, then $ran(P - Q) = (ran Ρ) \cap (ran Q)^{\perp}$ and $ker(P - Q) = ran Q + ker P$.

My attempts :

1. Because $(ran Q)^{\perp}=ker Q$ then if $x \in (ran Ρ) \cap (ran Q)^{\perp}$ then $Px=x$ and $Qx=0$ then $(P - Q)x=x$ but this is not enough to conclude that $x \in ran (P-Q)$ to be a projection.

2. If $x \in ran (P - Q)$ then $Px = Qx + x$ then $Px = PQx + Px$ thus $PQx=0$ but how this implies $Qx=0$?!

3. If $x \in ker(P - Q)$ then $Px=Qx$, how that implies $x=x_1 + x_2$ for $x_1 \in ran Q$ and $x_2 \in ker P$?

Answer 1

Let us prove that:

If $P$, $Q$ and $P-Q$ are projections, then $ \operatorname{ran}(P-Q) = ( \operatorname{ran}P)\cap ( \operatorname{ran}Q)^\perp$ and $\ker(P-Q)= \operatorname{ran}Q + \ker P$.

We are going to use what is already proved in: On equivalence of P-Q being a projection with the other statements.

Let first prove that $ \operatorname{ran}(P-Q) = ( \operatorname{ran}P)\cap ( \operatorname{ran}Q)^\perp$. Since $Q$ is a projection, $( \operatorname{ran}Q)^\perp = \ker Q$. So, we must prove that $ \operatorname{ran}(P-Q) = ( \operatorname{ran}P)\cap (\ker Q)$.

If $y \in \operatorname{ran}(P-Q) $ then there is $x \in H$ such that $y=Px-Qx$. So $$ Py = P^2 x - PQ x = Px -Qx =y $$ So $y\in \operatorname{ran}P$. On the other hand, $$ Qy = QPx - Q^2 x= Qx - Qx =0$$ So $y \in \ker Q$. So $y \in ( \operatorname{ran}P)\cap (\ker Q)$.

If $y \in ( \operatorname{ran}P)\cap (\ker Q)$, then $$ (P-Q)y = Py-Qy= Py =y$$ So $y \in \operatorname{ran}(P-Q) $.

So we have proved that $ \operatorname{ran}(P-Q) = ( \operatorname{ran}P)\cap (\ker Q) = ( \operatorname{ran}P)\cap ( \operatorname{ran}Q)^\perp$.

Finally, we have \begin{align*} \ker(P-Q) & = (\operatorname{ran}(P-Q))^\perp = \\ & = \left( \operatorname{ran}P \cap \ker Q \right)^\perp = \\ & =( \operatorname{ran}P)^\perp + (\ker Q)^\perp = \\ & = \ker P + \operatorname{ran}Q \end{align*} $\square$

Remark: It is also possible to prove directly that $\ker(P-Q)= \operatorname{ran}Q + \ker P$.

Proof: If $x \in \ker(P-Q)$ then $Px = Qx$. So, we have $$x = Qx + (x-Qx) = Qx + (x-Px)$$ But $(x- Px) \in \ker P$, so we have that $x \in \operatorname{ran}Q + \ker P$. So, $\ker(P-Q) \subseteq \operatorname{ran}Q + \ker P$.

If $x \in \operatorname{ran}Q + \ker P$ then there is $u, v \in H$ such that $x = Qu + v$ and $Pv=0$. So, \begin{align*} (P-Q)x & = PQu - Q^2 u + Pv -Qv = \\ & = PQu - Qu - Qv = & \text{ because } Pv=0 \\ & = -Qv = & \text{ because } PQ=Q \\ & = -QPv = & \text{ because } QP=Q \\ & = 0 & \text{ because } Pv=0 \end{align*}
So $x \in \ker(P-Q)$. So $\operatorname{ran}Q + \ker P \subseteq \ker(P-Q)$. So, we have proved that $\ker(P-Q)= \operatorname{ran}Q + \ker P$. $\square$