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I am stuck with the following exercise about projections in Rudin 12.26.

Let $H$ be a Hilbert space $P,Q\in B(H)$ self-adjoint projections (A projection has the property that $P^2=P$), then the following are equivalent.

(a) $P\geq Q$

(b) $R(P)\supset R(Q)$

(c) $PQ=Q$

(d) $QP=Q$

Theorem 12.4 says the following about a projection : $P$ self-adjoint is equivalent with $P$ is normal, with $R(P) = N(P)^{\perp}$, and with $(Px,x) = \left\| Px\right\|^2$.

If we just assume (a).$ P\geq Q$ then $(Px,x)\geq (Qx,x)$ which is then the same as $\left\|Px\right\|^2\geq \left\|Qx\right\|^2 $. Can we conclude (b)? Implementing my intuition about projections in $\mathbb{R}^n$ I can see why these things would be equivalent, but working straight from the definitions I can't really work it out.

Some ideas or suggestions?

DinkyDoe
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2 Answers2

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The equivalence $(b)\iff (c)$ is true in general for "projections" $P, Q$ (i.e. idempotency $P^2=P$ and $Q^2=Q$ suffices). Hint: $PQ=Q \iff (I-P)Q=0$ and $N(I-Q)=R(Q)$.

The equivalence $(c)\iff (d)$ is trivial. Hint: adjoint.

It remains to show that $(a)$ is equivalent to, say, $(b)$. You were off to a good start. One way to do that, is to prove the following useful fact. The equivalence follows immediately given what you already observed.

Fact If $P$ is a self-adjoint projection, then $\|Px\|\leq \|x\|$ for every $x\in H$ and $$R(P)=\{x\in H\;;\; \|Px\|=\|x\|\}$$

Hint Pythagoras.

Julien
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  • Can this be done algebraically? I want it as part of a proof in http://math.stackexchange.com/questions/632521/detail-in-the-algebraic-proof-of-the-polar-decomposition/633248?noredirect=1#comment1336468_633248? – Noix07 Jan 10 '14 at 20:09
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$(a)\ \Leftrightarrow (c),(d)$ can actually be done for an abstract von Neumann algebra, however it is not as instructive as one could have hoped.

$(a)\ \Rightarrow (c) $ :

  • a theorem caracterizing positive elements has that $P\geq Q\ \Leftrightarrow\ \exists R,\ P-Q= R^*R$. (Let's also write it shorter as $S:=R^*R$ but remember that it is positive)
  • Writing that $P= Q + S$ is a projection yields (after one simplification) $$S= QS+SQ+ S^2 $$
  • One deduce the (trivial, when written in terms of $P,Q$) relation $$(1-Q)S=S(Q+S)\quad \Longrightarrow\quad Q(1-Q)S=0=QS(Q+S) $$ $$\Longrightarrow\quad QSQ+ QS^2=0 $$ But by pluging explicitly the relation $S=P-Q$, one sees that $QS^2=QS^2Q$. Hence we have now a sum of two positive elements that vanish, each is 0. We obtain $$QPQ=Q $$ (in particular $PQPQ=PQ$ and $QP$ are projections, doesnt show orthogonality yet ) and even better $$QSQ=0=QR^*RQ\quad\Longrightarrow\quad RQ=0\quad \Longrightarrow SQ=0 $$ which is what we want.

$(c)\ \Rightarrow (a) $ :

  • By definition of the (left/right) support of an element in a von Neumann algebra, namely smallest projection such that $s(q)q=q$ /$qs(q)=q$ (cf. Sakai ) $ p\geq s(q)$ where $s(q)$ is the suppot of $q$.
  • But one also has $s(q)\leq q$. Hence by the first part of the proof $$ s(q)=s(q)q=q$$ "q projection is its own support" as one could have guessed.

This equivalence means that the order relation defined (as a consequence of positivity) for self-adjoint element in a von Neumann algebra agrees with the one defined by $(c)$ or $(d)$ when restricted to projections.

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Noix07
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