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Let $U,V$ be closed subspaces of the Hilbert space $H$ and $P_U, P_V$ the corresponding orthogonal projections on $U$ and $V$, respectively. I need to show: $$ U\subset V \Leftrightarrow P_U=P_VP_U=P_UP_V.$$

Let $x_0\in H,\, \, $and $U\subset V.$ Then, $\,y_0:=P_U(x_0)\in U.$ Also, $P_VP_U(x_0)=P_V(y_0)=y_0$ since $y_0 \in V.$ Since this is true for all $x_0 \in H$, the direction $\Rightarrow$ is proved. Is this correct, what I have done ?

I have difficulties into proving the other direction. Can somebody provide a proposal or give me a hint ?

Many thanks in advance.

user249018
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1 Answers1

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$\Rightarrow$ Let $x\in H$.

  • $P_U=P_VP_U$: Proved in the question.
  • $P_U=P_UP_V$: Consider $v=P_V(x)-x$. Then $v$ is orthogonal to every element of $V$, and in particular, to every element of $U$. This means that $v$ is in the null space of $P_U$ (because $H=\mathrm{ran}(P_U)\oplus\mathrm{ker}(P_U)$, or equivalently, by self-adjointness of $P_U$, $\forall y\in H,\ \langle y,P_U v \rangle=\langle P_U y,v\rangle = 0$, where the last inequality is because $P_Uy\in U$ and is therefore orthogonal to $v$.) In other words, $P_U v = 0$.

Therefore, \begin{align} P_U(P_V(x)-x) &= P_U(v) \\ P_U(P_V(x)-x)&= 0\\ P_U P_V(x) &= P_U(x). \end{align}

Since the result holds for all $x\in H$, $P_UP_V=P_U$.

$\Leftarrow$: Proof of contrapositive:

If $U$ is not a subset of $V$ then there is a $u\in U\setminus V$. Then

$$P_U(u)=u,$$ but

\begin{align} P_VP_U(u)&=P_V(u)\in V \end{align} so that $P_VP_U(u)\ne u$ because $u\notin V$.

Therefore $P_U\ne P_VP_U$.

cantorhead
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  • Thanks. I do have a problem concerning the commutativity of the orthogonal projections. I guess the implication $\Rightarrow$ is incomplete. Do you have any suggestions how to procede ? – user249018 Feb 06 '18 at 18:26
  • You are correct. I believe I have added the proof for $P_U=P_UP_V$. – cantorhead Feb 07 '18 at 00:49