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Let $\mathcal{S}$ and $\mathcal{T}$ be two subspaces of $\mathbb{R}^n$, let $P$ be the orthogonal projection of $\mathbb{R}^n$ on $\mathcal{S}$ and let $Q$ be the orthogonal projection of $\mathbb{R}^n$ onto $\mathcal{T}$.

  1. Show that if $P$ and $Q$ commute, then $PQ$ is a projection and $PQ$ is the projection onto $\mathcal{S}\cap \mathcal{T}$.
  2. Is the converse assertion true? Suppose $PQ$ is the orthogonal projection of $\mathbb{R}^n$ onto the intersection $\mathcal{S}\cap \mathcal{T}$. Must $P$ commute with $Q$.

Anybody has advice on how i should start proving this assertion?

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1 Answer 1

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(1)

If $P,Q$ commute, then $(PQ)(PQ) = (QP)((PQ) = Q P^2Q = QP Q = (QP)Q=(PQ)Q = P Q^2 = P Q$, hence $PQ$ is a projection.

Suppose $PQ x = x$, then $x \in $S (since $Px = P^2Q x= PQ x = x$), and since $QP x = PQ x = x$ we have $x \in T$. Hence $ x\in S \cap T$. Now suppose $x \in S \cap T$. Then $Px=x$ and $Qx = x$. Hence $PQ x = Px = x$, and so we have $x \in S \cap T$ iff $PQ x = x$.

(2) A projection is orthogonal iff it is self adjoint.

Since $P,Q,PQ$ are orthogonal, $(PQ)^* = PQ$, and so $(PQ)^* =Q^* P^* = Q P = PQ$.

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    $\begingroup$ An interesting addition to this would be: If $R$ was the projection on to $\mathcal{S}\cap\mathcal{T}$, when can we say $R=PQ$ (and further also $=QP$). For that we need the notion of orthogonality of $\mathcal{S}$ and $\mathcal{T}$ $\endgroup$ Dec 7, 2017 at 17:00
  • $\begingroup$ @JamesNichols: The orthogonal projection is unique, so, if I understand your comment, the answer is yes if $R$ is an orthogonal projection. $\endgroup$
    – copper.hat
    Dec 7, 2017 at 17:08
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    $\begingroup$ Yes the orthogonal projection is unique, but there is every chance that $PQ$ is not an orthogonal projection in general. Obviously from your answer $P$ and $Q$ commute iff $PQ$ is an orthogonal projection. But the further point would be, when exactly is $PQ$ an orthogonal projection? My point is a bit disconnected to the discussion in this question, as the question clearly states "Suppose $PQ$ is the orthogonal projection"... but I made this point because of this question. $\endgroup$ Dec 8, 2017 at 11:43

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