# Orthogonal projection and subspaces proof?

Let's let $M$ be a subspace of $\mathbb{R}^n$ and let $N$ be a subspace of $M$. Let $m$ and $n$ denote the orthogonal projection matrices onto $M$ and $N$. Show that $mn = nm = n$.

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I'm thinking about inner products of $n$ and $m$, but they would only be zero in the area in which $N$ is a subspace of $M$. I'm a bit lost. Any advice/pointers?

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This cannot be true. Take $m$ to be the identity and $n$ to be a projection onto a line. Then you cannot have $nm = m$. – copper.hat Sep 24 '13 at 22:07
I believe he meant to say $mn=nm=n$ instead of $m$. This would make sense. – laughing_man Sep 24 '13 at 22:07
Probably, but best to start with the correct statement! – copper.hat Sep 24 '13 at 22:08
Downvotes have little effect on reputation for either party. I generally reserve them for what I consider to be rubbish or misleadingly incorrect. If the downvoter leaves no comments, they are meaningless, if a little aggravating because of their puerile nature. – copper.hat Sep 25 '13 at 1:07
I completely agree. I feel like it's just a matter of etiquette more than anything else. If my post is wrong, I want to know why it's wrong. I don't care if I get a downvote, I just want some explanation (if possible), especially when I spent 15 or so minutes posting the damn thing. – St Vincent Sep 25 '13 at 1:09

Denoting matrices with small letters is confusing. I'll denote projections on $M$ and $N$ as $P_M$ and $P_N$ respectively. I will also denote $V = \mathbb{R}^n$, as this proof is good for any finite inner product space.

Note that we can write $V$ as a direct sum of three orthogonal spaces (see Wikipedia article on orthogonal complements):

$$V = M \oplus M^\perp = (V \cap M) \oplus M^\perp = ((N \oplus N^\perp) \cap M) \oplus M^\perp = N \oplus (N^\perp \cap M) \oplus M^\perp.$$

Let $v = x + y + z$, be any vector in $V$, where $x \in N$, $y \in N^\perp \cap M$, $z \in M^\perp$. Recall (Lemma 4.8 here) that for each $v \in V$ there exist (unique) $x,y,z$, so we can do this for any $v \in V$. Then

\begin{gather*} P_N x = x, \quad P_M x = x, \\ P_N y = 0, \quad P_M y = y, \\ P_N z = 0, \quad P_M z = 0, \end{gather*}

so

\begin{align*} P_M P_N v &= P_M P_N (x + y + z) = x, \\ P_N P_M v &= P_N P_M (x + y + z) = x, \\ P_N v &= P_N (x + y + z) = x, \end{align*}

which proves the statement.

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+1: Looks good to me. Don't get the downvote. – copper.hat Sep 25 '13 at 1:08
@copper.hat Thanks. It's two downvotes, actually. I don't understand why either. I see you got the same. – Vedran Šego Sep 25 '13 at 1:25

Presumably the statement was meant to be: Show $mn =nm = n$.

Note that if $P$ is a projection, then $x \in {\cal R} P=$ iff $Px = x$. (One direction is trivial, the other follows from $x = Py = P(Py + (I-P) y) = P(x + (I-P)y) = P x$.)

Suppose $N = {\cal R} n \subset M = {\cal R} m$.

First show $mn=n$. Choose $x$. Then $nx \in {\cal R} n \subset{\cal R} m$, hence $nx = mnx$, from which the result follows. This is true for all projections, orthogonal or not.

Now show $nm = n$. Here we need $m$ and $n$ to be orthogonal. It is straightforward to show that a projection $m$ is orthogonal iff $m$ is self-adjoint.

Then we have $m^*(I-m) = m(I-m) = 0$. Since $mn=n$, we have $n^* m^* = n^*$, and so $n(I-m) = n^*(I-m) = n^* m^* (I-m) = n^* m (I-m) = 0$.

Now we have $nx = n(mx + (I-m)x) = nm x + n(I-m)x = nmx$, which shows $n = nm$.

(The result is true for any closed subsets $M,N$ of a Hilbert space.)

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