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I am looking for an intuitive reason for a projection matrix of an orthogonal projection to be symmetric. The algebraic proof is straightforward yet somewhat unsatisfactory.

Take for example another property: $P=P^2$. It's clear that applying the projection one more time shouldn't change anything and hence the equality.

So what's the reason behind $P^T=P$?

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In general, a projection matrix has $P^2=P$. However, only orthogonal projection matrices are symmetric. –  robjohn Jul 31 '13 at 13:03
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up vote 7 down vote accepted

In general, if $P = P^2$, then $P$ is the projection onto $\operatorname{im}(P)$ along $\operatorname{ker}(P)$, so that $$\mathbb{R}^n = \operatorname{im}(P) \oplus \operatorname{ker}(P),$$ but $\operatorname{im}(P)$ and $\operatorname{ker}(P)$ need not be orthogonal subspaces. Given that $P = P^2$, you can check that $\operatorname{im}(P) \perp \operatorname{ker}(P)$ if and only if $P = P^T$, justifying the terminology "orthogonal projection."

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In the simple case where $P$ is projection onto a line with unit vector $\vec{e}$, you can look at it like this: if $\theta$ is the angle between $x$ and $\vec{e}$, and $\varphi$ is the angle between $y$ and $\vec e$, we have:

$$\langle Px,y \rangle = \langle \|x\| \cos{\theta \vec{e}},y \rangle = \|x\|\cos{\theta}\langle \vec{e},y \rangle = \|x\|\cos{\theta}\|y\|\cos{\varphi} \\ \langle x, Py \rangle = \langle x,\|y\|\cos{\varphi}\vec{e}\rangle = \|y\| \cos {\varphi} \langle x,\vec{e} \rangle = \|y\|\cos{\varphi}\|x\| \cos{\theta}.$$

(Recall that $P$ is symmetric iff $\langle Px, y \rangle = \langle x, Py \rangle$ for all $x,y \in V$.)

Another reason: Given a projection (orthogonal or not), we can describe it as the projection onto $A$ along $B$ (or $\rho_{A,B}$), where $A$ and $B$ are complementary subspaces. That is, if $V = A \oplus B$, take a basis for the vector space passing through the subspaces $A$ and $B$. Then the action of $P = \rho_{A,B}$ is to set the $B$-coordinates equal to zero, and leave the $A$ coordinates untouched. Every projection is of this form, and knowing $A$ and $B$ completely characterizes the projection. A projection is orthogonal iff $B = A^\perp$.

It is an exercise in the algebra of inner products to show that if $P = \rho_{A,B}$, then $P^T = \rho_{B^{\perp}, A^{\perp}}$ (no matter if $P$ is orthogonal or not). So if $P$ is an orthogonal projection and $A = B^{\perp}$, we can see that $\rho_{B^{\perp}, A^{\perp}} = \rho_{A,B}$, and $P$ is symmetric.

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