Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $P^T=P$?

share|cite|improve this question
In general, a projection matrix has $P^2=P$. However, only orthogonal projection matrices are symmetric. – robjohn Jul 31 '13 at 13:03
up vote 8 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."

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.