Why is a projection matrix of an orthogonal projection symmetric? I am quite new to this, so I hope you forgive me if the question is naïve. (Context: I am learning econometrics from Davidson & MacKinnon's book "Econometric Theory and Methods", and they do not seem to explain this; I've also looked at Luenberger's optimization book that deals with projections at an a bit more advanced level, but with no luck).
Suppose that I have an orthogonal projection $\mathbb P$ with is associated projection matrix $\bf P$. I am interested in projecting each vector in $\mathbb{R}^n$ into some subspace $A \subset \mathbb{R}^n$.
Question: why does it follow that $\bf{P}=P$$^T$, that is, $\bf P$ is symmetric? What textbook could I look at for this result?
 A: An attempt at geometrical intuition... 
Recall that:


*

*A symmetric matrix is self adjoint.

*A scalar product is determined only by the components in the mutual linear space (and independent of the orthogonal components of any of the vectors). 


What you want to "see" is that a projection is self adjoint thus symmetric-- following (1). Why is this so?
Consider the scalar product of a vector $x$ with the projection $A$ of a second vector $y$: $ \langle x,Ay \rangle$. Following (2), the product will depend only on the components of $x$ in the span of the projection of $y$. So the product should be the same as $\langle Ax,Ay \rangle$, and also $\langle Ax,y\rangle $ following the same argument. 
Since $A$ is self adjoint- it is symmetric. 
A: This is a fundamental results from linear algebra on orthogonal projections. A relatively simple approach is as follows. If $u_1, \ldots, u_m$ are orthonormal vectors spanning an $m$-dimensional subspace $A$, and $\mathbf{U}$ is the $n \times p$ matrix with the $u_i$'s as the columns, then 
$$\mathbf{P} = \mathbf{U}\mathbf{U}^T.$$
This follows directly from the fact that the orthogonal projection of $x$ onto $A$ can be computed in terms of the orthonormal basis of $A$ as
$$\sum_{i=1}^m u_i u_i^T x.$$
It follows directly from the formula above that $\mathbf{P}^2 = \mathbf{P}$ and that $\mathbf{P}^T = \mathbf{P}.$
It is also possible to give a different argument. If $\mathbf{P}$ is a projection matrix for an orthogonal projection, then, by definition, for all $x,y \in \mathbb{R}^n$ 
$$\mathbf{P}x \perp y-\mathbf{P}y.$$
Consequently,
$$0 = (\mathbf{P} x)^T (y - \mathbf{P}y) = x^T \mathbf{P}^T (I - \mathbf{P}) y = x^T (\mathbf{P}^T - \mathbf{P}^T \mathbf{P}) y $$
for all $x, y \in \mathbb{R}^n$. This shows that $\mathbf{P}^T = \mathbf{P}^T \mathbf{P}$, whence 
$$\mathbf{P} = (\mathbf{P}^T)^T = (\mathbf{P}^T \mathbf{P})^T = \mathbf{P}^T \mathbf{P} = \mathbf{P}^T.$$
