TITLE: The Projection Matrix is Equal to its Transpose
QUESTION [6 upvotes]: By inspecting its formula one can see easily that the matrix for projection onto a subspace is equal to its transpose.
But what is the underlying "geometric" reason for this equality?  I have hard time figuring out why it has to be so?
EDIT:
After reviewing the suggested links, it became more clear, but the reasoning was still escaping my intuition.    However, I think that I can (based on the formal arguments in the links) put forward a rough argument.  Basically, for any operator $A$ and two vectors $x$, $y$ $$<Ax,y>=<x,A^Ty>$$
For a vector $y$ in the orthogonal complement to the subspace S we're projecting onto using the $P^S$ projection operator, $$<P^Sx,y>=0=<x,(P^S)^Ty>$$ Because we took any vector $x$, it means that $(P^S)^Ty=0$.  As $y$ is in the orthogonal complement, $P^Sy=0$ this means that $$(P^S)^Ty=0=P^Sy$$
(in plain English, for any vector in the orthogonal complement to $S$ the action of $P^S$ makes it 0, which is a symmetric action).
For $y$ in $S$ and any $x$, $<P^Sx,y>=<x,y>$ and $P^Sy=y$ (by the properties of the projection operation), but also $<P^Sx,y>=<x,(P^S)^Ty>$ (true for any operator). Putting these together that yields that:$$(P^S)^Ty=y=P^Sy$$.
(in plain English, for any vector in $S$ the action of $P^S$ is the identity, which is a symmetric action).
To sum it up, this shows that the operator $P^S$ acts symmetrically on both elements of S and on elements in its orthogonal complement.  Being that it is a linear operator, then it acts symmetrically on all vectors $x$.
A big thing why this works (I think) is that the concept of projection implies (finite) dimension inner product space and that is much more structured than a simple (finite) dimensional vector space.
LAST EDIT:
Even more clear after digesting the answer of Trial and Error. Given a subspace M to project onto, any vector $z_1$ can be written as the sum of two orthogonal components:
$$z_1= (z_1-P_{\mathcal{M}}z_1) + P_{\mathcal{M}}z_1,$$
But the key geometric thing is that these two components belong to orthogonal subspaces, M and $\perp M$ (any vector from one subspace is perpendicular on any vector from the other subspace).
The projection operator will leave the  component in M unchanged and annihilate the component in $\perp M$.
Because of the fact that the subspaces themselves are orthogonal the product of ANY two arbitrary vectors $z_1, z_2$
$$<z_1,z_2>=<(z_1-P_{\mathcal{M}}z_1) + P_{\mathcal{M}}z_1,(z_2-P_{\mathcal{M}}z_2) + P_{\mathcal{M}}z_2>$$
becomes (by the orthogonality of M and $\perp M$:
$$<z_1,z_2>=<(z_1-P_{\mathcal{M}}z_1),(z_2-P_{\mathcal{M}}z_2)>+<P_{\mathcal{M}}z_1,P_{\mathcal{M}}z_2>$$
But applying $P_M$ annihilates/zeroes the first bracket and leaves unchanged the second bracket.  This holds whether we apply it to $z_1$ or to $z_2$. 
Ergo the effect of $P_M$ is the same whether applied to the first of second term of the scalar product.  So its matrix should be equal to its transpose.
The bolded text indicates key issues that I was not fully appreciating before. Thank you for your patience!

REPLY [6 votes]: As you learned in Calculus, the orthogonal projection $P$ of a vector $x$ onto a subspace $\mathcal{M}$ is obtained by finding the unique $m \in \mathcal{M}$ such that
$$
            (x-m)\perp \mathcal{M}. \tag{1}
$$
So the orthogonal projection operator $P_{\mathcal{M}}$ has the defining property that $(x-P_{\mathcal{M}}x)\perp \mathcal{M}$. And $(1)$ also gives
$$
                (x-P_{\mathcal{M}}x) \perp P_{\mathcal{M}}y,\;\;\; \forall x,y.
$$
Consequently,
$$
        \langle P_{\mathcal{M}}x,y\rangle=\langle P_{\mathcal{M}}x,(y-P_{\mathcal{M}}y)+P_{\mathcal{M}}y\rangle= \langle P_{\mathcal{M}}x,P_{\mathcal{M}}y\rangle
$$
From this it follows that 
$$
     \langle P_{\mathcal{M}}x,y\rangle=\langle P_{\mathcal{M}}x,P_{\mathcal{M}}y\rangle = \langle x,P_{\mathcal{M}}y\rangle.
$$
That's why orthogonal projection is always symmetric, whether you're working in a real or a complex space.