Confusion Around Orthogonal Project Matrices The link here describes an orthogonal projection matrix. In particular, I'm interested in the two properties:
i) An $n \times n$ matrix $P$ is a projection matrix if $P^2 = P$.
ii) A projection matrix $P$ is orthogonal if and only if $P = P^T$ (dealing explicitly with real matrices here).
Now consider an $n \times n$ real matrix $Q$ that so happens to be orthogonal as well as a projection matrix. Then it satisfies both properties above, as well as a property of orthogonal matrices in general: $Q^TQ = I$ where $I$ is the $n \times n$ identity matrix. But then it appears that from the second property $Q = Q^T$, we have $Q^2 = I$ and then immediately by the first property $Q = I$. But clearly it's not true that all $n \times n$ real matrices that are orthogonal projection matrices are the $n \times n$ identity matrices. Where has my reasoning gone wrong?
 A: You’re absolutely right: the word “orthogonal” refers to two different things here. An orthogonal matrix is one with orthonormal columns. 
On the other hand, an orthogonal projection $P$ is one for which the projection direction is orthogonal to its image, i.e., $\ker(P) = \operatorname{im}(P)^\perp$. This is a property of the projection whether or not you represent it as a matrix. The matrix of an orthogonal projection of $\mathbb R^n$ expressed relative to the standard basis is indeed symmetric per your definition, but that’s not generally true for other bases. For example, the orthogonal projection onto the $x$-$y$ plane in $\mathbb R^3$ has the matrix $\operatorname{diag}(1,1,0)$ relative the standard basis, but relative to the ordered basis $((1,2,3)^T, (0,1,0)^T, (3,2,1)^T)$ it is $$\begin{bmatrix}-\frac18&0&-\frac38\\\frac32&1&\frac12\\\frac38&0&\frac98\end{bmatrix},$$ which is clearly not symmetric. The definition that you’ve quoted in your question has some unstated assumptions about the bases that the matrix is expressed in. It’s a moderately interesting exercise to work out the conditions on the basis for the matrix of an orthogonal projection to be symmetric.  
Generally speaking, when you see the phrase “orthogonal projection matrix,” it means the latter: in other words, you should parse the phrase as “(orthogonal projection) matrix”—the matrix of an orthogonal projection—instead of “orthogonal (projection matrix).”
A: I suppose there are really three different classes of matrices here:


*

*Projection Matrices, satisfying $P^2 = P$,

*Orthogonal Matrices, satisfying $Q^TQ = QQ^T = I$, and 

*Orthogonal Projection Matrices, satisfying $P^2 = P$ and $P = P^T$.


The language may imply that the third group should be the intersection of the first two, but it is in fact not so.  The one feature that implies that the word "orthogonal" is justified in this context is that if we take a vector space and break it up into a direct sum of subspaces:
$$
V \;\; =\;\; W_1 \oplus W_2 \oplus \ldots \oplus W_k
$$
Then the sum of the orthogonal projections onto each subspace should add up to the identity matrix:
$$
P_{W_1} + P_{W_2} + \ldots + P_{W_k} \;\; =\;\; I.
$$
A Baby Example
Suppose we pick the subspace of $W_1\subset \mathbb{R}^3$ given by the following span:
$$
W_1 \;\; =\;\; \text{Span} \left \{\left [\begin{array}{c}
1/\sqrt{2}\\ 0\\ 1/\sqrt{2}
\end{array} \right ], \; \left [\begin{array}{c}
-1/\sqrt{2}\\ 0\\ 1/\sqrt{2}
\end{array} \right ]   \right \}.
$$
The orthogonal projection corresponding to this can easily be seen as 
$$
P_{W_1} \;\; =\;\; \left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right ].
$$
The orthogonal complement $W_2$ to this subspace will be $W_2$ which is simply spanned by the y-axis, so can be represented by the vector $[0 \; 1\; 0]^T$ with orthogonal projection:
$$
P_{W_2} \;\; =\;\; \left [ \begin{array}{ccc}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array} \right ].
$$
Clearly $P_{W_1} + P_{W_2} = I$.
