Fact: $P$ is a projection matrix iff $P^2 = P$.
So, we need to show that $P^2 = P \implies (I-P)^2 = I-P$.
Do you see how to do this?
EDIT: As mentioned by Vedran Šego in the comments below, the above only shows that $P$ is a projection matrix, not necessarily an orthogonal projection matrix. To show that $P$ is an orthogonal projection matrix, we also need to show that $P$ is symmetric $\implies$ $I-P$ is symmetric.
Your proof that $Q$ is an orthogonal projection matrix is correct. One geometric interpretation of the vectors $z=Pb$ and $v=Qb$ is exactly as user84413 stated: orthogonal projections onto the range of P and onto the orthogonal complement of the range of P.
Another geometric interpretation I use for students just starting to learn this topic is as follows.
1. P and Q divide our space into two orthogonal spaces
Any vector in $P$ is orthogonal to any vector in $Q$.
For example, if our space was the Cartesian X-Y plane, perhaps P is analogous to the X axis and Q is the Y axis. The X and Y axes are orthogonal, so any vector on the X axis (e.g., $(3,0)$) is orthogonal to any vector on the Y axis (e.g., $(0,7)$).
2. Any vector in our space can be written as a sum of a vector in P and a vector in Q
For any vector $b$, there are two 'component' vectors in $P$ and $Q$ whose sum is vector $b$. Let $z$ be the vector in space $P$ and $v$ be the vector in space $Q$, then
$$b = z + v$$
which is
$$b = Pb + Qb$$
For example, any vector in the X-Y plane can be represented as a sum of:
- a vector that lies solely on the X axis, and
- a vector that lies solely on the Y axis
In the X-Y plane, the vector $b=(3,7)$ can be written as the sum of $z=(3,0)$ (which is on the X-axis) and $v=(0,7$) (which is on the Y-axis). The vector $b=(3,7)$ is like the hypotenuse of a right triangle that has the two legs $z=(3,0)$ and $v=(0,7)$.
The vectors $z$ and $v$ are unique. In other words, once we are given the orthogonal subspaces P and Q there is only one way to get the two components of the vector $b$ that lie in $P$ and $Q$.
3. The 2-norm of any vector in our space can be calculated from the norms of the component vectors in P and Q
$$ \left \lVert b \right \rVert^2 = \left \lVert z \right \rVert^2 + \left \lVert v \right \rVert^2 $$
which is
$$ \left \lVert b \right \rVert^2 = \left \lVert Pb \right \rVert^2 + \left \lVert Qb \right \rVert^2 $$
For example, in the X-Y plane the vector $b=(3,7)$ has a 'length' that can be found by the Pythogorean theorem:
$$ \left \lVert (3,7) \right \rVert^2 = \left \lVert (3,0) \right \rVert^2 + \left \lVert (0,7) \right \rVert^2 $$
$$ \left \lVert (3,7) \right \rVert^2 = 3^2 + 7^2 $$
Best Answer
The answer to the body of your question is much quicker than the answer to the title.
Note that for any vector $x$, we have $$ Ax = vv^Tx = v\langle x,v \rangle = \langle x,v \rangle v $$ By definition, this is the projection of $x$ onto the vector $v$.
Yes, we could prove that in general, a matrix is an orthogonal projection if it is idempotent and symmetric. However, doing so is not necessary in answering this particular question.