Orthogonal projection matrix proof 
Let $P\in \mathbb{M}_m(\mathbb{R})$ a orthogonal projection matrix.
  Show that the matrix $Q=I-P$ is a orthogonal projection matrix. Make a
  geometric interpretation of the elements $z=Pb$ and $v=Qb$, where
  $b\in\mathbb{R}^n$.

I do not know if it's right, but to prove it is orthogonal projection matrix enough to show that is symmetric and idempotent?
I did
$Q^2=(I-P)^2=(I-P)(I-P)=I^2-2IP+P^2=I-2P+P=I-P$
$Q^T=(I-P)^T=I^T-P^T=I-P$ I use the fact that $I$ and $P$ is symmetric.
Can someone help me make a geometric interpretation and correct if I'm wrong the first part?
 A: 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:
$$ \left \lVert (3,7) \right \rVert^2 = \left \lVert (3,0) \right \rVert^2 + \left \lVert (0,7) \right \rVert^2 $$
$$ \left \lVert (3,7) \right \rVert^2 = 3^2 + 7^2 $$
