A projection is a mapping of a mathematical structure into a substructure such that the mapping is idempotent. In a familiar sense, 'structures' could be sets, topological spaces, metric spaces, vector spaces, inner product spaces, groups, graphs, hypergraphs, …, and whatever else.
I'm interested in a classic of linear algebra: vector projection.
$$\vec{a_1} = \frac{\vec{a} \cdot \vec{a}}{||\vec{b}||^2}\vec{b} = \frac{\vec{a} \cdot \vec{a}}{\vec{b} \cdot \vec{b}}\vec{b} = \left( \vec{a} \cdot \hat{b}\right) \hat{b}$$
It is natural to me that a projection of two vectors in $\mathbb{R}$ is a vector within the original space. What confused me is how to think about the projection as idempotent. The operation maps two vectors into one vector, so how could I apply the projection a second time if the second application of the map needs two vectors but the last projection only 'produced' one?
Perhaps I've confused two uses of the word 'projection', or that I'm just looking at the composition of functions the wrong way. Please show whether the vector projection is a projection in the sense defined above.
Best Answer
You're right, vector projection is a special case of "abstract" projection:
Let $V$ be a vector space. In linear algebra, a linear map $P: V \to V$ is a projection if it satisfies the equation $P^2 = P$. (This is your first usage of the word.)
Now, fix a vector $\mathbf b \in V$ and consider the linear map $\text{proj}_{\mathbf b} : V \to V$ given by the formula $\mathbf a \mapsto \text{proj}_{\mathbf b} \mathbf a$. In your question you were considering the case of one vectors undergoing projection; here, we're projecting the entire space onto $\mathbf b$. We now calculate that this map is in fact a projection, i.e. that it's idempotent: $$\text{proj}_{\mathbf b}(\text{proj}_{\mathbf b} \mathbf a) = (\text{proj}_{\mathbf b} \mathbf a \cdot \hat{\mathbf b})\hat{\mathbf b} = [(\mathbf a \cdot \hat{\mathbf b})\hat{\mathbf b}\cdot \hat{\mathbf b}]\hat{\mathbf b} = (\mathbf a \cdot \hat{\mathbf b})(\hat{\mathbf b} \cdot \hat{\mathbf b})\hat{\mathbf b} = (\mathbf a \cdot \hat{\mathbf b})\hat{\mathbf b} = \text{proj}_{\mathbf b} \mathbf a. $$
As a bonus, you could choose a basis for $V$ and write down the matrix $P$ of $\text{proj}_{\mathbf b}$ in that basis. You'll find that $P^2 = P$ holds.