I am told:
Let $(V,\langle\cdot,\cdot\rangle)$ be a inner product space (not necessarily a finite-dimensional one). Let $W \subset V$ be a finite-dimensional subspace with orthogonal basis $\{e_1,...,e_n\} $. Let the linear map $P_W \in \mathcal{L}(V)$ be:
$$P_W(x)=\sum_{j=1}^{n} \dfrac{\langle x,e_j\rangle}{\langle e_j,e_j\rangle}e_j $$
Show that $P_W$ is a projection.
I believe that a map is a projection if it is idempotent, but I'm not absolutely sure.
Additionaly, I was given the hint of computing $P_W(e_k)$ first, which gave me $P_W(e_k)=e_k$. So it seems that $P_W$ is idempotent regarding the vectors of the orthogonal basis. However, $x \in V$ so it cannot be expressed as a linear combination of those vectors, as in such case then it is clear that the map is idempotent. Any help is highly appreciated.
Thanks
Starting from $$ P_W(x)=\sum_{j=1}^{n} \dfrac{\langle x,e_j\rangle}{\langle e_j,e_j\rangle}e_j $$
You need to show that $P_{W}$ is a projection, that is:
$$P_{W}\circ P_{W} =P_{W}$$
in others owrds:
$$P_{W}(P_{W}(x))=P_{W}(x)$$
Try this:
Compute the following scalar product using the fact that your basis is orthogonal that is, you will need first to show that :
$$\{e_1,\ldots,e_n\}\ \text{is orthogonal}\ \Longleftrightarrow\langle{e_i,e_j}\rangle=\delta_{i,j} \Longrightarrow \langle P_{W}(x),e_j\rangle=\cdots = \langle x,e_j\rangle$$ replacing the sum inside and then try to replace the result here:
$$P_W\left(P_{W}(x)\right)=\sum_{j=1}^{n} \dfrac{\langle P_{W}(x),e_j\rangle}{\langle e_j,e_j\rangle}e_j=\cdots $$