Let $\{u_{1},u_{2},\cdots,u_{n}\}$ e an orthonormal basis for a subspace $U$ in an inner product space $X$.
Define the orthogonal projection of $X$ onto $U$, $P:X \to U$, to be $Px = \sum_{i=1}^{n}\langle x, u_{i} \rangle u_{i}$, where $\langle \cdot, u_{i} \rangle$ represents the inner product.
I need to prove that $P = P^{2}$; i.e., that $P$ is idempotent. I have already proven that $P$ is linear, and am therefore free to use it.
So far, I set up what I am trying to show as follows:
$P^{2}x = \sum_{i=1}^{n} \langle Px, u_{i}\rangle u_{i} =\sum_{i=1}^{n}\langle \sum_{i=1}^{n}\langle x, u_{i} \rangle u_{i},u_{i}\rangle u_{i}$
Then, I thought that perhaps expanding out the inner sum might be helpful, and then somewhere along the line I might be able to use linearity to get $\sum_{i=1}^{n}\langle x, u_{i}\rangle u_{i}$ eventually on the RHS.
This is about as far as I got playing around with the sums:
$\sum_{i=1}^{n}\langle \langle x, u_{1}\rangle u_{1}+\langle x, u_{2}\rangle u_{2} + \cdots + \langle x, u_{n} \rangle u_{n}, u_{i} \rangle u_{i} = \sum_{i=1}^{n}\left(\langle \langle x, u_{1} \rangle u_{1}, u_{i} \rangle + \langle \langle x, u_{2}\rangle u_{2}, u_{i} \rangle + \cdots + \langle \langle x, u_{n} \rangle u_{n}, u_{i} \rangle \right)u_{i} = \sum_{i=1}^{n}\left[\left(\langle \langle x, u_{1} \rangle u_{1}, u_{i}\rangle u_{i}\right) + \left(\langle \langle x, u_{2} \rangle u_{2}, u_{i} \rangle u_{i}\right) + \cdots + \left(\langle \langle x, u_{n} \rangle u_{n}, u_{i} \rangle u_{i} \right)\right] = \sum_{i=1}^{n} \langle \langle x, u_{1} \rangle u_{1}, u_{i}\rangle u_{i} + \sum_{i=1}^{n}\langle \langle x, u_{2} \rangle u_{2}, u_{i} \rangle u_{i} + \cdots + \sum_{i=1}^{n}\langle \langle x, u_{n} \rangle u_{n}, u_{i} \rangle u_{i}$
But, it's still not looking anywhere closer to where I need to be.
Could somebody please help me finish this?
Thank you.
Best Answer
It is simpler to start with $P(u_i)=\sum_{j=1}^n\langle u_i,u_j \rangle u_j$ and the only non vanishing term corresponds to $j=i$. So $P(u_i)=u_i$. Then we have by linearity
$$\begin{align}P(P(x))&=\sum_{i=1}^n\langle x,u_i \rangle P(u_i)\\&=\sum_{i=1}^n\langle x,u_i \rangle u_i\\&=P(x)\end{align}$$