# Prove that the orthogonal projection operator is idempotent

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.

Thank you.

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}

Just a hint: You should make use of the orthonormal condition of the basis, i.e. $\langle u_i,u_j\rangle=\delta_{ij}$.

Applying $P$ to both sides of the equation $$P(x) = \sum\limits_{i=1}^n \langle x, u_i \rangle u_i$$ and using linearity, you get $$P(P(x)) = \sum\limits_{i=1}^n \langle x,u_i \rangle P(u_i)$$ Now, what is $P(u_i)$?

• $P(u_{i}) = \sum_{i=1}^{n} \langle u_{i}, u_{i} \rangle u_{i}$. How does that help me?
– user100463
Dec 14, 2015 at 5:59
• That's not quite right..
– D_S
Dec 14, 2015 at 6:00
• For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
– D_S
Dec 14, 2015 at 6:01
• enlighten me. Obviously, I don't know what I'm doing.
– user100463
Dec 14, 2015 at 6:01
• Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = \langle u_j, u_1 \rangle u_1 + \langle u_j, u_2 \rangle u_2 + \cdots + \langle u_j, u_n \rangle u_n$$
– D_S
Dec 14, 2015 at 6:02

An approach might be: let's show that $P|_U=id_U$.

This is sufficient, because, since $\operatorname{im}P\subseteq U$, $$P^2=P\circ P=P|_U\circ P=id_U\circ P=P$$

Indeed, since $\{u_1,\cdots,u_n\}$ is a basis, you only need to show that $P(u_i)=u_i$. But $$P(u_i)=\sum_{j=1}^n\langle u_i,u_j\rangle u_j=\langle u_i,u_i\rangle u_i=u_i$$ $\square$