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Let $\{x_1,\dots,x_n\}$ be a finite orthonormal set in a Hilbert Space $H$. Prove that for any $x\in H$ the vector $$x-\sum_{k=1}^n \langle x,x_k\rangle x_k$$ is orthogonal to $x_k$ for every $k=1,\dots,n$.

Since $\{ x_1,\dots,x_n\}$ is an orthonormal set, we know that $\langle x_i,x_j\rangle = 0$ for $i\neq j$. We will show that $\langle x-\sum_{k=1}^n \langle x,x_k\rangle x_k , x_k \rangle = 0$ for all $k=1,\dots,n$. Then,

\begin{align} \langle x-\sum_{k=1}^n \langle x,x_k\rangle x_k , x_k \rangle &= \langle x,x_k\rangle -\langle \sum_{k=1}^n \langle x,x_k\rangle x_k , x_k\rangle \\ &= \langle x,x_k\rangle - \langle \langle x,x_k\rangle x_k, x_k \rangle \\ &= \langle x, x_k \rangle - \langle x,x_k\rangle \langle x_k,x_k\rangle \\ &= \langle x,x_k\rangle - \langle x,x_k\rangle \| x_k\| \end{align} So I messed up somewhere. Any help would be appreciated.

Dragonite
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    Your set is orthonormal, i.e., $|x_k| = 1$. Also note that it should read $|x_k|^2$ in the last line. – gerw Feb 20 '19 at 16:02
  • Besides the fact that you hadn't noticed the fact that the norm of your vectors is $1$, read your formula as $x-P_V(x)$ where $P_V(x)$ is the orthogonal projection on the subspace defined by the $x_k$. This will give an easier view... See https://math.stackexchange.com/q/112728 and the answers in there. – Jean Marie Feb 21 '19 at 08:15

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You didn't mess up, you forgot to notice another fact. The $x_i$ are not only orthogonal, but in fact orthonormal

Hagen von Eitzen
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