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There is a theorem that states:

Let $u_1, \dots, u_n$ be an orthogonal basis for a subspace $U$ in an inner product space. The orthogonal projection of any vector $x$ onto $U$ is the point $\displaystyle p=\sum_{i=1}^{n}\left \langle x,\hat{u}_i \right \rangle\hat{u}_i$.

Could someone assists me with the geometric interpretation?

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    $\begingroup$ I guess that $\hat{u}_i= \frac{u_i}{|u_i|}$. $\endgroup$ – Silvia Ghinassi Oct 30 '15 at 16:48
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    $\begingroup$ That is correct. It's the normalized version of the 'original' vector $\endgroup$ – Mathematicing Oct 30 '15 at 16:49
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    $\begingroup$ Write $\vec{x}$ as a sum of a vector in $U$ and a vector ortogonal to $U$. The projection is merely the vector in $U$. $\endgroup$ – Michael Burr Oct 30 '15 at 16:49
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A geometric interpretation is that if a subspace $U$ is the orthogonal direct sum of subspaces $U_1, \dots ,U_n$, then you can prove

  1. That the projector on $U$ is the sum of the projectors on $U_1, \dots,U_n$.
  2. That the projector on a vector line defined by a normalized vector $u$ is $q(x)=\langle x,u \rangle u$

Using those two facts (that are good to try to prove!), you'll get the desired result.

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