# Geometric interpretation for projection of a vector $x$ onto a subspace $U$.

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?

• I guess that $\hat{u}_i= \frac{u_i}{|u_i|}$. – Silvia Ghinassi Oct 30 '15 at 16:48
• That is correct. It's the normalized version of the 'original' vector – Mathematicing Oct 30 '15 at 16:49
• Write $\vec{x}$ as a sum of a vector in $U$ and a vector ortogonal to $U$. The projection is merely the vector in $U$. – Michael Burr Oct 30 '15 at 16:49

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 $q(x)=\langle x,u \rangle u$