Vector space/K/Inner product/Orthogonal projection/Introduction/Section

For a finite-dimensional ${}{\mathbb {K} }$-vector space, endowed with an inner product, and a linear subspace ${}U\subseteq V$, there exists an orthogonal complement ${}U^{\perp }$, and the space has the direct sum decomposition

${}V=U\oplus U^{\perp }\,.$

The projection

$p_{U}\colon V\longrightarrow U$

along ${}U^{\perp }$ is called the orthogonal projection onto ${}U$. This projection depends only on ${}U$, because the orthogonal complement is uniquely determined. Often, also the mapping ${}V\rightarrow U\rightarrow V$ is called the orthogonal projection onto ${}U$. An orthogonal projection is also described as dropping a perpendicular onto ${}U$.

Lemma

Let ${}V$ be a finite-dimensional ${}{\mathbb {K} }$-vector space, endowed with an inner product, and let ${}U\subseteq V$ denote a linear subspace with an orthonormal basis ${}u_{1},\ldots ,u_{m}$ of ${}U$. Then the orthogonal projection onto ${}U$ is given by

${}p_{U}(v)=\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}\,.$

Proof

We extend the basis to an orthonormal basis ${}u_{1},\ldots ,u_{n}$ of ${}V$. The orthogonal complement of ${}U$ is

${}U^{\perp }=\langle u_{m+1},\ldots ,u_{n}\rangle \,.$

Due to fact, we have

${}v=\sum _{i=1}^{n}\left\langle v,u_{i}\right\rangle u_{i}={\left(\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}\right)}+{\left(\sum _{i=m+1}^{n}\left\langle v,u_{i}\right\rangle u_{i}\right)}\,.$

Therefore, ${}\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}$ is the projection onto ${}U$ along ${}U^{\perp }$.

$\Box$