Inner product/K/Orthogonality/Introduction/Section

When an inner product is given, then we can express the property of two vectors to be orthogonal to each other.

Definition

Let ${}V$ denote a vector space over ${}{\mathbb {K} }$, endowed with an inner product ${}\left\langle -,-\right\rangle$. We say that two vectors ${}v,w\in V$ are orthogonal (or perpendicular) to each other if

${}\left\langle v,w\right\rangle =0\,$

holds.

Remark

The following reasoning shows that the orthogonality, defined via the inner product, corresponds to the intuitively given orthogonality . For orthogonal vectors ${}u,v\in V$, we get that ${}v$ has the same distance to the points ${}u$ and ${}-u$. This holds because of

${}{\begin{aligned}\Vert {v-u}\Vert ^{2}&=\left\langle v-u,v-u\right\rangle \\&=\left\langle v,v\right\rangle +\left\langle u,u\right\rangle \\&=\left\langle v+u,v+u\right\rangle \\&=\Vert {v+u}\Vert ^{2}.\end{aligned}}$

The reverse statement holds as well; see exercise.

Pythagoras of Samos lived in the sixth century a.D. However, "his“ theorem was already known thousand years earlier in Babylon.

Let us recall the Theorem of Pythagoras.

The following theorem is the Theorem of Pythagoras; more precisely, it is the version in the context of an inner product, and it is trivial. However, the relation to the classical, elementary-geometric Theorem of Pythagoras is difficult, because it is not clear at all whether our concept of orthogonality and our concept of a length, both introduced via the inner product, coincide with the corresponding intuitive concepts. That our concept of a norm is the true length concept rests itself on the Theorem of Pythagoras in a Cartesian coordinate system, which presupposes the classical theorem.

Theorem

Let ${}V$ be a ${}{\mathbb {K} }$-vector space, endowed with an inner product ${}\left\langle -,-\right\rangle$. Let ${}u,v\in V$ be vectors that are orthogonal to each other. Then

${}\Vert {u+v}\Vert ^{2}=\Vert {u}\Vert ^{2}+\Vert {v}\Vert ^{2}\,$

holds.

Proof

We have

${}\Vert {u+v}\Vert ^{2}=\left\langle u+v,u+v\right\rangle =\left\langle u,u\right\rangle +2\left\langle u,v\right\rangle +\left\langle v,v\right\rangle =\Vert {u}\Vert ^{2}+\Vert {v}\Vert ^{2}\,.$

$\Box$

Definition

Let ${}V$ be a ${}{\mathbb {K} }$-vector space, endowed with an inner product, and let ${}U\subseteq V$ denote a linear subspace. Then

${}U^{\perp }={\left\{v\in V\mid \left\langle v,u\right\rangle =0{\text{ for all }}u\in U\right\}}\,$

is called the orthogonal complement of ${}U$.

The orthogonal complement of a linear subspace is again a linear subspace, see exercise. If a generating system of ${}U$ is given, then a vector ${}v\in V$ belongs to the orthogonal complement of ${}U$ if it is orthogonal to all the vectors of the generating system, see exercise.

Example

Let ${}V=\mathbb {R} ^{n}$, endowed with the standard inner product. For the one-dimensional linear subspace ${}\mathbb {R} e_{i}$, generated by the standard vector ${}e_{i}$, the orthogonal complement consits of all vectors ${}{\begin{pmatrix}x_{1}\\\vdots \\x_{i-1}\\0\\x_{i+1}\\\vdots \\x_{n}\end{pmatrix}}$, where the ${}i$-th entry is ${}0$. For a one-dimensional linear subspace ${}\mathbb {R} v$, generated by a vector

${}v={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}\neq 0\,,$

the orthogonal complement can be found by determining the solution space of the linear equation

${}a_{1}x_{1}+\cdots +a_{n}x_{n}=0\,.$

The orthogonal space

${}U=(\mathbb {R} v)^{\perp }={\left\{{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\mid a_{1}x_{1}+\cdots +a_{n}x_{n}=0\right\}}\,$

has dimension ${}n-1$; this is a so-called hyperplane. The vector ${}v$ is called a normal vector for the hyperplane ${}U$.

For a linear subspace ${}U\subseteq \mathbb {R} ^{n}$ that is given by a basis (or a generating system) ${}v_{i}={\begin{pmatrix}a_{i1}\\\vdots \\a_{in}\end{pmatrix}}$, ${}i=1,\ldots ,k$, the orthogonal complement is the solution space of the system of linear equations

${}A{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}=0\,,$

where ${}A={\left(a_{ij}\right)}$ is the matrix formed by the vectors ${}v_{i}$ as rows.