# Vector space/Dual space/Linear subspace/Introduction/Section
## Definition

For a linear subspace $U\subseteq V$ in a $K$-vector space, we call
    $U^{\perp }={\left\{f\in {V}^{*}\mid f(u)=0{\text{ for all }}u\in U\right\}}\subseteq {V}^{*}\,$
the orthogonal space
$U$
These orthogonal spaces are linear subspaces of ${V}^{*}$; see exercise. Whether a linear form $f$ belongs to $U^{\perp }$ can be checked on a generating system of $U$; see exercise. The property $f\in U^{\perp }$ is equivalent with $U\subseteq \operatorname {kern} f$.

## Example

We consider the linear subspace
    $U=\langle {\begin{pmatrix}8\\6\\5\end{pmatrix}},{\begin{pmatrix}4\\7\\-3\end{pmatrix}}\rangle \subseteq \mathbb {R} ^{3}\,.$
The orthogonal space of $U$ consists of all linear forms
    $f\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R}$
with $f{\begin{pmatrix}8\\6\\5\end{pmatrix}}=0$ and $f{\begin{pmatrix}4\\7\\-3\end{pmatrix}}=0$. Because a linear form is described by a row matrix $\left(a,\,b,\,c\right)$ with respect to the standard basis, we are dealing with the solution set of the linear system
    $8a+6b+5c=0\,,$
    $4a+7b-3c=0\,.$
The solution space is
    $U^{\perp }={\left\{s\left({\frac {17}{4}},\,1,\,-8\right)\mid s\in K\right\}}\,.$

## Example

Let $V$ be a finite-dimensional $K$-vector space with a basis $v_{i}$, $i\in I$, and the corresponding dual basis $v_{i}^{*}$, $i\in I$. Let
    $U=\langle v_{j},\,j\in J\rangle \,$
for a subset $J\subseteq I$. Then
    $U^{\perp }=\langle v_{i}^{*},\,i\not \in J\rangle \,.$

## Definition

Let $V$ be a $K$-vector space, and $F\subseteq {V}^{*}$ be a linear subspace of the dual space ${V}^{*}$ of $V$. Then
    $F^{\perp }={\left\{v\in V\mid f(v)=0{\text{ for all }}f\in F\right\}}\subseteq V\,$
is called the orthogonal space
$F$

## Example

Let a homogeneous linear system
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
over a field $K$ be given. We consider the $i$-th equation as a kernel condition for the linear form
    $L_{i}\colon K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{n}\right)\longmapsto \sum _{j=1}^{n}a_{ij}x_{j}.$
Let
    $F=\langle L_{1},\ldots ,L_{m}\rangle \,$
denote the linear subspace of the dual space ${K^{n}}^{*}$ generated by these linear forms. Then $F^{\perp }$ is the solution space of this linear system.
In general, we have the relation
    $F^{\perp }=\bigcap _{f\in F}\operatorname {kern} f\,.$
In particular,
    $\langle f\rangle ^{\perp }=\operatorname {kern} f\,.$

## Lemma

Let $V$ be a $K$-vector space with dual space
${V}^{*}$
* For linear subspaces $U\subseteq U'\subseteq V$, we have
    $U^{\perp }\supseteq U'^{\perp }\,.$
* For linear subspaces $F\subseteq F'\subseteq {V}^{*}$, we have
    $F^{\perp }\supseteq F'^{\perp }\,.$
* Let $V$ be finite-dimensional. Then and
    ${\left(U^{\perp }\right)}^{\perp }=U\,$
    ${\left(F^{\perp }\right)}^{\perp }=F\,.$
* Let $V$ be finite-dimensional. Then and
    $\dim _{K}{\left(U^{\perp }\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,$
    $\dim _{K}{\left(F^{\perp }\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(F\right)}\,.$

### Proof

(1) and (2) are clear. (3). The inclusion
    $U\subseteq {\left(U^{\perp }\right)}^{\perp }\,$
is also clear. Let $v\in V$, $v\notin U$. Then we can choose a basis $u_{1},\ldots ,u_{r}$ of $U$ and extend it to a basis $u_{1},\ldots ,u_{r},v,v_{1},\ldots ,v_{\ell }$ of $V$. The linear form $v^{*}$ vanishes on $U$, therefore, it belongs to $U^{\perp }$. Because of
    $v^{*}(v)=1\neq 0\,,$
we have $v\notin {\left(U^{\perp }\right)}^{\perp }$.
The inclusion
    $F\subseteq {\left(F^{\perp }\right)}^{\perp }\,$
holds immediately. Let $f\in {\left(F^{\perp }\right)}^{\perp }$, that is,
    $F^{\perp }\subseteq \operatorname {kern} f\,.$
Let $f_{1},\ldots ,f_{m}$ be a generating system of $F$. Due to exercise we have that $f$ is a linear combination of the $f_{i}$; therefore, $f\in F$.
(4). We first prove the second part. Let $f_{1},\ldots ,f_{r}$ be a basis of $F$, and let
    $\varphi \colon V\longrightarrow K^{r}$
denote the mapping where these linear forms are the components. Here, we have
    $F^{\perp }=\operatorname {kern} \varphi \,.$
Assume that the mapping $\varphi$ is not surjective. Then $\operatorname {Im} \varphi$ is a strict linear subspace of $K^{r}$ and its dimension is at most $r-1$. Let $W$ be a $(r-1)$-dimensional linear subspace with
    $\operatorname {Im} \varphi \subseteq W\subseteq K^{r}\,.$
Due to fact, there is a linear form
    $g\colon K^{r}\longrightarrow K,$
whose kernel is exactly $W$. Write $g=\sum _{i=1}^{r}a_{i}p_{i}$, where $p_{i}$ is the $i$th projection. Then
    $\sum _{i=1}^{r}a_{i}f_{i}=g\circ \varphi =0\,,$
contradicting the linear independence of the $f_{i}$. Moreover, $\varphi$ is surjective and the statement follows from fact.
The first part follows by using $U={\left(U^{\perp }\right)}^{\perp }$ and applying the second part to $F=U^{\perp }$.
$\Box$