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

of ${}U$.

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

of ${}F$.

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

${}{V}^{*}$. Then the following statements hold.

1. For linear subspaces ${}U\subseteq U'\subseteq V$, we have

   ${}U^{\perp }\supseteq U'^{\perp }\,.$

2. For linear subspaces ${}F\subseteq F'\subseteq {V}^{*}$, we have

   ${}F^{\perp }\supseteq F'^{\perp }\,.$

3. Let ${}V$ be finite-dimensional. Then

   ${}{\left(U^{\perp }\right)}^{\perp }=U\,$

   and

   ${}{\left(F^{\perp }\right)}^{\perp }=F\,.$

4. Let ${}V$ be finite-dimensional. Then

   ${}\dim _{K}{\left(U^{\perp }\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,$

   and

   ${}\dim _{K}{\left(F^{\perp }\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(F\right)}\,.$

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 ${}F=U^{\perp }$.

$\Box$