Linear subspace/Dual space/Orthogonal space/Correspondence/Fact/Proof

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

To fact