# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 15

    Linear subspaces and dual space
Linear subspaces of some $K$-vector space $V$ are in direct relation with linear subspaces of the dual space ${V}^{*}$.

## 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 again linear subspaces, see Exercise 15.4 . Whether a linear form $f$ belongs to $U^{\perp }$ can be checked on a generating system of $U$, see Exercise 15.5 . In the second semester, when we have inner products at hand, there will also be an orthogonal space for $U\subseteq V$ in $V$ itself.

## 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 Lemma 14.6 , 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 Theorem 11.5 .
The first part follows by using $U={\left(U^{\perp }\right)}^{\perp }$ and applying the second part to $F=U^{\perp }$.
$\Box$

## Lemma

Let $V$ be a finite-dimensional $K$-vector space and $U\subseteq V$ a linear subspace. Then the following hold.
a) There exist linear forms $L_{1},\ldots ,L_{r}$ on $V$ such that
    $U=\bigcap _{i=1}^{r}\operatorname {kern} L_{i}\,.$
b) $U\subseteq V$ is the kernel of a linear mapping on $V$ to some $K^{r}$.
c) Every linear subspace $U\subseteq K^{n}$ is the solution space of a

### Proof

$\Box$
    The dual mapping

## Definition

Let $K$ denote a field, let $V$ and $W$ denote $K$-vector spaces, and let
    $\varphi \colon V\longrightarrow W$
denote a $K$-linear mapping. Then the mapping
    $\varphi ^{*}\colon \operatorname {Hom} _{K}{\left(W,K\right)}={W}^{*}\longrightarrow \operatorname {Hom} _{K}{\left(V,K\right)}={V}^{*},f\longmapsto f\circ \varphi ,$
$\varphi$
This assignment arises from just considering the composition
    $V{\stackrel {\varphi }{\longrightarrow }}W{\stackrel {f}{\longrightarrow }}K.$
The dual mapping is a special case of the situation described in Lemma 13.8  (1). In particular, the dual mapping is again linear.

## Lemma

Let $U,V,W$ denote vector spaces over a field $K$ and let
    $\psi \colon U\longrightarrow V$
and
    $\varphi \colon V\longrightarrow W$
be
* For the dual mapping, we have
    ${\left(\varphi \circ \psi \right)}^{*}=\psi ^{*}\circ \varphi ^{*}\,.$
* For the identity on $V$, we have
    ${\operatorname {Id} _{V}}^{*}=\operatorname {Id} _{{V}^{*}}\,.$
* If $\psi$ is surjective then $\psi ^{*}$ is injective.
* If $\psi$ is injective then $\psi ^{*}$ is surjective.

### Proof

(1). For $f\in {W}^{*}$, we have
    ${\left(\varphi \circ \psi \right)}^{*}(f)=f\circ {\left(\varphi \circ \psi \right)}={\left(f\circ \varphi \right)}\circ \psi =\varphi ^{*}(f)\circ \psi =\psi ^{*}(\varphi ^{*}(f))\,.$
(2) follows directly from $f\circ \operatorname {Id} _{V}=f$.
(3). Let $f\in {V}^{*}$ and
    $\psi ^{*}(f)=0\,.$
Because of the surjectivity of $\psi$, there exist for every $v\in V$ a $u\in U$ such that $\psi (u)=v$. Therefore
    $f(v)=f(\psi (u))=(\psi ^{*}(f))(u)=0\,,$
and $f$ is itself the zero mapping. Due to Lemma 11.4 , $\psi ^{*}$ injective.
(4). The condition means that we may consider $U\subseteq V$ as a linear subspace. Because of Lemma 9.12 , we can write
    $V=U\oplus U'\,$
with another $K$-linear subspace $U'\subseteq V$. A linear form
    $g\colon U\longrightarrow K$
can always be extended to a linear form
    ${\tilde {g}}\colon V\longrightarrow K,$
for example, by defining ${\tilde {g}}$ on $U'$ to be the zero form. This means the surjectivity.
$\Box$

## Lemma

Let $K$ be a field and let $V$ and $W$ be vector spaces over $K$, where $W$ is finite-dimensional. Let
    $\varphi \colon V\longrightarrow W$
denote a linear mapping. Then there exist vectors $w_{1},\ldots ,w_{n}\in W$ and linear forms $f_{1},\ldots ,f_{n}$ on $V$ such that
    $\varphi =f_{1}w_{1}+f_{2}w_{2}+\cdots +f_{n}w_{n}\,$

### Proof

Let $w_{1},\ldots ,w_{n}$ be a basis of $W$ and $w_{1}^{*},\ldots ,w_{n}^{*}$ the corresponding dual basis. We set
    $f_{i}=\varphi ^{*}(w_{i}^{*})=w_{i}^{*}\circ \varphi \,.$
Then, for every vector $v\in V$, we have
    ${\begin{aligned}{\left(\sum _{i=1}^{n}f_{i}w_{i}\right)}(v)&=\sum _{i=1}^{n}f_{i}(v)w_{i}\\&=\sum _{i=1}^{n}{\left(w_{i}^{*}\circ \varphi \right)}(v)w_{i}\\&=\sum _{i=1}^{n}w_{i}^{*}(\varphi (v))w_{i}\\&=\varphi (v),\end{aligned}}$
where the last equation rests on Lemma 14.13 .
$\Box$

## Lemma

Let $K$ be a field, let $V$ denote an $n$-dimensional $K$-vector space with a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$, and let $W$ be an $m$-dimensional vector space with a basis ${\mathfrak {w}}=w_{1},\ldots ,w_{m}$. Let $v_{1}^{*},\ldots ,v_{n}^{*}$ and $w_{1}^{*},\ldots ,w_{m}^{*}$ be the corresponding dual bases. Let
    $\varphi \colon V\longrightarrow W$
be a linear mapping, and suppose that it is described by the $m\times n$-matrix
    $M=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )=(a_{ij})_{ij}\,$
with respect to the given bases. Then the dual mapping
    $\varphi ^{*}\colon {W}^{*}\longrightarrow {V}^{*}$
is described by the transposed matrix ${{\left(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\right)}^{\text{tr}}}$ with respect to the dual bases of
${W}^{*}$
${V}^{*}$

### Proof

The claim means the equality
    $\varphi ^{*}(w_{j}^{*})=\sum _{i=1}^{n}a_{ji}v_{i}^{*}\,$
in ${V}^{*}$. This can be checked on the basis $v_{k}$, $k=1,\ldots ,n$. On one hand, we have
    ${\begin{aligned}{\left(\varphi ^{*}(w_{j}^{*})\right)}(v_{k})&=w_{j}^{*}{\left(\varphi (v_{k})\right)}\\&=w_{j}^{*}{\left(\sum _{r=1}^{m}a_{rk}w_{r}\right)}\\&=\sum _{r=1}^{m}a_{rk}w_{j}^{*}(w_{r})\\&=a_{jk},\end{aligned}}$
on the other hand, we have
    ${\left(\sum _{i=1}^{n}a_{ji}v_{i}^{*}\right)}(v_{k})=a_{jk}\,.$
$\Box$
    The bidual

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. Then the dual space of the dual space ${V}^{*}$, that is,
    ${({V}^{*})}^{*},$
$V$

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space. Then there exists a natural injective linear mapping
    $\Psi \colon V\longrightarrow {({V}^{*})}^{*},v\longmapsto {\left(f\mapsto f(v)\right)}.$
If $V$ has finite dimension, then $\Psi$ is an

### Proof

Let $v\in V$ be fixed. First of all, we show that $\Psi (v)$ is a linear form on the dual space ${V}^{*}$. Obviously, $\Psi (v)$ is a mapping from ${V}^{*}$ to $K$. The additivity follows from
    $(\Psi (v))(f_{1}+f_{2})=(f_{1}+f_{2})(v)=f_{1}(v)+f_{2}(v)=(\Psi (v))(f_{1})+(\Psi (v))(f_{2})\,,$
where we have used the definition of the addition on the dual space. The compatibility with the scalar multiplication follows similarly from
    $(\Psi (v))(sf)=(sf)(v)=s(f(v))=s((\Psi (v))(f))\,.$
In order to prove the additivity of $\Psi$, let $v,w\in V$ be given. We have to show the equality
    $\Psi (v+w)=\Psi (v)+\Psi (w)\,.$
This is an equality inside of ${\left({V}^{*}\right)}^{*}$, in particular, it is an equality of mappings. So let $f\in {V}^{*}$ be given. Then, the additivity follows from
    $(\Psi (v+w))(f)=f(v+w)=f(v)+f(w)=(\Psi (v))(f)+(\Psi (w))(f)\,.$
The scalar compatibility follows from
    $(\Psi (sv))(f)=f(sv)=s(f(v))=s((\Psi (v))(f))\,.$
In order to prove injectivity, let $v\in V$ with $\Psi (v)=0$ be given. this means that for all linear forms $f\in {V}^{*}$, we have $f(v)=0$. But then, due to Fact *****, we have
    $v=0\,.$
By the criterion for injectiviy, $\Psi$ is injective.
In the finite-dimensional case, the bijectivity follows from injectivity and from Corollary 13.12 .
$\Box$
Thus, the mapping $\Psi$ sends a vector $v$ to the evaluation (or evaluation mapping) which evaluates a linear form $f$ an the point $v$.
    Footnotes
* ↑ Here, $fw$ is to be understood in the sense of Remark 14.4 .
* ↑ In $W$, the relation $\varphi (v_{k})=\sum _{r=1}^{m}a_{rk}w_{r}$ hold. Note that here, the running index is the first index; in the equation claimed, the running index is the second index, corresponding to transposing.