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

of ${}U$.

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

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 ${}v\notin {\left(U^{\perp }\right)}^{\perp }$.

(4). 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 14.5 , there is a linear form

$g\colon K^{r}\longrightarrow K,$

whose kernel is exactly ${}W$. Write ${}g=\sum _{i=1}^{r}a_{i}p_{i}$. 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 .

$\Box$

Corollary

Let ${}V$ be a finite-dimensional ${}K$-vector space and ${}U\subseteq V$ a linear subspace. Then there exist linear forms ${}L_{1},\ldots ,L_{r}$ on ${}V$ such that

${}U=\bigcap _{i=1}^{r}\operatorname {kern} L_{i}\,.$

Every linear subspace

${}U\subseteq V$ is the kernel of a linear mapping and every linear subspace of ${}K^{n}$ is the solution space of a

linear system.

Proof

See Exercise 15.8 .

$\Box$

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

$\varphi ^{*}\colon \operatorname {Hom} _{K}{\left(W,K\right)}={W}^{*}\longrightarrow \operatorname {Hom} _{K}{\left(V,K\right)}={V}^{*},f\longmapsto f\circ \varphi ,$

is called the dual mapping of ${}\varphi$.

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

linear mappings. Then the following hold.

1. For the dual mapping, we have

   ${}{\left(\varphi \circ \psi \right)}^{*}=\psi ^{*}\circ \varphi ^{*}\,.$

2. For the identity on ${}V$, we have

   ${}{\operatorname {Id} _{V}}^{*}=\operatorname {Id} _{{V}^{*}}\,.$

3. If ${}\psi$ is surjective then ${}\psi ^{*}$ is injective.
4. 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

${}\varphi =f_{1}w_{1}+f_{2}w_{2}+\cdots +f_{n}w_{n}\,$

holds.

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 14.12 .

$\Box$

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

${}{W}^{*}$ and ${}{V}^{*}$.

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,

${({V}^{*})}^{*},$

is called the Bidual of ${}V$.

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

isomorphism.

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

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