07. Dual Spaces

Dual

Let $V$ be a vector space over $\mathbf{F}$

The dual $V^{\prime }$ is defined as the vector space of linear maps from $V$ to $\mathbf{F}$

$$V^{\prime }=\mathrm{Hom}(V,\mathbf{F})$$

The elements of $V^{\prime }$ are called linear functionals

Link to original

17 - Dual Basis Theorem

Dual Basis

Let $V$ be finite dimensional with basis $\mathcal{B}=\{e_1,\cdots ,e_n\}$

Define the Dual of $e_i^{\prime }$ of $e_i$ (relative to $\mathcal{B}$) by

$$e_i^{\prime }(e_j)={\delta }_{ij}=\{\begin{array}{ll}1& \text{ if }i=j\\ 0& \text{ if }i\ne j\end{array}$$

So

$${\mathcal{B}}^{\prime }:=\{e_1^{\prime },\cdots ,e_n^{\prime }\}\text{ is a basis for }{V}^{\prime }$$

where ${\mathcal{B}}^{\prime }$ is the dual basis

The assignment $e_i\mapsto e_i^{\prime }$ defines an isomorphism of vector spaces
Hence

$$\dim V=\dim V^{\prime }$$

Proof

Assume for some $a_i\in \mathbf{F}$ such that $\sum a_i{e}_i^{\prime }=0$ then for all $j$ then

$$0=0(e_j)=\left(\sum a_i{e}_i^{\prime }\right)(e_j)=\sum a_i{e}_i^{\prime }(e_j)=a_j$$

Hence ${\mathcal{B}}^{\prime }$ is linearly independent

Let $f\in V^{\prime }$ and define $a_i:=f(e_i)$
So

$$f=\sum a_j{e}_j^{\prime }$$

As $f$ evaluates to $a_i$ on $e_i$ and a linear map is determined by values on any basis then

$${\mathcal{B}}^{\prime }\text{ is spanning}$$

So ${\mathcal{B}}^{\prime }$ is a basis for $V^{\prime }$

Link to original

18 - Evaluation Map

Evaluation Map

Let $V$ be a finite dimensional vector space
Then

$$V^{\prime \prime }:=V\to (V^{\prime }{)}^{\prime }$$

is defined by $v\mapsto E_v$ which is a natural linear isomorphism
where $E_v$ is the evaluation map at $v$ defined by

$$E_v(f):=f(v)\text{ for }f\in V^{\prime }$$

Note: Natural means independent of a choice of basis

Proof

First $E_v$ is a linear map from $V^{\prime }$ to $\mathbf{F}$ since

$$E_v(f+\lambda g):=(f+\lambda g)(v):=f(v)+\lambda g(v)=:E_v(f)+\lambda E_v(g)$$

for all $f,g\in V^{\prime }$, $v\in V$ and $\lambda \in \mathbf{F}$
Hence assignment $v\mapsto E_v$ is well-defined

Map $v\mapsto E_v$ itself is linear as

$$E_{\lambda v+w}=\lambda E_v+E_w\text{ for all }v,w\in V\text{ and }\lambda in\mathbf{F}$$

as each functional $f$ is linear

Map is also injective, since

$$E_v=0⟹E_v(f)=f(v)=0\text{ for all }f\in V^{\prime }$$

If $v\ne 0$ then extend $\{e_1=v\}$ to a basis $\mathcal{B}$ of $V$
For $f=e_1^{\prime }$ then

$$E_v(e_1^{\prime })=e_1^{\prime }(e_1)=1$$

This contradicts the fact that $E_v(e_1)=0$
Hence $v=0$ which shows that the assignment $v\mapsto E_v$ is injective

By Dual Basis Theorem then

$$\dim V=\dim (V^{\prime }))=\dim (V^{\prime }{)}^{\prime }$$

So by injectivity and the Rank-Nullity Theorem then assignment is surjective

Link to original

Hyperplane

Let $V$ be a vector space with dimension $n$
Then the kernel of a non-zero linear function $f:V\to \mathbf{F}$ has dimension $n-1$

Preimage $f^{-1}(\{c\})$ for constant $c\in \mathbf{F}$ is called a hyperplane with dimension $n-1$

Case when $V={\mathbf{F}}^n$ (column vectors) Every hyperplane is defined by equation

$$a_1{b}_1+a\cdots +a_n{b}_n=c$$

for fixed scalar $c$ and fixed $\underline{b}=(b_1,\cdots ,b_n)\in ({\mathbf{F}}^n{)}^T$ (row vectors)

When $c=0$ then different choices of $\underline{b}$ can define the same hyperplane
E.g. scaling $\underline{b}$
So different functions can have same kernel

Link to original

---

7.1 Annihilators

Annihilator

Let $U\subseteq V$ be a subspace of $V$

The annihilator of $U$ is defined to be

$$U^0=\{f\in V^{\prime }:f(u)=0\text{ for all }u\in U\}$$

Hence $f\in V^{\prime }$ lies in $U^0$ if $f{\mid }_U=0$

Link to original

Subspace Property of Annihilators

Let $U\subseteq V$ be a subspace of $V$
then

$$U^0\text{ is a subspace of }{V}^{\prime }$$

Proof

Let $f,g\in U^0$ and $\lambda \in \mathbf{F}$

$$(f+\lambda g)(u)=f(u)+\lambda g(u)=0+\lambda \cdot 0=0$$

So $f+\lambda g\in U^0$

Also trivially, $0\in U^0$ so $U^0\ne \varnothing$

Link to original

19 - Dimension of Annihilators

Dimension Of Annihilators

Let $V$ be finite dimensional and $U\subseteq V$ be a subspace then

$$\dim (U^0)=\dim (V)-\dim (U)$$

Proof

Let $\{e_1,\cdots ,e_m\}$ be a basis for $U$ and extend it to $\{e_1,\cdots ,e_n\}$ for $V$
Let $\{e_1^{\prime },\cdots ,e_n^{\prime }\}$ be the dual basis

As $e_j^{\prime }(e_i)=0$ for $j=m+1,\cdots ,n$ and $i=1,\cdots ,m$ then

$$e_j^{\prime }\in U^0\text{ for }j=m+1,\cdots ,n$$

Hence

$$⟨e_{m+1}^{\prime },\cdot ,e_n^{\prime }⟩\subseteq U^0$$

Conversely for $f\in U^0$ then there exists $a_i\in \mathbf{F}$ such that $f={\sum }_{i=1}^n{a}_i{e}_i^{\prime }$
As $e_i\in U$ for $i=1,\cdots ,m$ then

$$f(e_i)=0\text{ and hence }{a}_i=0$$

Hence

$$f\in ⟨e_{m+1}^{\prime },\cdots ,e_n^{\prime }⟩$$

Thus

$$U^0=⟨e_{m+1}^{\prime },\cdots ,e_n^{\prime }⟩$$

where the set of vectors is a subset of the dual basis, spanning and is linearly independent so a basis of $U^0$

$$\dim (U^0)=n-m=\dim (V)-\dim (U)$$

Link to original

20 - Properties of Annihilators

Properties of Annihilators

Let $U,W$ be subspaces of $V$ then

$$\begin{array}{rl}& (1)U\subseteq W⟹W^0\subseteq U^0\\ & (2)(U+W{)}^0=U^0\cap W^0\\ & (3)U^0+W^0\subseteq (U\cap W{)}^0\end{array}$$

Note: for $(3)$ it is equal if $\dim (V)$ is finite

Proof

$$\begin{array}{rlrl}& (1)v\in W^0& & ⟹\forall w\in W:f(w)=0\\ & & & ⟹\forall u\in U\subseteq W:f(u)=0\\ & & & ⟹f\in U^0\\ & & & \\ & (2)f\in (U+W{)}^0& & ⟺\forall u\in U,\forall w\in W:f(u+w)=0\\ & & & ⟺\forall u\in U:f(u)=0\text{ and }\forall w\in W:f(w)=0\\ & & & ⟺f\in U^0\cap W^0\\ & & & \\ & (3)f\in U^0+W^0& & ⟹\exists g\in U^0\text{ and }\exists h\in W^0:f=g+h\\ & & & ⟹\forall x\in U\cap W:f(x)=g(x)+h(x)=0\\ & & & ⟹f\in (U\cap W{)}^0\end{array}$$

For $(3)$ then it follows that $U^0+W^0\subseteq (U\cap W{)}^0$
If $V$ is finite-dimensional then

$$\begin{array}{rl}\dim (U^0& +W^0)=\dim (U^0)+\dim (W^0)-\dim (U^0\cap W^0)\\ & =\dim (U^0)+\dim (W^0)-\dim (U+W{)}^0\\ & =(\dim (V)-\dim (U))+(\dim (V)-\dim (W))-(\dim (V)-\dim (U+W{)}^0)\\ & =\dim (V)-\dim (U)-\dim (W)+(\dim (U)+\dim (W)-\dim (U\cap W))\\ & =\dim (V)-\dim (U\cap W)\\ & =\dim ((U\cap W{)}^0)\end{array}$$

Note that the first step is from Dimension Formula

Link to original

21 - Isomorphism of the Natural Mapping

Isomorphism of the Natural Mapping

Let $U$ be a subspace of a finite dimensional vector space $V$
Under the natural map

$$V\to V^{\prime \prime }(:=(V^{\prime }{)}^{\prime })$$

which is given by

$$v\mapsto E_V$$

Then $U$ is mapped isomorphically to

$$U^{00}:=(U^0{)}^0$$

Proof

Let us here write $E:V\to V^{\prime \prime }$ for the natural isomorphism $v\mapsto E_v$

For $v\in V$ the functional $E_v$ is in $U^{00}$ if and only if

$$\forall f\in U^0\text{ we have }{E}_v(f)=f(v)=0$$

Hence if $v\in U$ then $E_v\in U^{00}$ thus

$$U\cong E(U)\subseteq U^{00}$$

When $V$ is finite dimensional then by Dimension of Annihilators

$$\begin{array}{rl}\dim (U^{00})& =\dim (V^{\prime })-\dim (U^0)\\ & =\dim (V)-(\dim (V)-\dim (U))\\ & =\dim (U)\end{array}$$

Hence

$$U\cong E(U)=U^{00}$$

Link to original

22 - Isomorphism of Quotient Space Dual

Isomorphism of Quotient Space Dual

Let $U\subseteq V$ be a subspace
Then there exists a normal isomorphism

$$U^0\simeq (V/U{)}^{\prime }$$

given by $f\mapsto \overline{f}$, where $\overline{f}(v+U):=f(v)$ for $v\in V$

Proof

Let $f\in U^0$
Note that $\overline{f}$ is well-defined because $f{\mid }_U=0$

Map is linear as

$$\overline{\lambda f+h}=\lambda \overline{f}+\overline{h}\text{ for all }f,h\in U^0\text{ and }\lambda \in \mathbf{F}$$

Map is injective as

$$\overline{f}=0⟹\overline{f}(v+U)=f(v)=0\text{ for all }v\in V⟹f=0$$

Note that for the finite dimensional case, considering dimensions of both sides gives result

In general, we can construct an inverse for $f\mapsto \overline{f}$ as follows.

Let $g\in (V/U{)}^{\prime }$ then define $\hat{g}\in V^{\prime }$ by $\hat{g}(v):=g(v+U)$, then $\hat{g}$ is linear in $v$
As the map $g\mapsto \hat{g}$ is linear in $g$, with the image in $U^0$
Finally then $\hat{\overline{f}}=f$ and $\overline{\hat{g}}=g$

Link to original

---

7.2 Dual Maps

Dual Map

Let $T:V\to W$ be a linear map of vector spaces

The dual map is defined by

$$T^{\prime }:W^{\prime }\to V^{\prime },f\mapsto f\circ T$$

Note that $f\circ T:V\to W\to \mathbf{F}$ is linear, so $f\circ T\in V^{\prime }$

Link to original

Linear Property of the Dual Map

Let $T:V\to W$ be a linear map of vector spaces then

$$T^{\prime }\text{ is a linear map }$$

Proof

Let $f,g\in W^{\prime }$, $\lambda \in \mathbf{F}$
For $v\in V$ then

$$\begin{array}{rl}{T}^{\prime }(f+\lambda g)(v)& =((f+\lambda g)\circ T)(v)\\ & =(f+\lambda g)(Tv)\\ & =f(Tv)+\lambda g(Tv)\\ & =T^{\prime }(f)(v)+\lambda T^{\prime }g(v)\\ & =(T^{\prime }(f)+\lambda T^{\prime }(g))(v)\end{array}$$

Link to original

23 - Dual Map Isomorphism Theorem

Dual Map Isomorphism Theorem

Let $V,W$ be two finite dimensional vector space
Assignment $T\mapsto T^{\prime }$ is a natural isomorphism from

$$\mathrm{hom}(V,W)\text{ to }\mathrm{hom}(W^{\prime },V^{\prime })$$

Proof

We need to check that the assignment $T\mapsto T^{\prime }$ linear in $T$
Let $T,S\in \mathrm{hom}(V,W)$ and $\lambda \in \mathbf{F}$
For $f\in W^{\prime }$ then

$$\begin{array}{rl}((T+\lambda S{)}^{\prime })(f))(v)& =f((T+\lambda S)(v))\\ & =f(T(v)+\lambda S(v))\\ & =f(T(v))+\lambda f(S(v))\\ & =T^{\prime }(f)(v)+\lambda S^{\prime }(f)(v)\\ & =(T^{\prime }(f)+\lambda S^{\prime }(f))(v)\\ & =((T^{\prime }+\lambda S^{\prime })(f))(v)\end{array}$$

Hence

$$(T+\lambda S{)}^{\prime }=T^{\prime }+\lambda S^{\prime }$$

To prove injectivity assume $T^{\prime }=0$ then
For all $f\in W^{\prime }$ then

$$T^{\prime }(f)=0$$

which is an identity of functionals on $V$ so
For all $f\in W^{\prime }$ , and for all $v\in V$ we have $T^{\prime }(f)(v)=0$ so

$$T^{\prime }(f)(v):=f(Tv)=:E_{Tv}(f)=0$$

But then $E_{Tv}=0$ hence $Tv=0$ by Evaluation Map applied to $W$

As this holds for all $v\in V$ then $T=0$

When the vector spaces are finite dimensional, we have

$$\dim (\mathrm{hom}(V,W))=\dim (V)\dim (W)=\dim W^{\prime }\dim V^{\prime }=\dim (\mathrm{hom}(W^{\prime },V^{\prime }))$$

Hence the map is also surjective

Link to original

24 - Dual Basis Matrix Transpose Theorem

Dual Basis Matrix Transpose Theorem

Let $V$ and $W$ be finite dimensional
Let ${\mathcal{B}}_W$ and ${\mathcal{B}}_V$ be bases for $W$ and $V$ then

For any linear map $T:V\to W$

$$({}_{{\mathcal{B}}_W}[T{]}_{{\mathcal{B}}_V}{)}^T={}_{{\mathcal{B}}_V^{\prime }}[T{]}_{{\mathcal{B}}_W^{\prime }}$$

where ${\mathcal{B}}_W^{\prime }$ and ${\mathcal{B}}_V^{\prime }$ are the dual bases

Proof

Let ${\mathcal{B}}_V=\{e_1,\cdots ,e_n\}$ and ${\mathcal{B}}_W=\{x_1,\cdots ,x_m\}$ and

$${}_{{\mathcal{B}}_W}[T{]}_{{\mathcal{B}}_V}=A=(a_{ij})$$

then

$$T(e_j)=\sum _{i=1}^m{a}_{ij}{x}_i\text{ and hence }{x}_i^{\prime }(T(e_j))=a_{ij}$$

Let

$${}_{{\mathcal{B}}_V^{\prime }}[T^{\prime }{]}_{{\mathcal{B}}_W^{\prime }}=B=(b_{ij})$$

then

$$T^{\prime }(x_i^{\prime })=\sum _{j=1}^n{b}_{ji}{e}_j^{\prime }\text{ and hence }(T^{\prime }(x_i^{\prime }))(e_j)=b_{ji}$$

As by definition then

$$(T^{\prime }(x_i^{\prime }))(e_j)=x_i^{\prime }(T(e_j))$$

Hence

$$a_{ij}=b_{ji}⟹A^T=B$$

Link to original