ALG1_L20

Dual Spaces

Preliminaries

For vector spaces $V$ and $W$ over a field $\mathbb{F}$, we define

$${\mathrm{hom}}_{\mathbb{F}}(V,W)\equiv \{\text{All linear transformations V→W}\}.$$

This is a vector space over $\mathbb{F}$ with point-wise addition and scalar multiplication.

Important

The field over which a vector space is defined can change its dimension. For example, consider the complex numbers $\mathbb{C}$ as a vector space over the field $\mathbb{C}$. The dimension of this vector space is 1, i.e, ${\dim }_{\mathbb{C}}\mathbb{C}=1$. However, if we consider $\mathbb{C}$ as a vector space over the field $\mathbb{R}$, its dimension is 2, i.e, ${\dim }_{\mathbb{R}}\mathbb{C}=2$.

Theorem 1(Theorem ).

If ${\dim }_{\mathbb{F}}V=m$ and ${\dim }_{\mathbb{F}}W=n$ then ${\dim }_{\mathbb{F}}{\mathrm{hom}}_{\mathbb{F}}(V,W)=mn$

Proof
Let ${\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_m$ be a basis of $V$.
Let ${\mathbf{w}}_1,{\mathbf{w}}_2,\dots ,{\mathbf{w}}_n$ be a basis of $W$.
For a vector $\mathbf{v}\in V$, $\mathbf{v}={\lambda }_1{\mathbf{v}}_1+{\lambda }_2{\mathbf{v}}_2+\dots {\lambda }_m{\mathbf{v}}_m$ for some ${\lambda }_i\in \mathbb{F}$.

For $1\le i\le m$ and $1\le j\le n$, define $T_{ij}:V\to W$ such that $\mathbf{v}\mapsto {\lambda }_i{\mathbf{w}}_j$. Observe that

$$T_{ij}({\mathbf{v}}_k)=\{\begin{array}{ll}0& k\ne i\\ {\mathbf{w}}_j& k=i\end{array}$$

We want to show that $\{T_{ij}\}$ is a basis for ${\mathrm{hom}}_{\mathbb{F}}(V,W)$.

1) $\{T_{ij}\}$ is a spanning set
Let $S\in {\mathrm{hom}}_{\mathbb{F}}(V,W)$.

$$S({\mathbf{v}}_i)\in W⟹S({\mathbf{v}}_i)={\alpha }_{i1}{\mathbf{w}}_1+{\alpha }_{i2}{\mathbf{w}}_2+\cdots +{\alpha }_{in}{\mathbf{w}}_n.$$

Define

$$S_0\equiv \begin{array}{cccccccc}{\alpha }_{11}{T}_{11}& +& {\alpha }_{12}{T}_{12}& +& \dots & +& {\alpha }_{1n}{T}_{1n}& +\\ {\alpha }_{21}{T}_{21}& +& {\alpha }_{22}{T}_{22}& +& \dots & +& {\alpha }_{2n}{T}_{2n}& +\\ ⋮& & ⋮& & \ddots & & ⋮& \\ {\alpha }_{m1}{T}_{m1}& +& {\alpha }_{m2}{T}_{m2}& +& \dots & +& {\alpha }_{mn}{T}_{mn}& \end{array}$$

Now,

$$\begin{array}{rl}{S}_0({\mathbf{v}}_k)& ={\alpha }_{11}{T}_{11}({\mathbf{v}}_k)+\cdots +{\alpha }_{mn}{T}_{mn}({\mathbf{v}}_k)\\ & ={\alpha }_{k1}{\mathbf{w}}_1+{\alpha }_{k2}{\mathbf{w}}_2+\cdots +{\alpha }_{kn}{\mathbf{w}}_n\\ & =S({\mathbf{v}}_k).\end{array}$$

That is, $S$ and $S_0$ agree on the basis of $V$. Since a linear map is characterized by the values it takes on a basis, this implies $S$ and $S_0$ are equal. Therefore $S\in \text{Span}\left(\{T_{ij}\}\right)$.

2) $\{T_{ij}\}$ is a linearly independent set
Suppose there exist ${\beta }_{ij}$ such that

$${\beta }_{11}{T}_{11}+\cdots +{\beta }_{mn}{T}_{mn}=0.$$

Now for $1\le k\le m$, we can say

$$\begin{array}{rl}& ({\beta }_{11}{T}_{11}+\cdots +{\beta }_{mn}{T}_{mn})({\mathbf{v}}_k)=0\\ ⟹& {\beta }_{k1}{\mathbf{w}}_1+\cdots +{\beta }_{kn}{\mathbf{w}}_n=0\\ ⟹& {\beta }_{k1}={\beta }_{k2}=\cdots ={\beta }_{kn}=0\end{array}$$

Therefore all ${\beta }_{ij}=0$. ❏

Theorem 2(Corollary).

A basis for the set of all $m\times n$ matrices is given by matrices $\{M_{ij}\}$ with $1$ in the $(i,j{)}^{th}$ position and $0$s everywhere else.

Theorem 3(Corollary).

$${\dim }_{\mathbb{F}}{\mathrm{hom}}_{\mathbb{F}}(V,\mathbb{F})={\dim }_{\mathbb{F}}V.$$

So, if $V$ is a finite dimensional vector space, then $V\cong \mathrm{hom}(V,\mathbb{F})$, since $V\cong {\mathbb{F}}^{\dim V}\cong \mathrm{hom}(V,\mathbb{F})$. There is no universal construction for an isomorphism $\varphi :V\to \mathrm{hom}(V,\mathbb{F})$ though, because defining such an isomorphism requires us to pick a basis. More on this in a bit.

Notation

Definition 4(Definition).

If $V$ is a vector space over field $\mathbb{F}$, then its dual space is $\hat{V}\equiv {\mathrm{hom}}_{\mathbb{F}}(V,\mathbb{F})$.

Elements of $\hat{V}$, $f:V\to \mathbb{F}$ are called linear functionals.

If $V$ has basis ${\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_n$ then define ${\hat{v}}_i\in \hat{V}$ as

$$\widehat{v_i}({\mathbf{v}}_k)\equiv \{\begin{array}{ll}0& i\ne k\\ 1& i=k\end{array}$$

These are the $T_{ij}$‘s from above ($j$ is moot since $\dim \mathbb{F}=1$). It is evident that ${\hat{v}}_1,{\hat{v}}_2,\dots ,{\hat{v}}_n$ is a basis of $\hat{V}$.

Remark

If $V$ is a finite dimensional vector space and $\exists \mathbf{v}\in V$ such that $\mathbf{v}\ne 0$ then there is an $f\in \hat{V}$ such that $f(\mathbf{v})\ne 0$.

Incidentally, this is also true for infinite dimensional vector spaces.

The double dual

Definition 5(Definition).

We will denote the dual space of the dual space of $V$ (aka the double dual of $V$) by $\hat{\hat{V}}\equiv {\mathrm{hom}}_{\mathbb{F}}(\hat{V},\mathbb{F})$.

For ${\mathbf{v}}_0\in V$, define corresponding $T_{{\mathbf{v}}_0}\in \hat{\hat{V}}$ ($T_{{\mathbf{v}}_0}:\hat{V}\to \mathbb{F}$) such that $f\stackrel{T_{{\mathbf{v}}_0}}{\mapsto }f({\mathbf{v}}_0)$. Define $\psi :V\to \hat{\hat{V}}$ such that $\mathbf{v}\stackrel{\psi }{\mapsto }{T}_{\mathbf{v}}$.

Now, observe that $T_{\mathbf{v}}$ is linear over $\mathbf{v}$:

$$T_{\alpha \mathbf{v}+\beta \mathbf{w}}(f)=f(\alpha \mathbf{v}+\beta \mathbf{w})=\alpha f(\mathbf{v})+\beta f(\mathbf{w})=\alpha T_{\mathbf{v}}(f)+\beta T_{\mathbf{w}}(f).$$

Hence,

$$\psi (\alpha \mathbf{v}+\beta \mathbf{w})=T_{\alpha \mathbf{v}+\beta \mathbf{w}}=\alpha T_{\mathbf{v}}+\beta T_{\mathbf{w}}=\alpha \psi (\mathbf{v})+\beta \psi (\mathbf{w}).$$

So, $\psi$ is a linear map!

Is psi an isomorphism?

If we can show that $\psi (\mathbf{v})=0⟹\mathbf{v}=0$ then we can say that $\psi$ is injective.

$$\begin{array}{lllc}& \psi (\mathbf{v})& =0& \\ ⟹& T_{\mathbf{v}}& =0& \\ ⟹& T_{\mathbf{v}}(f)& =0& \forall f\in \hat{V}\\ ⟹& f(\mathbf{v})& =0& \forall f\in \hat{V}\\ ⟹& \mathbf{v}& =0& \because \text{ if }\mathbf{v}\ne 0,\text{ then }\exists f\in \hat{V}\text{ s.t }f(\mathbf{v})\ne 0\end{array}$$

Now, from the rank nullity theorem, $\dim \mathrm{Im}\psi =\dim V$. Note that $\dim V=\dim \hat{V}=\dim \hat{\hat{V}}$. Hence, $\psi$ is surjective. Therefore, $\psi$ is a bijective linear map, i.e, an isomorphism.

$$\psi :V\stackrel{\sim }{⟶}\hat{\hat{V}}$$

Notice that we did not have to make any arbitrary “choices” (like choosing a basis, etc) when defining $\psi$. Such an isomorphism is called a canonical isomorphism. Contrast with an isomorphism $\varphi :V\to \hat{V}$, which requires us to pick a basis to define. Thus, $V\cong \hat{V}$ is not canonical, while $V\cong \hat{\hat{V}}$ is.

This renders all higher dual spaces moot, since we have canonical isomorphisms

$$V\cong \hat{\hat{V}}\cong \hat{\widehat{\hat{\hat{V}}}}\cong \dots ,$$ $$\hat{V}\cong \widehat{\hat{\hat{V}}}\cong \hat{\hat{\widehat{\hat{\hat{V}}}}}\cong \dots .$$