12 - Duals

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

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

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 $f{\mid }_U=0$

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 $U^0\ne \varnothing$