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Linear Functionals, Annihilator and Double Dual

Linear Functionals, Annihilator and Double Dual








Definition:


If $V$ is a vector space over the field $F$ and $S$ be a subset of $V$, the annihilator of $S$ is $S^0$ and $S^0$ is the set of linear functionals $f$ on $V$ such that $f(alpha)=0$ for each $alpha in S$.
Thus, $S^0={f in V^* | f(alpha)=0, forall alpha in S}$
Claim: $S^0$ is a subspace of $V^*$.
Let $f_1, f_2 in S^0$ and $c in F$.
Then for each $alpha in S$,
Consider $(cf_1+f_2)(alpha)$ 
               $=(cf_1)(alpha)+f_2(alpha)$ 
               $=cf_1(alpha)+f_2(alpha)$ 
               $=0+0$ 
Thus, $(cf_1+f_2)(alpha)=0$              $forall alpha in S$ 
Therefore, $S^0$ is a subspace of $V^*$.

Remark:

(i) $S^0$ is a subspace of $V^*$.
(ii) If $S={0}$  then $S^0=V^*$ 
(iii) If $S=V$ then $S^0={0}$

Theorem

Let $V$ be a finite dimensional vector space over the field $F$ and $W$ be a subspace of $V$. The $dim W + dim W^0=dim V$ 

Proof:

Let $dim W=k$ and $W={alpha_1, alpha_2, … , alpha_k}$ is a basis for $W$. Choose vectors $alpha_{k+1}, alpha_{k+2}, … , alpha_n$ such that ${alpha_1, alpha_2, … , alpha_k, alpha_{k+1}, alpha_{k+2}, … , alpha_n}$ is a basis for $V$. Then there exists a unique dual basis ${f_1, f_2, …, f_k, f_{k+1}, …, f_n}$ for $V^*$ and dual to $B$ such that $f_i(alpha_j)=delta_{ij}$. 
Claim that ${f_{k+1}, f_{k+2}…, f_n}$ is a basis for $W^0$. 
Note that all $f_i in W^0$ because $f_i(alpha_j)=delta_{ij}=0$,      for $ige k+1, jle k$ 
$implies f_i(alpha)=0$, whenever $alpha$ is a linear combination of $alpha_1, alpha_2, …, alpha_n$ and $i ge k+1$.
Hence, ${f_{k+1}, f_{k+2}…, f_n}$ is linearly independent. 
Now to show that ${f_{k+1}, f_{k+2}…, f_n}$ spans $W^0$. 
Since, for each linear functional $f in V^*$,
$f=displaystylesum_{i=1}^{n}f(alpha_i)f_i$ 
$f=displaystylesum_{i=1}^{k}f(alpha_i)f_i+displaystylesum_{i=k+1}^{n}f(alpha_i)f_i$ 
$f=0+displaystylesum_{i=k+1}^{n}f(alpha_i)f_i$ 
$f=displaystylesum_{i=k+1}^{n}f(alpha_i)f_i$ 
Thus, $B^*={f_{k+1}, f_{k+2}…, f_n}$ is the basis for $W^0$.
Now $dim W=k, dim V=n$.
So, $dim W^0=n-k$
                        $=dim V-dim W$ 
Hence, $dim W + dim W^0=dim V$ 


Corollary:

Let $W_1, W_2$ be subspaces of a finite dimensional vector space $V$ over the field $F$. Then $W_1=W_2$ iff $W_1^0=W_2^0$. 

Proof

For any subspaces $W_1, W_2$, If $W_1=W_2$ then $W_1^0=W_2^0$. 
Suppose that $W_1ne W_2$ 
Therefore, there is some $alpha in W_2$ such that $alpha notin W_1$. 
Then by dual basis theorem, there is a linear functional $f(beta)=0,   forall beta in W_1$ and $f(alpha)ne 0$ 
$implies f in W_1^0$ but $fnotin W_2$ 
$implies W_1^0ne W_2^0$ 
Thus, if $W_1ne W_2$ then $W_1^0ne W_2^0$ 
i.e. if $W_1^0=W_2^0$ then $W_1=W_2$ 
Hence, $W_1=W_2 iff W_1^0=W_2^0$ 

Double Dual

Let $V$ be a vector space over the field $F$. Then $V^**$ is called the double dual of $V$. ${V^*=L(V, F), V^{**}=L(V^*, F)}$ 
If $alpha in V$ then $alpha$ includes a linear functional $L_alpha$ on $V$ defined by $L_alpha(f)=f(alpha)$,          for $fin V^*$ 
For linearity of $L_alpha$ , let $f, g in V^{**}$ and $c in F$ 
Then,
$L_alpha(cf+g)=(cf+g)(alpha)$ 
                         $=cf(alpha)+g(alpha)$
                         $=cL_alpha(f)+L_alpha(g)$ 

Remark


If $V$ is a finite dimensional vector space and $alphane 0 in V, L_alpha ne 0$ then $f(alpha)ne 0$ for $f in V^*$.

 
Theorem

Let $V$ be a finite dimensional vector space over the field $F$. Then $alpha mapsto L_alpha$ is an isomorphism from $V$ onto $V{**}$ where $L_alpha(f)=f(alpha)$, for all $f in V^*$. 

Proof

Let $theta: V rightarrow V^{**}$ defined by $theta(alpha)=L_alpha$, where $L_alpha(f)=f(alpha), forall f in V^*$. 
Claim that $theta$ is an isomorphism. 
Let $alpha, beta in V$ be any two vectors and $c in F$. 
To show that $theta(calpha+beta)=ctheta(alpha)+theta(beta)$ 
For this, write $gamma=calpha+beta$
Then $L_gamma(f)=L_(calpha+beta)(f),                  forall f in V^*$ 
$=f(calpha+beta)$ 
$=cf(alpha)+f(beta)$ 
$=cL_alpha(f)+L_beta(f)$ 
$=(cL_alpha+L_beta)(f)$ 
Therefore, $L_gamma(f)=(cL_alpha+L_beta)(f)$
$implies L_gamma=cL_alpha+L_beta$ 
$implies theta(gamma)=ctheta(alpha)+theta(beta)$ 
Therefore, $theta$ is linear (homomorphism).
By above remark, $L_alpha=0$ iff $alpha=0$ 
This shows that $theta$ is non-singular.
Thus, $theta$ is a non-singular linear transformation. 
Since, $dim V=dim V^*= dim V^{**}$ 
Therefore, $theta$ is invertible linear transformation. 
Hence, $theta$ is an isomorphism. 
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