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.
