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Friday, February 3, 2023

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 $V^*$.

(i) $S^0$ is a subspace of $V^*$.

(ii) If $S={0}$ then $S^0=V^*$

(iii) If $S=V$ then $S^0={0}$

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

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, $dim W + dim W^0=dim V$

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

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, $W_1=W_2 iff W_1^0=W_2^0$

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,

$L_alpha(cf+g)=(cf+g)(alpha)$

$=cf(alpha)+g(alpha)$

$=cL_alpha(f)+L_alpha(g)$

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

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 $f in V^*$.

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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