# 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 $i\ge k+1, j\le k$
$\implies f_i(\alpha)=0$, 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 $f \in V^*$,
$f=\displaystyle\sum_{i=1}^{n}f(\alpha_i)f_i$
$f=\displaystyle\sum_{i=1}^{k}f(\alpha_i)f_i+\displaystyle\sum_{i=k+1}^{n}f(\alpha_i)f_i$
$f=0+\displaystyle\sum_{i=k+1}^{n}f(\alpha_i)f_i$
$f=\displaystyle\sum_{i=k+1}^{n}f(\alpha_i)f_i$
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 $W_1^0=W_2^0$.

### Proof:

For any subspaces $W_1, W_2$, If $W_1=W_2$ then $W_1^0=W_2^0$.
Suppose that $W_1\ne W_2$
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 $f\notin W_2$
$\implies W_1^0\ne W_2^0$
Thus, if $W_1\ne W_2$ then $W_1^0\ne W_2^0$
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 $f\in V^*$
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 $\alpha\ne 0 \in V, L_\alpha \ne 0$ 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 $f \in V^*$.

### 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 $\theta(c\alpha+\beta)=c\theta(\alpha)+\theta(\beta)$
For this, write $\gamma=c\alpha+\beta$
Then $L_\gamma(f)=L_(c\alpha+\beta)(f), \forall \ f \in V^*$
$=f(c\alpha+\beta)$
$=cf(\alpha)+f(\beta)$
$=cL_\alpha(f)+L_\beta(f)$
$=(cL_\alpha+L_\beta)(f)$
Therefore, $L_\gamma(f)=(cL_\alpha+L_\beta)(f)$
$\implies L_\gamma=cL_\alpha+L_\beta$
$\implies \theta(\gamma)=c\theta(\alpha)+\theta(\beta)$
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.