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Linear Functionals, Annihilator and Double Dual

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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, ​\( dim \ W + dim \ W^0=dim \ V \)

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 ​\( W_1\ne W_2 \)​ 
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 ​\( f \in V^* \)​.

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 ​\( \theta(c\alpha+\beta)=c\theta(\alpha)+\theta(\beta) \)
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. 
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