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 (1):
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 (2):
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.
