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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 â€‹\( S^0 \)â€‹ and â€‹\( S^0 \)â€‹ is the set of linear functionals f on V such that â€‹\( f(\alpha)=0 \)â€‹ for each â€‹\( \alpha \in S \)â€‹.

Thus, â€‹\( S^0=\{f \in V^* | f(\alpha)=0, \forall \ \alpha \in S\} \)â€‹

Claim: â€‹\( S^0 \)â€‹ is a subspace of â€‹\( V^* \)â€‹.

Let â€‹\( f_1, f_2 \in S^0 \)â€‹ and â€‹\( c \in F \)â€‹.

Then for each â€‹\( \alpha \in S \)â€‹,

Consider â€‹\( (cf_1+f_2)(\alpha) \)â€‹Â

Â Â Â Â Â Â Â â€‹\( =(cf_1)(\alpha)+f_2(\alpha) \)â€‹

Â Â Â Â Â Â Â â€‹\( =cf_1(\alpha)+f_2(\alpha) \)â€‹

Â Â Â Â Â Â Â â€‹\( =0+0 \)â€‹

Thus, â€‹\( (cf_1+f_2)(\alpha)=0 \ \Â Â Â Â Â Â Â \forall \ \alpha \in S \)â€‹

Therefore, â€‹\( S^0 \)â€‹ is a subspace of â€‹\( V^* \)â€‹.

__Remark__:

(i) â€‹\( S^0 \)â€‹ is a subspace of â€‹\( V^* \)â€‹.

(ii) If â€‹\( S=\{0\} \)â€‹ Â then â€‹\( S^0=V^* \)â€‹

(iii) If â€‹\( S=V \)â€‹ then â€‹\( S^0=\{0\} \)â€‹

__Theorem__:Â

**Let V be a finite dimensional vector space over the field F and W be a subspace of V. The â€‹\( dim \ W + dim \ W^0=dim \ V \)â€‹**Â

__Proof__:

Let â€‹\( dim \ W=k \ \ and \ \ W=\{\alpha_1, \alpha_2, … , \alpha_k\} \)â€‹ is a basis for W. Choose vectors â€‹\( \alpha_{k+1}, \alpha_{k+2}, … , \alpha_n \)â€‹ such that â€‹\( \{\alpha_1, \alpha_2, … , \alpha_k, \alpha_{k+1}, \alpha_{k+2}, … , \alpha_n\} \)â€‹ is a basis for V. Then there exists a unique dual basis â€‹\( \{f_1, f_2, …, f_k, f_{k+1}, …, f_n\} \)â€‹ for â€‹\( V^* \)â€‹ and dual to B such that â€‹\( f_i(\alpha_j)=\delta_{ij} \)â€‹.Â

Claim that â€‹\( \{f_{k+1}, f_{k+2}…, f_n\} \)â€‹ is a basis for â€‹\( W^0 \)â€‹.Â

Note that all â€‹\( f_i \in W^0 \)â€‹ because â€‹\( f_i(\alpha_j)=\delta_{ij}=0 \)â€‹,Â Â Â for â€‹\( i\ge k+1, j\le k \)â€‹

â€‹\( \implies f_i(\alpha)=0 \)â€‹, whenever â€‹\( \alpha \)â€‹ is a linear combination of â€‹\( \alpha_1, \alpha_2, …, \alpha_n \)â€‹ and â€‹\( i \ge k+1 \)â€‹.

Hence, â€‹\( \{f_{k+1}, f_{k+2}…, f_n\} \)â€‹ is linearly independent.Â

Now to show that â€‹\( \{f_{k+1}, f_{k+2}…, f_n\} \)â€‹ spans â€‹\( W^0 \)â€‹.Â

Since, for each linear functional â€‹\( f \in V^* \)â€‹,

â€‹\( f=\displaystyle\sum_{i=1}^{n}f(\alpha_i)f_i \)â€‹

â€‹\( f=\displaystyle\sum_{i=1}^{k}f(\alpha_i)f_i+\displaystyle\sum_{i=k+1}^{n}f(\alpha_i)f_i \)â€‹

â€‹\( f=0+\displaystyle\sum_{i=k+1}^{n}f(\alpha_i)f_i \)â€‹

â€‹\( f=\displaystyle\sum_{i=k+1}^{n}f(\alpha_i)f_i \)â€‹

Thus, â€‹\( B^*=\{f_{k+1}, f_{k+2}…, f_n\} \)â€‹ is the basis for â€‹\( W^0 \)â€‹.

Now â€‹\( dim \ W=k, dim \ V=n \)â€‹.

So, â€‹\( dim \ W^0=n-k \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =dim \ V-dim \ W \)â€‹

Hence, â€‹\( dim \ W + dim \ W^0=dim \ V \)â€‹

__Corollary__:

**Let â€‹\( W_1, W_2 \)â€‹ be subspaces of a finite dimensional vector space V over the field F. Then â€‹\( W_1=W_2 \)â€‹ iff â€‹\( W_1^0=W_2^0 \)â€‹.Â **

__Proof__:Â

For any subspaces â€‹\( W_1, W_2 \)â€‹, If â€‹\( W_1=W_2 \)â€‹ then â€‹\( W_1^0=W_2^0 \)â€‹.Â

Suppose that â€‹\( W_1\ne W_2 \)â€‹Â

Therefore, there is some â€‹\( \alpha \in W_2 \)â€‹ such that â€‹\( \alpha \notin W_1 \)â€‹.Â

Then by dual basis theorem, there is a linear functional â€‹\( f(\beta)=0,Â Â \forall \ \beta \in W_1 \)â€‹ and â€‹\( f(\alpha)\ne 0 \)â€‹

â€‹\( \implies f \in W_1^0 \)â€‹ but â€‹\( f\notin W_2 \)â€‹Â

â€‹\( \implies W_1^0\ne W_2^0 \)â€‹Â

Thus, if â€‹\( W_1\ne W_2 \)â€‹ then â€‹\( W_1^0\ne W_2^0 \)â€‹

i.e. if â€‹\( W_1^0=W_2^0 \)â€‹ then â€‹\( W_1=W_2 \)â€‹

Hence, â€‹\( W_1=W_2 \iff W_1^0=W_2^0 \)â€‹

__Double Dual__:Â

Let V be a vector space over the field F. Then â€‹\( V^{**} \)â€‹ is called the double dual of V. â€‹\( \{V^*=L(V, F), V^{**}=L(V^*, F)\} \)â€‹

If â€‹\( \alpha \in V \)â€‹ then â€‹\( \alpha \)â€‹ includes a linear functional â€‹\( L_\alpha \)â€‹ on V defined by â€‹\( L_\alpha(f)=f(\alpha), \ for \ f\in V^* \)â€‹Â

For linearity of â€‹\( L_\alpha \)â€‹ , let â€‹\( f, g \in V^{**} \)â€‹ and â€‹\( c \in F \)â€‹Â

Then,

â€‹\( L_\alpha(cf+g)=(cf+g)(\alpha) \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =cf(\alpha)+g(\alpha) \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =cL_\alpha(f)+L_\alpha(g) \)â€‹

__Remark__:Â

If V is a finite dimensional vector space and â€‹\( \alpha\ne 0 \in V, L_\alpha \ne 0 \)â€‹ then â€‹\( f(\alpha)\ne 0 \)â€‹ for â€‹\( f \in V^* \)â€‹.

__Theorem__:Â

**Let V be a finite dimensional vector space over the field F. Then â€‹\( \alpha \mapsto L_\alpha \)â€‹ is an isomorphism from V onto â€‹\( V^{**} \)â€‹ where â€‹\( L_\alpha(f)=f(\alpha) \)â€‹, for all â€‹\( f \in V^* \)â€‹.Â **

__Proof__:Â

Let â€‹\( \theta: V \rightarrow V^{**} \)â€‹ defined by â€‹\( \theta(\alpha)=L_\alpha \)â€‹, where â€‹\( L_\alpha(f)=f(\alpha), \forall \ f \in V^* \)â€‹.Â

Claim that â€‹\( \theta \)â€‹ is an isomorphism.Â

Let â€‹\( \alpha, \beta \in V \)â€‹ be any two vectors and â€‹\( c \in F \)â€‹.Â

To show that â€‹\( \theta(c\alpha+\beta)=c\theta(\alpha)+\theta(\beta) \)â€‹

For this, write â€‹\( \gamma=c\alpha+\beta \)â€‹

Then â€‹\( L_\gamma(f)=L_(c\alpha+\beta)(f),Â Â Â Â Â Â Â Â Â \forall \ f \in V^* \)â€‹Â

â€‹\( =f(c\alpha+\beta) \)â€‹

â€‹\( =cf(\alpha)+f(\beta) \)â€‹

â€‹\( =cL_\alpha(f)+L_\beta(f) \)â€‹

â€‹\( =(cL_\alpha+L_\beta)(f) \)â€‹

Therefore, â€‹\( L_\gamma(f)=(cL_\alpha+L_\beta)(f) \)â€‹

â€‹\( \implies L_\gamma=cL_\alpha+L_\beta \)â€‹

â€‹\( \implies \theta(\gamma)=c\theta(\alpha)+\theta(\beta) \)â€‹

Therefore, â€‹\( \theta \)â€‹ is linear (homomorphism).

By above remark, â€‹\( L_\alpha=0 \)â€‹ iff â€‹\( \alpha=0 \)â€‹

This shows that â€‹\( \theta \)â€‹ is non-singular.

Thus, â€‹\( \theta \)â€‹ is a non-singular linear transformation.Â

Since, â€‹\( dim \ V=dim \ V^*= dim \ V^{**} \)â€‹

Therefore,\( \theta \)is invertible linear transformation.Â

Hence, â€‹\( \theta \)â€‹ is an isomorphism.Â

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