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HomeMathematicsAlgebraLinear Functionals, Annihilator and Double Dual

# 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, ​$$dim \ W + dim \ W^0=dim \ V$$

### 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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