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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 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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