annihilator of vector subspace

If $V$ is a vector space, and $S$ is any subset of $V$,the annihilator of $S$, denoted by $S^0$,is the subspace of the dual space $V^{*}$that kills every vector in $S$:

$$S^0=\{\varphi \in V^{*}:\varphi (v)=0\text{ for all }v\in S\}.$$

Similarly, if $\mathrm{\Lambda }$ is any subset of $V^{*}$, the annihilated subspaceof $\mathrm{\Lambda }$ is

$${\mathrm{\Lambda }}^{-0}=\{v\in V:\varphi (v)=0\text{ for all }\varphi \in \mathrm{\Lambda }\}=\bigcap _{\varphi \in \mathrm{\Lambda }}\ker \varphi .$$

(Note: this may not be the standard notation.)

1 Properties

Assume $V$ is finite-dimensional.Let $W$ and $\mathrm{\Phi }$ denote subspaces of $V$ and $V^{*}$, respectively,and let $\widehat{}$ denote the natural isomorphism from $V$ to its double dual $V^{**}$.

1. $S^0={\left(\mathrm{span}S\right)}^0$

2. ${\mathrm{\Lambda }}^{-0}={\left(\mathrm{span}\mathrm{\Lambda }\right)}^{-0}$

3. $W^{00}=\widehat{W}$

4. ${\left({\mathrm{\Phi }}^{-0}\right)}^0=\mathrm{\Phi }$

5. ${\left(W^0\right)}^{-0}=W$

6. $dimW+dim{W}^0=dimV$ (a dimension theorem)

7. $dim\mathrm{\Phi }+dim{\mathrm{\Phi }}^{-0}=dim{V}^{*}=dimV$

8. ${(W_1+W_2)}^0=W_1^0\cap W_2^0$, where $W_1+W_2$ denotesthe sum of two subspaces of $V$.

9. If $T:V\to V$ is a linear operator, and $W=\ker T$,then the image of the pullback $T^{*}:V^{*}\to V^{*}$ is $W^0$.