Exercise 26 page 115 of Linear Algebra Done Right by Sheldon Axler is the following:
Suppose $V$ is finite-dimensional and $\Gamma$ is a subspace of $V'$. Show that $$\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0$$
where $V'$ is the dual space of $V$ and, for any $S\subset V$, $S^0$ is the annihilator of $S$.
Attempt:
Let $S=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}$. Clearly, $\Gamma\subset S^0$.I tried to show that $S^0\subset\Gamma$ using some bases of $V$ and $V'$, but I failed.
I also tried to show that $\dim{\Gamma}=\dim{V}-\dim{S}=\dim{S^°}$. If $s_1,\dots,s_n$ is a basis of $S$ and $s_1',\dots,s_n'$ its dual, and if $\psi_1,\dots,\psi_p$ is a basis of $\Gamma$, it's easy to see that $s_1',\dots,s_n',\psi_1,\dots,\psi_p$ is a linearly independent list of $V'$. Also, if you extend $s_1,\dots,s_n$ to a basis $s_1,\dots,s_n,v_1,\dots,v_m$ of $V$, then its dual $s_1',\dots,s_n',v_1',\dots,v_m'$ is a basis of $V'$; in fact $S^0=\text{span}\{v_1',\dots,v_m'\}$. Two remarkable facts:$$\forall i\in[1,m],\,\exists j\in[1,p],\,\psi_j(v_i)\neq0$$ because $v_j\notin S$, and $$\forall i\in[1,p],\,\exists j\in[1,m],\,\psi_i(v_j)\neq 0$$ because $\psi_i\neq 0$; in other words the matrix of the inclusion map from $\Gamma$ to $V'$ with respect to the basis of $\Gamma$ and the dual base of the chosen basis of $V$ has no $0$ row nor $0$ column. But this doesn't seem to provide a way to prove that any linear comination of the $(v_i')_{1\le i\le m}$ is a linear combination of the elements of the basis of $\Gamma$. I believe that $v_1,\dots, v_m$ should be chosen more carefuly but I fail to.
Could you please help me? Thank you in advance!
Best Answer
For a subspace $T\subseteq V'$ define $T^0 := \{v\in V\,|\, \varphi(v) = 0\text{ for all $\varphi\in T$}\}$ (which is the same as your definition under the identification $V = V''$). In your case, we have $\Gamma^0 = S$.
Then we have $\dim V = \dim U + \dim U^0$ for each subspace $U\subseteq V$: Take a basis $u_1,\dotsc,u_d$ of $U$ and complete it to a basis $u_1,\dotsc,u_n$ of $V$. Denoting by $u'_1,\dotsc,u'_n$ the dual basis, it is easy to see that $U^0$ is spanned by $u'_{d+1},\dotsc,u'_n$, which proves the claim.
Likewise, we have $\dim V' = \dim T + \dim T^0$ for any subspace $T\subseteq V'$. More precisely, let $\varphi_1,\dotsc,\varphi_d$ be a basis of $T$ and complete it to a basis $\varphi_1,\dotsc,\varphi_n$ of $V'$. Then it follows directly from the definitions that $T^0 = \bigcap_{i=1}^d \ker\varphi_i$. It remains to check that $\dim \bigcap_{i=1}^d \ker \varphi_i = n-d$. Indeed, putting $W_j = \bigcap_{i=1}^j \ker\varphi_i$, for $j=1,\dotsc,d$, we obtain an increasing chain $$ 0 = W_n \subset W_{n-1} \subset\dotsb \subset W_1 \subset V. $$ Since obviously $\dim W_j \ge \dim W_{j+1} \ge \dim W_j-1$, for all $j$, and $\dim W_1 = n -1$, we must in fact have $\dim W_j = n-j$. In particular, $$ \dim T^0 = \dim \bigcap_{i=1}^d\ker\varphi_i = \dim W_d = n-d = \dim V' - \dim T. $$
Now, it follows that $T^{00} = T$ for $T\subseteq V'$: $T\subseteq T^{00}$ is clear and both have the same dimension, because of $$\dim T = \dim V' - \dim T^0 = \dim V - \dim T^0 = \dim T^{00}.$$
With this, we conclude $\Gamma = \Gamma^{00} = S^0$ (using $\Gamma^0 = S$ for the last equality).