Proof for $(Z^\perp)^{\perp} = Z$

Question

Let $V$ be a vector space of dimension n over an arbitrary field $F$, and let $V^{*}$ be the dual space of $V$. The definition of annihilator is $Z^{\perp} = \{f \in V^{*} | f(v) = 0 \; \forall v \in Z\}$

So far, I have an outline for a double containment proof for the proposition that $(Z^\perp)^{\perp} = Z$ for every subspace $Z\subset V$ for dim$V<\infty$.

For the direction $Z \subset (Z^{\perp})^{\perp}$:

1. Let $v_1, \dots, v_k$ be a basis of $Z$ and extend to a bases $v_1, \cdots, v_n$ of $V$. Let $f_1, \cdots, f_n$ be the dual bases of $V^{*}$. Then $f_i(v_j) = \delta^{j}_{i}$
2. $Z^{\perp} = $ span$\{f_{k+1}, \cdots, f_n\}$
3. Since $f_i(v_j) = 0$ for $i = k+1, \cdots, n$ and $ j = 1, \cdots, k$, we have $v_1, \cdots, v_k \in (Z^{\perp})^{\perp}$ So, $Z = $ span $(v_1, \cdots, v_k) \subset (Z^{\perp})^{\perp}$

For the other direction, I have figured it out.

However, I am not sure why #2 holds. Can anyone help me?

Answer 1

You want to write down equations that say $f \in Z^\perp$ in terms of the coefficients of $f$. That is, if $f = a_1 f_1 + a_2 f_2 + \dots + a_n f_n$ then $f \in Z^\perp$ if and only if $f(v_1) = 0, f(v_2) = 0, \dots, f(v_k) = 0$. But what is the equation $f(v_i) = 0$ in terms of the coefficients of $f$? It reduces to $a_i = 0$. So $f \in Z^\perp$ means that $a_1, \dots, a_k = 0$ means that $f \in \operatorname{span}\{f_{k+1},\dots,f_n\}$.

But there is a simpler approach: $f \in Z^\perp$ means $f(v) = 0$ for every $v \in Z$. Now rewrite this statement in such a way that $f$ and $v$ are introduced next to each other: $\forall f \in Z^\perp, \forall v \in Z, f(v) = 0$. Now swap the "for all"s: $\forall v \in Z, \forall f \in Z^\perp, f(v) = 0$. Finally, rewrite this in English: $v \in Z$ means that for all $f \in Z^\perp$, we have $f(v) = 0$. This is exactly what it means to say $v \in (Z^\perp)^\perp$.

If you pay attention, we can prove that $Z \subseteq (Z^\perp)^\perp$ without talking about bases. Which means this statement holds in infinite dimensions as well. It is the other direction: $(Z^\perp)^\perp \subseteq Z$ which does not always hold in infinite dimensions.