$W^0\subset U^0\implies U\subset W$

Question

Is the following Proof Correct? Do please refer to the note below.

Theorem. Given that $V$ is a finite-dimensional vector space and $U$ and $W$ are subspaces of $V$ such that $W^0\subset U^0$ then $U\subset W$

Proof. Let $w_1,w_2,...,w_n$ be the basis for $W$ and let us extend it to yield the basis $w_1,w_2,...,w_n,w_{n+1},w_{n+2},...,w_n$ for $V$ furthermore let $\phi_1,\phi_2,...,\phi_n,\phi_{n+1},\phi_{n+2},...,\phi_{m}$ be the corresponding dual basis for $V'$.

Having dealt with the preliminaries, assume now that $u\in U$, which when expressed as a linear combination of our chosen basis yields the following equation. $$u = c_1w_1+c_2w_2+\cdot\cdot\cdot+c_nw_n+c_{n+1}w_{n+1}+c_{n+2}w_{n+2}+\cdot\cdot\cdot+c_mw_{m}$$

It is not difficult to see that $W^{0} = \operatorname{span}(\phi_{n+1},\phi_{n+2},...,\phi_{m})$ see example $\textbf{3.104}$, and since $W^0\subset U^0$ then in particular $\phi_{n+1}\in U^{0},\phi_{n+2}\in U^{0},...,\phi_{m}\in U^{0}$ consequently $\phi_{n+1}(u) = c_{n+1} = \phi_{n+2}(u) = c_{n+2} =\cdot\cdot\cdot = \phi_{m}(u) = c_m = 0$ which together with the above equation implies that $u\in\operatorname{span}(w_1,w_2,...,w_n) = W$.

$\blacksquare$

NOTE

• $U^0$ denotes the annihilator of the subspace $U$.

• Example $3.104$ basically demonstrates the if $v_1,v_2,...,v_n,v_{n+1},v_{n+2},...,v_m$ is basis for $V$ and $\phi_1,\phi_2,...,\phi_n,\phi_{n+1},\phi_{n+2},...,\phi_m$ is the corresponding dual basis then if $v_1,v_2,...,v_n$ is a basis for $U$ then $\phi_{n+1},\phi_{n+2},...,\phi_m$ is the basis for $U^0$.

Answer 1

I think that it is easier to start with showing that if $A $ is a subspace of a finite dimensional $V$, then $x \in A$ iff $w(x) = 0$ for all $w \in A^0$.

One direction is immediate.

Suppose $w(x) = 0 $ for all $w \in A^0$.

Let $a_1,...,a_d$ be a basis for $A$, and extend to a basis of $V$ with $a_{d+1},..,a_n$.

Let $x = \sum_k\alpha_k a_k$. Define the functional $w$ by $w(a_i) = \delta_{ik}$ with $k > d$ and note that $w \in A^0$ and so $w(x)= \alpha_k = 0$. Hence $x = \sum_{k=1}^d \alpha_k a_k \in A$.

Back to the question:

Suppose $W^0 \subset U^0$ and $x \in U$. Then $w(x) = 0$ for all $w \in U^0$ and so $w(x) = 0$ for all $w \in W^0$ and so $x \in W$.

Answer 2

You're overthinking this, in my humble opinion....

The annihilator of $S\subset V$ is the set of $f^*\in V^*$ such that $[f,v]=0$ for all $s\in S$.

In other words, this is some set with a distinguishing property that is defined by how it relates to another set..... now, if we start adding more elements to the set we have to satisfy it would seem that we could only possibly lose elements that have the distinguishing property because our requirement becomes more strong.

Therefore, if $S\subset T$ then $T^0 \subset S^0$

Really even the definition of "annihilator" doesn't matter. It's a property such that as you add more elements to the set, it becomes more difficult to satisfy. That's about all that matters.

EDIT: No need to dualize (though that works in the FINITE dimensional case).... here's the idea.... repeat the argument with the point of view inverted: "If more things satisfy some new set than they did before, it's because that set became smaller"

If that doesn't make sense to you, then yes, in a finite dimensional space it is very easy to show that $E^{00}=E$ merely invoke the definition of "annihilator" then notice that if $T^0\subset S^0$ then $S=S^{00}\subset T^{00}=T$