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Suppose $V$ is finite-dimensional and $\Gamma$ is a subspace of $V'$. Show that $$\Gamma = \{v\in V: \phi(v) = 0\ \forall \phi\in\Gamma\}^0$$

It's straightforward to show that $\Gamma\subset \{v\in V: \phi(v) = 0\ \forall \phi\in\Gamma\}^0$, but I can't come up with the reverse direction.

vvvvv
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  • My edit should clarify things. Γ is a subspace of $V'=L(V,F)$, the space of linear maps from V to some field. – vvvvv Oct 02 '16 at 00:53
  • The book uses V' to denote the dual space of V. If V is a vector space over a field F, V' consists of all linear functions from V to F. So, the meaning of f(v) for f in V' and v in V is regular function application. The application notation isn't given any notation for arbitrary vectors. – vvvvv Oct 02 '16 at 01:03
  • I've described the use of the apostrophe in the previous comment; V' contains all linear functions from V to its underlying field. The 0 superscript is used to refer to the annihilator of a subspace U of V. $U^0={ \phi\in V' s.t. \phi(u) = 0 \forall u\in U }$ – vvvvv Oct 02 '16 at 01:07
  • @shaihorowitz: please read the comment above yours. – Martin Argerami Oct 02 '16 at 01:24
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    The most straightforward approach, I think, is to consider the dimensions of the spaces involved. It will suffice to find that the set on the right has the same dimension as $\Gamma$. – Ben Grossmann Oct 02 '16 at 01:42

3 Answers3

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Since $V$ is finite-dimensional, we have that the double annihilator of a subspace is again the subspace (in general, it is the weak$^*$-closure).

So, since $$ \Gamma^0=\{v\in V: \phi(v) = 0\ \forall \phi\in\Gamma\},$$ taking annihilators again we get $$ \Gamma = \{v\in V: \phi(v) = 0\ \forall \phi\in\Gamma\}^0. $$

Martin Argerami
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Let $\phi_1, \cdots, \phi_m$ be a basis of $\Gamma$.

Then by Exercise 3.F.30, $$\dim(\text{null } \phi_1 \cap \cdots \cap \text{null } \phi_m) = \dim V - m.$$

Let $U := \{v \in V | \phi(v) = 0 \text{ for every }\phi \in \Gamma\}$.

It is easy to check that $U$ is a subspace of $V$.

By 3.106 on p.106,

$$\dim U + \dim U^0 = \dim V.$$

It is easy to check that $U = \text{null } \phi_1 \cap \cdots \cap \text{null } \phi_m$.

So, $\dim U = \dim(\text{null } \phi_1 \cap \cdots \cap \text{null } \phi_m) = \dim V - m$.
So, $\dim U^0 = \dim V - \dim U = \dim V - (\dim V - m) = m = \dim \Gamma$.
So, $\Gamma \subset U^0$ and $\dim \Gamma = \dim U^0$.
So, $\Gamma = U^0$.

tchappy ha
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Let $U = \{ v \in V : \varphi v = 0, \forall \varphi \in \Gamma \}$. $U$ contains 0 and is closed under addition and scalar multiplication, thus a subspace of $V$.

By problem 25, $$ U = \{ v \in V : \varphi v = 0, \forall \varphi \in U^0 \} $$ Suppose $V$ has a basis $v_1, \ldots, v_n$ and the dual basis $\varphi_1, \ldots, \varphi_n$. If $\Gamma$ and $U^0$ disagree on $\varphi_i$ as an item of their basis, then the two definitions of $U$ will disagree on $v_i \in U$, which is a contradiction.

Thus $\Gamma$ and $U^0$ have the same basis.

amdyes
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