I have a question regarding the following definition of an annihilator of a finite dimensional vector space. I think I understand the two definitions but I don't really get the link implied by the last sentence.


If $A \subset V$, the annihilator of $A, A^{°}$, is the set of all $f$ in $V^*$ such that $f(a) = 0$ for all $a$ in $A$. Similarly, if $A \subset V^*$, then

$$ A^{°} = \{a\in V\colon\ f(a) = 0 \text{ for all } f\in A \}. $$

If we view $V$ as $(V^{*})^{*}$, the second definition is included in the first.

So now my questions are why may we view $V$ as $(V^{*})^{*}$, I know they're equivalent by the isomorphism but this doesn't mean that they're equal, does it? And how does the first definition include the second one?


1 Answer 1


They are not equal, but algebraically isomorphic, because a finite dimensional vector space is reflexive and the second definition is included in the first if You identify every $a\in V$ with it´s image $Ja$ under the canonical injection (in this case isomorphism) $J:V\rightarrow (V^*)^*,a\mapsto(Ja:V^*\rightarrow\mathbb{C},f\mapsto f(a))$,because the first definition then reads for $V^*$:

$ A^0=\{Ja\in V^{**}:Ja(f)=f(a)=0\}$

  • $\begingroup$ So it is because of this one to one correspondence (bijection) that you can identify with every element of $V^{**}$ an element of $V$ to reconstruct out of the set of the first definition with $A \subset V^*$ the set in the second definition under the canonical injection? $\endgroup$
    – Dylan_VM
    Commented Sep 18, 2015 at 12:41
  • $\begingroup$ Exactly, but Your doubts were justified, because the two definitions don´t really lead to the same sets, it´s just that there elements can be identified via the canonical isomorphism $\endgroup$ Commented Sep 18, 2015 at 13:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .