Hi everyone is my first time reading about dual spaces and in one part of the notes that I read, says: The dual of the quotient space $V/U$ is naturally a subspace of $V$, namely the annihilators of $U$ in $V$.

I have doubt about this when says a naturally subspace does not actually mean which there is a natural identification of $(V/U)^*$ to the set of all the annihilators $U$ in $V$ under a map more than a subspace?

Clearly exists an epimorphism $f: V \twoheadrightarrow V/U $ and we can associate the map $f^t: (V/U)^* \rightarrow V^*$ by $h\mapsto h\circ f$, where $h\in (V/U)^*$, so $f^t[\,(V/U)^*]= \{\text{the annihilators of U in V} \}$, or there exist a natural identification of $(V/U)^*$ to the set of all the annihilators $U$ in $V$ under $f^t$? Am I completely off track or my intuition is correct?

Claim: Let $f, f^t$ as define above then $f^t[\,(V/U)^*]= \{\text{the annihilators of U in V} \}$

Proof of claim: ($\Rightarrow$) $T\in f^t[\,(V/U)^*]$, so $T=h\circ f$ for some $h\in (V/U)^*$, i.e., $h: V/U \rightarrow \mathbb{F}$. If $x\in U$, so $f(x)= x+U= U$ and then $h(f(x))= h(U)=0$.

($\Leftarrow$) Let $T\in \{\text{the annihilators of U in V} \}$, i.e., $T(x)=0$ whenever $x\in U$. Let define $\overline{T}: V/U \rightarrow \mathbb{F}$ by the formula $\overline{T}(x+U)=T(x)$, we claim that $\overline{T}\circ f=T$, let $x\in V$ so $(\overline{T}\circ f) (x)=\overline{T}(f(x))=\overline{T}(x+U)=T(x)$. Then $\overline{T}\circ f= f^t(\overline{T})=T$, i.e., $T\in f^t[\,(V/U)^*]$.

One more thing is very natural in the literature read that if $V$ is a vector space (a finite dimensional vector space) then is the dual of other space, that doesn't really mean that $V$ is isomorphic to the dual of other space instead of is the dual of some other space because not all the vector spaces are the set of linear functionals?

Thanks in advance.

Edit: If $U$ is a subspace of $V$ (and $V$ is finite dimensional vector space), clearly we have $V= U \oplus U'$ and $U'\cong V/U$ ($U'\hookrightarrow V \twoheadrightarrow V/U$) then we can conclude that $(V/U)^* \cong (U')^*$ and is not difficult to show that $(U')^*$ contain all the annihilators of $U$ in $V$ is in that way in which as says in the book "the dual of the quotient space $V/U$ is naturally a subspace of $V$, namely the annihilators of $U$ in $V$" because we can identified naturally with $(U')^*$?

Edit: Other thing: In other part says 'If we choose a basis of $V$, and use it to identify elements of V with “column vectors” of length n, then elements of $V^*$ correspond to “row vectors” of the same length'. Why is this true?

Clearly if we set $\mathcal{B}= \{v_1,..v_n\}$ be a basis for $V$, any element of $V$, says $v\in V$ can be expresses uniquely as $v=\sum_i a_i v_i$, so we can associate $v\mapsto [V]^\mathcal{B}$ which is the vector column. Now if $\mathcal{B^*}= \{v_1^*..v_n^*\}$ is the dual basis for $V^*$, so for any $f\in V^*$ we have $f= \sum_i f(v_i)v_i^*$ but I can't see in which sense we can associate it to a row vector $[f]_{B^*}$, naturally?

Any comment it would be great. Thanks :)

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    $\begingroup$ A little bit about the last part; a finite dimensional vector space $V$ is isomorphic to the dual of any vector space $W$ with the same dimension as $V$, but this should not be thought of as an identification because there are lots of choices of isomorphism. However, $V$ is isomorphic to $(V^*)^*$ in a natural way, that involves no choices, and so it is more reasonable to treat this as an identification and say that $V$ is a dual space (it's the dual of its dual). $\endgroup$ – mdp Jan 28 '14 at 17:49
  • $\begingroup$ Ok I understand and indeed the proof is not very hard that $V\simeq (V^*)^*$, and clearly $(V^*)^*$ is the dual of $V^*$ so in that case we says that there exist a natural isomorphism which identify $V$ to $(V^*)^*$ which is the dual of $V^*$ but in almost all, at least the book which I read, says that $V$ is the dual of some space, isn't better to say is isomorphic to a space which is the dual of some other space? in other words $V\simeq (V^*)^*$ and $(V^*)^*$ is the dual of $V^*$. For beginners as me is a little confuse when sometimes "up to an isomorphism" and "genuine" equality are mix $\endgroup$ – Jose Antonio Jan 28 '14 at 18:00
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    $\begingroup$ I would argue it is better to say "is" (if you can) than "is isomorphic to". This is quite a subtle issue; when dealing with vector spaces, you don't often see genuine equality! Certainly $V$ is not equal to $(V^*)^*$, but the two are more than just isomorphic, because there is a unique "best" way of identifying them. A lot of people (maybe unfortunately) will say that one vector space "is" another one (without the word equal) when they are naturally isomorphic, but not when they are only unnaturally isomorphic. $\endgroup$ – mdp Jan 28 '14 at 18:06
  • $\begingroup$ Ok, I understand just for beginners as me is sometimes a little confuse (I suppose in this context natural isomorphism means free of the choice of basis) :P. I think is similar for the first part doesn't? Thanks for you help. $\endgroup$ – Jose Antonio Jan 28 '14 at 18:13
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    $\begingroup$ I agree with the answer given: as you write in your question, there is a natural injective linear map $(V/U)^* \hookrightarrow V^*$: namely, this realizes $(V/U)^*$ as the linear functionals which annihilate $U$. If the "$*$" is missing on the $V$ in your notes, it is most likely a typo. $\endgroup$ – Pete L. Clark Jan 29 '14 at 7:23

It seems more natural (to me at least) to view the dual of $V/U$ as naturally a subspace of $V^*$, not $V$. This arises from the exact sequence

$$0 \to U \to V \to V/U \to 0$$

which, upon taking duals (i.e. applying $\text{Hom}_k(\_, k)$, which is an exact functor for vector spaces), gives an exact sequence

$$0 \to (V/U)^* \to V^* \to U^* \to 0$$

It's not a priori clear how to view $(V/U)^* \subseteq V$. If $V$ is finite-dimensional, one could pick an isomorphism $V \xrightarrow{\sim} V^*$, but such a choice of isomorphism is equivalent to a choice of inner product on $V$, and there's no canonical choice. Furthermore, this fails in the infinite-dimensional case.

If by an "annihilator of $U$ in $V$" you mean a linear functional $\phi : V \to k$ such that $U \subseteq \ker \phi$, then it is true that the image of $(V/U)^*$ in $V^*$, under the canonical map above, is the set of annihilators of $U$ in $V$ (this can be seen by exactness of the dual sequence). Thus, I would suspect that your notes really say (or ought to say) that $(V/U)^*$ is naturally a subspace of $V^*$.

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  • $\begingroup$ Maybe is a typo. With respect to $(V/U)^*$ is naturally a subspace of $V^*$ is not completely clear to me if is actually $(V/U)*\subset V^*$ or if the injective map $(V/U)* \rightarrow V^*$ identifies $(V/U)*$ to a subspace of $V^*$? $\endgroup$ – Jose Antonio Jan 29 '14 at 7:29
  • $\begingroup$ Ok thanks for your answer very complete. I think I got it with your answer and the comment of Peter, the natural map $(V/U)^* \hookrightarrow V^*$ associate $(V/U)^*$ to the set of annihilators of $U$ in $V^*$, which is the set of all linear functionals that maps all vectors in $U$ to $\,0$, is not "strictly" speaking a subspace as a novice as me think at first is natural identification to a subspace of $V^*$. Am I right? $\endgroup$ – Jose Antonio Jan 29 '14 at 8:02
  • $\begingroup$ And of course this answer immediately what is the dimension of the set of annihilators of $U$ in $V^*$, the codimention of $U$ in $V$ (everything is in the finite case). $\endgroup$ – Jose Antonio Jan 29 '14 at 8:11
  • $\begingroup$ @JoseAntonio: Concerning your 2nd and 3rd comments, you are correct. However, it is worth noting that the embedding $(V/U)^* \hookrightarrow V^*$ is canonical (being the dual of the projection $V \to V/U$). In such cases it is often the case that one will identify $(V/U)^*$ with its image in $V^*$, even though they differ as sets $\endgroup$ – zcn Jan 29 '14 at 9:43
  • $\begingroup$ Thanks again for your answer. I suppose is similar when we thought $V/U$ as a natural complement of $U$. Since V is assume to be finite dimensional vector space, so $V= U\oplus U'$ and exist a natural map $U'\hookrightarrow V \twoheadrightarrow V/U$ $\endgroup$ – Jose Antonio Jan 29 '14 at 17:27

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