"Easy" (maybe not) question about dual spaces (Lineal Algebra). 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 :)
 A: 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 $V^*$.
