What is the significance of annihilator subspace? I am a mathematics student.I am studying linear algebra in which I have dual spaces,annihilators and other related concepts.I want to know what is the motivation behind defining annihilator of a subspace which is $W^0=\{\phi\in V^*: W\subset \operatorname{Ker}\phi\}$. I have seen the result $\dim W^0=\dim V-\dim W$,but I cannot understand what is the exact correlation between $W$ and its annihilator $W^0$. Can someone give me some motivation behind defining dual spaces and annihilators.One motivation of defining dual basis I think is to consider the transpose,but I am not sure.For annihilators the motivation is I think from orthogonal complement of inner product.I am also confused with the notions $V/W^0\simeq W$ or $(V/W)^*\simeq W^0$.
 A: Question:"I have seen the result $dim(W_0)=dim(V)−dim(W)$,but I cannot understand what is the exact correlation between $W$ and its annihilator $W_0$. Can someone give me some motivation behind defining dual spaces and annihilators.One motivation of defining dual basis I think is to consider the transpose,but I am not sure.For annihilators the motivation is I think from orthogonal complement of inner product."
Answer: Given a vector subspace $W \subseteq V$ and let $i: W \rightarrow V$ be the inclusion map. You get an exact sequence of $k$-vector spaces
$$0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$$
and when you dualize this sequence you get an exact sequence
$$0 \rightarrow (V/W)^* \rightarrow V^* \rightarrow^{i^*} W^* \rightarrow 0$$
where the map $i^*$ is the induced map: By definition $i^*(\phi):=\phi \circ i$.
From this you may prove that there are equalities $W^0 = ker (i^*) = (V/W)^*$ as subspaces of $V^*$ as follows: If $i^*(\phi):=
\phi \circ i=0$ it follows $\phi(W)=0$ hence $\phi \in W^0$. Conversely if $\phi \in W^0$ it follows $\phi(W)=0$ and hence $\phi \in i^*$.
From this it follows that
$$dim(W^0)=dim((V/W)^*)=dim(V^*)-dim(W^*)=dim(V)-dim(W).$$
You need to know what a quotient vector space is and what the dual vector space is. You do not need an inner product for this to hold.
Comment: I am also confused with the notions $V/W^0\simeq W$ or $(V/W)^*\simeq W^0$.
Note: Given any inclusion of vector spaces $i: W \rightarrow V$ you may construct the "quotient vector space" $V/W$. The sub space $W \subseteq V$ defines an equivalence relation on the set of vectors $v\in V$: two vectors $u,v$ are equivalent $u \cong v$ iff $u-v \in W$. The set of equivalence classes is denoted $V/W$ and given a vector $v\in V$ we denote by $[v]$ the equivalence class of $v$. This is the class $[v]:=\{v+u: u\in W\}$. The set of equivalence classes $V/W$ is naturally a $k$-vector space and there is a canonical map  $p:V \rightarrow V/W$ of $k$-vector spaces defined by $p(v):=[v]$. It has kernel $ker(p)=W$.
