Constructing a linear map from annihilator of a subspace to dual of the quotient space If W is a subspace of V, let V/W denote the quotient of V by W and let (V/W)* denote its linear dual. Construct a non zero linear map from Ann(W) to (V/W)*
(1)From Basis
(2) Canonical
I am weak in quotient space so can't understand how to proceed in this question. I have a general understanding of annihilators and dual spaces but because of the quotient space I am unable to proceed.
Not a homework question, stumbled online but was not able to solve it.
 A: One example of a "canonical map" as you've put it is to define $\Phi:\operatorname{Ann}(W) \to (V/W)^*$ by $\Phi(f) = g$, where for any $f : V \to \Bbb F$ in $\operatorname{Ann}(W)$, we define the map $g:V/W \to \Bbb F$ by
$$
g(v + W) = f(v)
$$
More succinctly, you could say that we have defined $\Phi$ by
$$
[\Phi(f)](v+W) = f(v), \quad v + W \in V/W
$$
It will take a bit of work and careful consideration of the definitions of $\operatorname{Ann}(W)$ and $V/W$ to show that this $\Phi$ makes sense (that is, to show that $\Phi$ is well-defined and linear).  It is notable that this map $\Phi$ is, in addition to being linear, an isomorphism of vector spaces since it is bijective.

On the other hand, non-canonical maps are easy to construct between any two vector spaces.  For your example, we can define a map as follows: let $\{f_1,\dots,f_m\}$ be a basis of $\operatorname{Ann}(W)$, and let $g$ be a non-zero element of $(V/W)^*$.  We can then define
$$
\Psi(\alpha_1 f_1 + \cdots + \alpha_m f_m) = \alpha_1\, g; \qquad \alpha_1,\dots,\alpha_m \in \Bbb F
$$
By the definition of a basis, this definition describes a linear map over the entirety of the domain $\operatorname{Ann}(W)$.  This map is non-zero since $\Psi(f_1) = g$ is a non-zero output.
Because we used a basis of $\operatorname{Ann}(W)$ to describe this map, there was no need to consider the structure of the vector spaces except to say that they are both vector spaces.
