Show that $V/U\cong (U^0)'$

Question

Let $U$ be a subspace of the finite dimensional vector space $V$ over a field $F$. Show that $V/U\cong (U^0)'$.

We have that $U^0=\{f\in V' \ | \ \forall u\in U: f(u)=0\}$ is the annihilator and $V/U=\{v+U \ | \ v \in V\}$ is the quotient space. So then $(U^0)' = \{T:U^0\rightarrow F \ | \ T \text{ is linear}\}$.

Consider the function $\phi: v+U \mapsto E_v$, where $E_v$ is the evaluation map $E_v(f) = f(v)$.

Is the map I'm looking for, and if so, how can I demonstrate that it is well-defined and injective?

(Linearity is obvious and I can also demonstrate that $\dim (V/U)=\dim((U^0)')$ to get bijectivity, once I have injectivity.)

Answer 1

You are defining $E_v:U^{0} \to F$. If $v_1+U=v_2+U$ then $v_1-v_2\in U$ so $f(v_1-v_2)=0$ or $f(v_1)=f(v_2)$. This proves that the map $v \to E_v$ is well defined. Now suppose $E_v=E_w$. Then $f(v)=f(w)$ for all $f \in U^{0}$. This implies that $v-w \in U$. This is because if $v-w \notin U$ we can always construct a linear map $f$ which has the value $0$ at every point of $U$ and has the value $1$ at $v-w$. [You can start with a basis for $U$, add $(v-w)$ to it and then extend to a basis for $V$. Once you do this it is easy to define a linear map with the required properties]. It should now be clearf that $v \to E_v$ is injective.