# inner product space , dual space, proof about isomorphism

Let $V$ be a vector space (not necessary being finite dimensional) and let $U,W$ be subspaces of $V$ such that $V = U\oplus W$.

Prove that $V^\ast/(W^0)$ is isomorphic to $W^\ast$.

Notation and Definitions:

• $W^0$ is the annihilator of $W$

• $W^0=\{f \in V^\ast\mid \text{f(v)=0, for all v in W}\}$

• $V^\ast$ is the dual space of $V$

Before doing this proof, I have a question: when we construct a mapping from $V^\ast$ to $W^\ast$, do we need $W^0$ to be the kernel of that mapping?

• If you plan on using the First Isomorphism theorem in order to establish the fact that $V^\ast/W^0\cong W^\ast$, then yes, you will need the linear transformation $T:V^\ast \rightarrow W^\ast$ to have kernel exactly $W^0$. – Hayden Apr 20 '14 at 13:49
• But how do we construct the mapping from V* to W*? – Kevin Lee Apr 20 '14 at 14:07
• @KevinLee : Just restrict a linear functional on $V$ to $W$ - this will give you the mapping $V^{\ast} \to W^{\ast}$ – Prahlad Vaidyanathan Apr 20 '14 at 14:26
• I atempt to prove (i)Im(ϕ)=W* (ii)ker(ϕ)=W^0 – Kevin Lee Apr 20 '14 at 14:28
• for (i), I can prove Im(ϕ) contained in W*, but the converse direction I have no ideas, i.e., W* contained in ker(ϕ). – Kevin Lee Apr 20 '14 at 14:30

Consider the following linear mapping, $$\phi:V^*/W^o \rightarrow W^*$$ $$[f]\longmapsto f|_W$$ Lets see if it's well defined, consider $f,g\in V^*$ such that $[f]=[g]$, this means $f-g\in W^o$. Now given any $w\in W$ we have that: \begin{align*} (f-g)(w)&=0\\ \Rightarrow f(w)&=g(w)\\ \Rightarrow f|_W&=g|_W\\ \end{align*}
Next we are going to prove that $\phi$ is inyective, if $f$ is such that $\phi(f)=0\in W^*$ we have that $$\forall w\in W:\quad f|_W(w)=0$$ \begin{align*} \Rightarrow f&\in W^o\\ \Rightarrow [f]&=0 \end{align*}
Finally lets see if $\phi$ is onto. Take any $g\in W^*$, we define $f\in V^*$ such that $f(v=u+w)=g(w)$ where $u+w$ is the unique decomposition of $v$ given by $V=U\oplus W$. $f$ is clearly linear and $f|W=g$.
• Well the $\ker(\phi)$ is $\in V^*/W^o$ since the map is inyective, but $=W^o$. – Julio Cáceres Apr 21 '14 at 3:21