# 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?

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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 at 13:49
Thanks for your guide~! –  Kevin Lee Apr 20 at 13:54
But how do we construct the mapping from V* to W*? –  Kevin Lee Apr 20 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 at 14:26
I atempt to prove (i)Im(ϕ)=W* (ii)ker(ϕ)=W^0 –  Kevin Lee Apr 20 at 14:28

## 1 Answer

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 $f|W=g$.

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Hello Julio C'aceres, thanks for your proof. But in your beginning of the proof, you define φ:V*/W^0-->W*, how do we verify that ker(φ)=W^0 ? –  Kevin Lee Apr 21 at 2:52
Here I find the same question of other's ask. He applies the first homomorphism theorm in the final step.math.stackexchange.com/questions/295064/… –  Kevin Lee Apr 21 at 2:56
Well the $\ker(\phi)$ is $[0]\in V^*/W^o$ since the map is inyective, but $[0]=W^o$. –  Julio Cáceres Apr 21 at 3:21
I didn't used the first isomorphism theorem, I constructed the map directly :P –  Julio Cáceres Apr 21 at 3:21
Thank you for help~! Mucho gracias~! –  Kevin Lee Apr 21 at 5:29