Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
@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 1

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 $f|W=g$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.