I'm stuck in the following question:
Let V be a finite-dimensional vector space over F. The dual space of V is defined to be $V^∗ = L(V, F)$ (i.e. the space of linear transforms from V to F).
(a) Prove that $dimV^∗ =dimV$.
(b) Let W be a subspace of V and define the annihilator of W to be $W^⊥ = {l ∈ V^∗ | l(w⃗ ) = 0, \forall w⃗ ∈ W }$. Prove that $W^⊥$ is a subspace of $V^∗$ and that $V^∗/W^⊥ ≃ W^∗ $.
(c) Prove that if $V = W1⊕W2$,then $V^∗ =W1^⊥⊕W2^⊥$.
What I did:
Item a is not hard, just assuming a basis for V and using linearity of linear transformations. Item b is my biggest problem. Given the 1st Isomorphisms Theorem, which relates the null space, range and the domain of a transformation, it suffices to show that $W^*$ is the range of a transformation and that $W^⊥$ is the null space of the same transformation. I haven't been able to do the latter and I'm not sure if my attempt at the first is correct:
Let $T\in L(V^*,W^*)$, because $W^*$ is a subspace of $V^*$, $T(w^*)\in W^*, \forall w^* \in W^*$. So $W^* \subseteq R_T$, but $R_T \subseteq W^*$, so $R_T=W^*$.
Haven't been able to do anything for item c.
For part b), as suggested by the statement, let $\gamma:V^*\to W^*$ be the restriction map: for any $w\in W$, $\gamma(f)=f|_W$. This is a linear map. And $$ \ker\gamma=\{f\in V^*:\ f(w)=0\ \text{ for all } w\in W\}=W^\perp. $$ So $W^\perp$ is a subspace, and by the First Isomorphism Theorem $V^*/W^\perp\simeq W^*$ (after checking that $\gamma$ is onto, which is equivalent to say that any $g\in W^*$ extends to $V^*$).
For part c), given $f\in V^*$, since any $v\in V$ can be written as $v=v_1+v_2$ in a unique way with $v_1\in W_1$, $v_2\in W_2$, define $f_1\in W_1^*$, $f_2\in W_2^*$ by $$ f_1(w)=f(w_1\oplus 0),\ \ f_2(w_2)=f(0\oplus w_2). $$ As $f$ is linear, $f=f_1+f_2$. We also have $f_1\in W_2^\perp$, $f_2\in W_1^\perp$. This shows that the map $\delta:W_2^\perp\oplus W_1^\perp\to V^*$ given by $$ \delta(f_1\oplus f_2)=f_1+f_2$$ is onto. Linearity and injectivity are easy to check.