Question on adjoint operator

Question

Theorem Let $V$ and $W$ be finite dimensional inner product spaces over $F$ and let $T\in L(V,W)$. Then there is a unique function $T^*:W\to V$, defined by the condition $<T(v),w>=<v,T^*(w)>$ for all $v\in V$ and $w\in W$. This function is in $L(W,V)$ and is called the adjoint of $T$.

Actually I stuck in the proof. I saw the proof in book .if $w$ is arbitrary fixed element of $W$. consider the function $\phi_w:V\to F$ defined by $\phi_w(v)=<T(v),w>$. The problem is that why $\phi_w(v)=<T(v),w>$ instead of $\phi_w(v)=<v,w>$.

Answer 1

Use Reisz Representation Theorem

As you define $\phi_w:V\to F$ such that $\phi_w(v)=<T(v),w>$ for fixed $w\in W$. $\phi_w$ is linear functional. you can start with $$\phi_w(au+bv)=<T(au+bv),w>$$. Now follow as @Ravi explained. Next you need to show that $T^*$ is linear so that $T^*\in L(V,W)$

Good luck.

Answer 2

We want to find $T^*(w)$ such that $<T(v),w>=<v,T^*(w)>$ for all $v$ and $w$. Start with the left side. If you verify that it is linear in the variable $v$ for fixed $w$ you can conclude that existence of $T^*(w)$. This is because any linear map $L: V \to F$ is of the form $L(v)= \langle v, z \rangle$ for some $z$. Verification of lineaity is straightforward.