Uniqueness of the Adjoint operator So I was just stuck in the middle of proving the uniqueness of the adjoint operator. 
Known theorem(I already know how to prove it): Assume V is a finite dimensional inner product space over a field F, and let $g: V \to F$ be linear transformations. Then there exists a unique $y \in V$ such that $g(x) = \langle x, y\rangle$ for all $x \in V$.    
I want to prove the following theorem: V, T are given above. prove there exist a unique $T^*: V\to V$ such that $\langle T(x), y \rangle = \langle x, T^*(y) \rangle$ $\forall x, y \in V$. 
Here is the sketch of my proof: $\exists! y' \in V$ such that  $g(x) = \langle x, y' \rangle = \langle T(x), y\rangle$. Now we define $T^*: V \to V$ as $T^*(y) = y'$ and claim $T^*$ is unique. Then I don't know what to do next because I only know when $y$ and $T$ are fixed, I can find a unique $y'$ in $\langle x, y' \rangle = \langle T(x), y\rangle$. When I choose different $y$, how can I ensure that the $T^*$ is unique? How to prove that $\textbf{for all x, y} \in V$, there exist a unique $T^*$? In other words, when I can find unique $T^*(y_1)$ for a given $y_1$ in $\langle x, T^*(y_1)\rangle = \langle T(x), y_1\rangle$ and  $T^*(y_2)$ for a given $y_2$ in $\langle x, T^*(y_2)\rangle = \langle T(x), y_2\rangle$. How I can ensure that those two $T^*$ are actually the same.
 A: Suppose there are maps $T'$ and $T''$ such that, for every $x,y\in V$,
$$
\langle T(x),y\rangle=\langle x,T'(y)\rangle=\langle x,T''(y)\rangle
$$
Fix $y\in V$; then, for every $x\in V$,
$$
\langle x,T'(y)-T''(y)\rangle=\langle x,T'(y)\rangle-\langle x,T''(y)\rangle
=\langle T(x),y\rangle-\langle T(x),y\rangle=0
$$
In particular, for $x=T'(y)-T''(y)$, we get
$$\langle T'(y)-T''(y),T'(y)-T''(y)\rangle=\langle x,x\rangle=0,
$$
so $x=T'(y)-T''(y)=0$. Since $y$ is arbitrary, we get $T'=T''$.
A: Fix $y \in Y$. Then $\langle T(-),y \rangle$ it a functional, for which is has to exist a unqiue vector $T^*(y)$ which verifies
$$
\langle T(x),y \rangle = \langle x, T^*(y) \rangle
$$
for all $x \in V$. Now, we have to see that $T^*$ is linear. By construction, 
$$
\begin{align}
\langle T(x),\alpha z+y \rangle &= \overline{\alpha} \langle T(x),z \rangle + \langle T(x),y \rangle = \overline{\alpha} \langle x,T^*(z) \rangle + \langle x,T^*(y) \rangle  \\ & = \langle x, \alpha T^*(z) + T^*(y)\rangle.
\end{align}
$$
and therefore since $\alpha T^*(z) + T^*(y)$ represents $\langle T(-),\alpha z+y \rangle$, by the uniqueness of such vector we have that
$$
\alpha T^*(z) + T^*(y) = T^*(\alpha z+y).
$$
Plugging $\alpha = 1$ or $y = 0$ proves each condition of linearity. 
Now, uniqueness: suppose that you have another transformation $S$ that verifies $\langle T(x) , y \rangle  = \langle x,  S(y) \rangle$ for all $x,y$ in $V$. It suffices to see that $T^*(y) = S(y)$ for each $y \in V$, so let's fix $y \in V$. For any $x \in V$,
$$
\langle x, T^*(y) -S(y) \rangle = \langle x, T^*(y) \rangle - \langle x,S(y) \rangle = \langle T(x),y \rangle - \langle T(x),y \rangle = 0
$$
and so $T^*(y) - S(y) = 0$, which concludes the proof: recall that a vector $v$ is zero if and only if $\langle v, z \rangle = 0$ for all $z \in V$.
