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.
Best Answer
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 $T'=T''$.