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.

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 $$z \in V$$.

• So for a fix $y \in V$, there exists a unqiue vector $T^*(y)$ which verifies$\langle T(x),y \rangle = \langle x, T^*(y) \rangle$ for all $x \in V$. But how to prove that for two different fixed $y_1, y_2 \in V$, $T^*_1 = T^*_2$? Because you need this equation to prove for $\textbf{for all x, y in V}$, there exists a unqiue vector $T^*(y)$ which verifies$\langle T(x),y \rangle = \langle x, T^*(y) \rangle$ . – YellowRiver Feb 25 at 7:27
• I don't quite understand your question: to define $T^*$ you only need to give an image $T^*(y)$ provided a vector $y \in V$. There exists a unique vector $T^*(y)$ which verifies the aforementioned condition, which by its convenient name it is taken to be the image of $y$ via $T^*$. Later on, we prove that if there is another transformation $S$ which verifies what is asked, then $T^* = S$. The 'for all' you emphasize is already implicit: take $x,y \in V$. By construction of $T^*(y)$, we have that $\langle T(x), y \rangle = \langle x,T^*(y) \rangle$. – Guido A. Feb 25 at 7:46
• Notice that when $T$ is fixed, $y$ and $T^*$ are bijective. That's being said, each $y$ corresponds to a unique$T^*$. Want I want to prove is those $T^*$ are the same. That is why I emphasize for all $y$. – YellowRiver Feb 25 at 8:08
• What do you mean by '$y$ and $T^*$ are bijective'? There is no need to have a bijective correspondence between $y$ and $T^*(y)$. If $T = 0$ for example, then $T^*(y) = 0$ regardless of $y$. Also, you are using $T^*$ to mean a vector and an operator simultaneously, I don't think I follow. I have proved that for a fixed $T$ there is a unique adjoint operator. Could you pinpoint what part of my argument does not convince you? – Guido A. Feb 25 at 16:50
• Thank you. I think I knew where I was wrong. – YellowRiver Feb 25 at 18:55

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 $$T'=T''$$.