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 $z \in V$.

  • $\begingroup$ 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$\langle T(x),y \rangle = \langle x, T^*(y) \rangle$ . $\endgroup$ – YellowRiver Feb 25 at 7:27
  • $\begingroup$ 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 $\langle T(x), y \rangle = \langle x,T^*(y) \rangle$. $\endgroup$ – Guido A. Feb 25 at 7:46
  • $\begingroup$ 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 $y$. $\endgroup$ – YellowRiver Feb 25 at 8:08
  • $\begingroup$ 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? $\endgroup$ – Guido A. Feb 25 at 16:50
  • $\begingroup$ Thank you. I think I knew where I was wrong. $\endgroup$ – 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 $T'=T''$.


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