Uniqueness of the Adjoint operator

Question

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.

Answer 1

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 $z \in V$.

Answer 2

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 $T'=T''$.

Answer 3

Suppose there are two adjoints $A,B$, so that

$$ \langle Tx, y \rangle = \langle x, Ay \rangle $$ $$ \langle Tx, y \rangle = \langle x, By \rangle .$$

Subtracting the two equations yields

$$ 0 = \langle x, Ay \rangle - \langle x, By \rangle $$ $$ 0 = \langle x, Ay \rangle + \langle x, -By \rangle $$ $$ 0 = \langle x, Ay - By \rangle $$ $$ 0 = \langle x, (A - B) y \rangle, \ \forall x, \forall y \in V.$$

Now, it may be tempting to use the theorem $\forall v \in V, \ \langle v,w \rangle = 0 \implies w = 0$. But that $\forall$ is quantifying over the entire vector space $V$, where as $(A - B) y$ is just a subset of $V$.

Luckily though, we can show it's more than just a subset; it's also a subspace. Since $A,B$ are linear then their ranges $R(A),R(B)$ are subspaces of $V$ so $R(A-B)$ is a subspace of $V$.

Now $$ 0 = \langle x, (A - B) y \rangle, \ \forall x, \forall y \in V $$ implies $$ 0 = \langle x, (A - B) y \rangle, \ \forall x, \forall y \in R(A-B) .$$

Now we can instantiate the theorem which give us $(A - B) y = 0$. Since $y$ is arbitrary then $A-B = 0$ so $A = B$.