I'm trying to prove the uniqueness of the Adjoint operator and I feel as my proof is a bit substandard; it is completely different from the suggested proof, can anyone tell me if I'm on the right track? I can prove that the adjoint is a linear operator, but proving the uniqueness of the adjoint is the step I'm having trouble with:
Assume V is a finite dimensional inner product space, and set $f^*(w) = w' $ we are trying to show that there exists a unique $w' \in V $ such that:
$$\langle f(v),w \rangle = \langle v, w'\rangle \space \forall \space v \in V$$
Assume that there exists a vector $w'' $ also satisfying this:
$$\langle f(v),w \rangle = \langle v, w''\rangle \space \forall \space v \in V$$
We want to prove that the difference, $w''-w'$ is zero.
$$ \langle v, w''\rangle – \langle v, w'\rangle = \langle v, w''-w'\rangle = 0 $$
So $w''-w'$ is in the orthogonal complement of v. But as this is true $ \forall \space v $, then $w'' – w'$ must be orthogonal for arbitary $ v \in V$ and hence is the zero vector? (This is the part I am unsure about).
The notes also had a proof outlined below which I could not follow, could anyone point me in the right path to understanding it?
Let $ (e_1,…e_n) $ be an orthonormal basis for $V$
Then $f^* $ must satisfy $ \langle e_j, f^*w\rangle = \langle fe_j, w\rangle \forall \space j$ (This part I understand)
I have trouble drawing the next conclusion:
Hence, $f^*w = \sum_{j=1}^n \langle fe_j, w\rangle e_j$ so $f^*$ must be unique.
Best Answer
If $\langle x, y \rangle = 0$ for all $x$, then we must have $y=0$. To see this, just choose $x=y$ which gives $\langle y, y \rangle = \|y\|^2 =0$ from which it follows that $y = 0$.
So, if $\langle v, w' \rangle = \langle v, w'' \rangle$ for all $v$, then you have $\langle v, w'-w'' \rangle = 0 $ for all $v$ from which it follows that $w'=w''$.
Existence is straightforward to establish if you have an orthonormal basis, say $e_k$:
Then, with $v= \sum_k v_k e_k$, we have $\langle f(v), w \rangle = \sum_k \overline{v_k} \langle f(e_k), w \rangle = \langle \sum_k v_k e_k, \sum_k \langle f(e_k), w \rangle e_k \rangle$, that is, $\langle f(v), w \rangle = \langle v, w' \rangle $, where $w'=f^*(w) = \sum_k \langle f(e_k), w \rangle e_k $.
It is straightforward to see that the function $ w \mapsto \sum_k \langle f(e_k), w \rangle e_k $ is linear and the comment above shows that the value $f^*(w)$ is unique, from which it follows that $w \mapsto f^*(w)$ is unique.