Proof of Uniqueness of the Adjoint operator 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.
 A: 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.
A: As a proof of (uniquely) the uniqueness of the adjoint operator your proof is correct, and more economical than the one in the notes (there is no need to choose any basis). Basically a vector is determined by its inner product with arbitrary vectors (because an inner product is non-degenerate by definition; this is the "being orthogonal to everything implies being zero" part you were unsure about), and the adjointness condition prescribes the inner product of $f^*(w)$ with arbitrary vectors$~v$, thereby determining $f^*(w)$ completely.
What this does not show is that $f^*(w)$ exists, that there is a vector at all that satisfies the conditions. However, the prescribed inner product values are linear in$~v$, and it is also true (in finite dimension) that every linear function $V\to\Bbb R$ can be realised as the inner product with some vector. In fact by non-degeneracy this vector is unique, and once you know this, it is obvious that the defining condition defines $f^*(w)$ uniquely.
What the proof in the notes adds is an explicit formula for $f^*(w)$. For this it uses that for any orthonormal basis $e_1,\ldots,e_n$, the coordinate functions are given by inner products: for every $v$ one has $v=\sum_{i=1}^n\langle e_i,v\rangle e_i$. Applying this to $v=f^*(w)$ it obtains the coordinates of $f^*(w)$ in the given basis explicitly.
