About positive operators being Hermitian... I have been asked to prove the following: 
If $V$ is a finite-dimensional inner product space over $\mathbb C$, and if $A:V→V$ satisfies $⟨Av,v⟩≥0$ for all $v∈V$, then $A$ is Hermitian.
A proof is available at the address Show that a positive operator is also hermitian but I have come with a different approach that seems a lot simpler (which suggests that it's likely very wrong).
Here's my reasoning: if $A$ is positive ($⟨Av,v⟩≥0$ for all $v∈V$), then the inner product is necessarily real, which means that it is equal to its conjugate, so we can say:
$⟨Av,v⟩ = ⟨Av,v⟩^* = ⟨v,Av⟩$
By definition of Hermitian conjugate:
$⟨v,Av⟩ = ⟨A^\dagger v,v⟩$
Therefore:
$⟨Av,v⟩ = ⟨A^\dagger v,v⟩$ for all $v∈V$
Is this enough to conclude that $A = A^\dagger$, proving that $A$ is Hermitian? 
What am I doing wrong?
Thanks!
J.
(My first post on math.stackexchange, so be gentle :))

Yes.  The polarization identity shows that if $⟨Bv,v⟩ = 0$ for all $v$ then $⟨Bv,w⟩ = 0$ for all $v,w$, and thus $B=0$.
