Its something that is in all basic algebra books that, a semipositive definite operator is Hermitian and the eigenvalues of this hermitian operator are positive.
But I couldn't find any place where this relation is explained. How can one demonstrate that a semipositive operator imply hermiticity?
$$\langle v |A |v\rangle \geq 0 \quad \forall |v\rangle \quad \stackrel{}{\Rightarrow} \quad A=A^{\dagger} $$
Where $A$ and $|v\rangle$ are elements of a Hilbert space of finite dimension i.e. complex matrix elements etc.
This is asumed in lot of books but I couldn't find it anywhere. If sameone could help me how to start, because I tried to prove the negation also but I don't arrive to any place.
Note that two operators $A,B$ are equal if and only if we have $\langle v|A |v\rangle = \langle v|B|v\rangle$ for all $|v\rangle \in H$. You can reach this conclusion by considering the expression $\langle v|(A - B)|v \rangle$ with $|v \rangle = |x\rangle + z|y \rangle$, taking both $z = 1$ and $z = i$. Alternatively, one could use the polarization identity.
Note also that in general, $\langle u|A|v\rangle^* = \langle v|A^\dagger |u \rangle$ (where $z^*$ denotes the complex conjugate of $z$).
Now, if $A$ is semipositive, then $\langle v|A|v\rangle \geq 0$ implies that $\langle v|A|v\rangle$ is real for all $|v\rangle$. Thus, $$ \langle v|A|v\rangle = \langle v|A|v\rangle^* = \langle v|A^\dagger|v\rangle. $$ Because this holds for all $|v\rangle$, we conclude that $A = A^\dagger$.