Non-negative operator & self-adjoint operator I am wondering how to show that:
if $A$ is a non-negative operator, then $A$ is self-adjoint.

Def. 1. $A$ is non-negative if $\langle Ax,x \rangle \geq 0$ for $\forall x\in H$, where $H$ is a Hilbert space.
Def. 2. $A$ is self-adjoint if $A = A^*$. 
 A: For any linear  $A:H \rightarrow H $, we have
$\langle A^{*}x, x\rangle = \langle x, Ax\rangle =\overline {\langle Ax, x \rangle},$
But, for $A $ non-negative, then  $\langle Ax, x\rangle$ is real, so
$\langle Ax, x\rangle = \overline {\langle Ax, x \rangle}$
i.e. $\langle Ax, x\rangle =\langle A^{*}x, x\rangle $
$\implies \langle (A-A^{*})x, x\rangle = 0, \forall x \in H$.
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Claim: If $\langle Tx, x\rangle = 0 \: \forall x $ in a complex Hilbert space, then $T=0$.
Proof: 
Pick any $u, v \in H $, and let $T:H \rightarrow H $ such that $\langle Tx, x\rangle = 0 \: \forall x \in H $.
Then
$ 0 = \langle T(u+v), u+v\rangle = \langle Tu, v\rangle + \langle Tv, u\rangle$
$ \implies - \langle Tu,v\rangle = \langle Tv,u\rangle $,
and
$0 = \langle T(u+iv), u+iv\rangle = i\langle Tv, u\rangle - i\langle Tu, v\rangle$
$ \implies \langle Tu,v\rangle =  \langle Tv,u\rangle$.
Then $\langle Tu, v\rangle = -  \langle Tu, v\rangle $
i.e. $\langle Tu, v\rangle = 0 \: \forall u,v \in H. $
$\implies T = 0. $
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Hence, $A=A^{*} $.
