Non-negative operator & self-adjoint operator

Question

I am wondering how to show that:

if $A$ is a non-negative operator, then $A$ is self-adjoint.

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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 $A = A^*$.

Answer 1

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^{*} $.