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This question already has an answer here:

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 $A = A^*$.

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marked as duplicate by Trevor Gunn, hardmath, Daniel W. Farlow, Shailesh, Namaste linear-algebra Jun 11 '17 at 0:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I'm guessing $H$ is a Hilbert space, and you mean $\langle Ax,x\rangle$? $\endgroup$ – Jonathan Michael Foonlan Tsang Jun 10 '17 at 10:53
  • $\begingroup$ @JonathanMichaelFoonlanTsang, yes. Thanks for mentioning that. $\endgroup$ – Amin Jun 10 '17 at 10:54
  • $\begingroup$ The key here is that $H$ must be a complex inner product space. For a real inner product space these properties are not equivalent; see the discussion under Does non-symmetric positive definite matrix have positive eigenvalues? $\endgroup$ – hardmath Jun 10 '17 at 16:01
  • $\begingroup$ Thanks @hardmath; how can prove for $H$ real inner product space? $\endgroup$ – Amin Jun 10 '17 at 16:14
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    $\begingroup$ Have a look at the example in the Accepted Answer for Non-symmetric positive-definite matrices. $\endgroup$ – hardmath Jun 10 '17 at 16:22
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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^{*} $.

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