# Non-negative operator & self-adjoint operator [duplicate]

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 $A = A^*$.

## marked as duplicate by Trevor Gunn, hardmath, Daniel W. Farlow, Shailesh, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 11 '17 at 0:38

• I'm guessing $H$ is a Hilbert space, and you mean $\langle Ax,x\rangle$? – Jonathan Michael Foonlan Tsang Jun 10 '17 at 10:53
• @JonathanMichaelFoonlanTsang, yes. Thanks for mentioning that. – Amin Jun 10 '17 at 10:54
• 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? – hardmath Jun 10 '17 at 16:01
• Thanks @hardmath; how can prove for $H$ real inner product space? – Amin Jun 10 '17 at 16:14
• Have a look at the example in the Accepted Answer for Non-symmetric positive-definite matrices. – hardmath Jun 10 '17 at 16:22

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. $\langle Ax, x\rangle =\langle A^{*}x, x\rangle$

$\implies \langle (A-A^{*})x, x\rangle = 0, \forall x \in H$.

$\\$

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. $\langle Tu, v\rangle = 0 \: \forall u,v \in H.$

$\implies T = 0.$

$\\$

Hence, $A=A^{*}$.