# Let $A$, $B$ be positive operators in a Hilbert space and $\langle Ax,x \rangle=\langle Bx,x \rangle$ for all $x$, show that $A=B$

Let $A$ and $B$ be positive operators in a Hilbert space $H$, and suppose that $\langle Ax,x\rangle=\langle Bx,x\rangle$ for every $x$ in $H$. Show that $A=B$.

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Hello and welcome to math.stackexchange. Is this a HW or exam question? What have you tried? –  Laura Balzano May 21 '13 at 21:02
If $H$ is real, you can use $T$ self-adjoint and $(Tx,x)=0$ for all $x$ implies $T=0$. If $H$ is complex, you can even remove the self-adjoint assumption. Two important lemmas which can be proved by polarization. In any case, try to show $(Tx,y)=0$. –  1015 May 21 '13 at 21:04
This is HW problem [] –  SHIBASHIS May 23 '13 at 10:09
yeah it is not needed that A, B should be positive that's why i am confused –  SHIBASHIS May 23 '13 at 10:13

If they are positive then they are selfadjoint and there exists a unique square root. from the fact that $\langle Ax,x \rangle = \langle Bx,x \rangle$ we get that also $A - B$ is positive and $$0 = \langle Ax - Bx,x\rangle = \|\sqrt{A - B}x\|^2$$ then $\sqrt{A - B} = 0$ and by uniquiness of the square root $A - B = 0$ proving the claim.