Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A=B$.

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
thanks for the solution – SHIBASHIS May 23 '13 at 10:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.