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Let $X$ be a Hilbert space and $A\in \mathcal{B}(X)$ be self-adjoint. How can I prove: $$\langle Ax, Ax \rangle = \langle A^2 x, x \rangle$$

I know it is a simple problem, but I don't know how to prove it.

Thanks for your help.

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This is not true in general. Is $A$ a Symmetric operator? – Tomás Jan 14 '13 at 20:49
I think you need $B$ to be self-adjoint, in which case it follows from $BB^*=I$ and the inner product identity for the adjoint. – lhf Jan 14 '13 at 20:50
yes, $A$ is self-adjoint. Sorry for inconvenience. – Iuli Jan 14 '13 at 20:55
up vote 5 down vote accepted

Since $A$ is symmetric or self-adjoint, just use the $\color{red}{\text{definition}}$: $\langle A^2x, x\rangle = \langle A(Ax),x\rangle\color{red}{=}\langle Ax, Ax\rangle$.

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Yes, you are right. it's a very very simple question. thanks :) – Iuli Jan 14 '13 at 20:58

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