# Why is $\|A\varphi_n\|^2=(\varphi_n,A^2\varphi_n)$ where $\{\varphi_n\}_{n\in\mathbb N}$ is an ONB?

I was reading the section on trace-class operators from Reed and Simon whence I encountered this formula. If $\{\varphi_n\}_{n\in\mathbb N}$ is an orthonormal basis of a Hilbert space $\mathcal H$ and $A \in \mathcal L(\mathcal H)$, then $$\|A\varphi_n\|^2=(\varphi_n,A^2\varphi_n)\,.$$

I understand that $\|A\varphi_n\|^2=\sum_{m\in\mathbb N} |(\varphi_m, A\varphi_n)|^2$ which means that I have to prove the following.

$$\sum_{m\in\mathbb N} |(\varphi_m, A\varphi_n)|^2 = (\varphi_n,A^2\varphi_n)\,.$$

I skimmed through all the previous pages of the book, but I cannot seem to find a proof of the above. Kindly help.

• Is $A$ self-adjoint? Since $A$ is trace-class, it is compact and, in particular, bounded. – Alex Ortiz Jul 26 '17 at 19:06
• @AOrtiz $\mathcal L(\mathcal H)$ is the set of bounded operators on $\mathcal H.$ I am not sure if $A$ is self-adjoint. Is it needed for trace-class operators? If you happen to have access to Reed and Simon, please check the proof of theorem $VI.18.$ – Nanashi No Gombe Jul 26 '17 at 19:15
• I have Reed and Simon here. Perhaps you could point to the line in the proof you are questioning? – Alex Ortiz Jul 26 '17 at 19:28
• @AOrtiz The second equality: $(\varphi_n,A \varphi_n) = \|A^{1/2} \varphi_n\|^2 .$ – Nanashi No Gombe Jul 26 '17 at 19:32

From Wikipedia: A bounded non-negative operator on a complex Hilbert space is self-adjoint. (This is also stated on pg. 195 of Reed and Simon following the definition of positive operator.) Thus $(A^{1/2})^2 = A$.
If $(\cdot,\cdot)$ is the inner product on $\mathcal H$, then we have $$(\varphi_n,A\varphi_n) = (\varphi_n,A^{1/2}A^{1/2}\varphi_n) = (A^{1/2}\varphi_n,A^{1/2}\varphi_n)=\lVert A^{1/2}\varphi_n\rVert^2,$$ where the first equality is using what we noted above, the second comes from the defining property of the adjoint, and the third is by definition of the norm induced by the inner product.
• If you would excuse my stupidity, I cannot see how this implies that $(\varphi_n,A \varphi_n) = \|A^{1/2} \varphi_n\|^2.$ Would you be kind enough to show the explicit steps to this end? Thanks. – Nanashi No Gombe Jul 26 '17 at 19:40
• $(\varphi_{n},A \varphi_{n}) = (\varphi_{n},(A^{\frac{1}{2}})^{2} \varphi_{n}) = (A^{\frac{1}{2}}\varphi_{n},A^{\frac{1}{2}}\varphi_{n}) = \|A^{\frac{1}{2}} \varphi_{n}\|^{2}$ since the square-root is self-adjoint. – fourierwho Jul 26 '17 at 19:42
$\Vert A\varphi\Vert^2=(A\varphi_n, A\varphi_n) = (\varphi_n, AA\varphi_n) = (\varphi_n,A^2\varphi_n)$