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Let $H$ be a Hilbert space and $T\in B(H)$ be a bounded linear operator on $H$, then $T=I$ $\Longleftrightarrow$ $\langle\psi,T\psi\rangle=1$ for every $\|\psi\|=1$.

It is easy to examine the "$\Longrightarrow$". But how to show the opposite implication?

Insomnia
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  • if $| \psi|=1$, then $\langle \psi, T\psi\rangle = 1\Leftrightarrow \langle \psi, (T-I)\psi\rangle = 0$. So you are asking if $A =0$ when $\langle \psi, A\psi\rangle = 0$. – Arctic Char Aug 02 '20 at 06:48

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This is false if the scalar field is $\mathbb R$. A counte-example is $T=I+S$ wheree $S$ is rotation by $90^{0}$ in $\mathbb R^{2}$.

In a complex Hilbert space it is well known that if $S$ is a bounded operator then$ \langle \psi, S (\psi) \rangle=0$ for all $\psi$ of norm $1$ (and hence for all $\psi$) implies that $S=0$. Apply this to $S=T-I$ and you get the conclusion easily.

Kavi Rama Murthy
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  • For the result I have used you can imitate the proof here: https://math.stackexchange.com/questions/2835911/langle-tf-f-rangle-0-implies-langle-tg-f-rangle-0?rq=1 – Kavi Rama Murthy Aug 02 '20 at 07:40