A characterization of identity operator on Hilbert space

Question

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?

Answer 1

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.