> [!theorem] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]] and $T \in L(\ch, \ch)$ be a [[Bounded Linear Operator|bounded linear operator]]. Denote $p(x) = \angles{Tx, x}$, then for any $x, y \in \ch$, > $ > \angles{Tx, y} = \frac{1}{4}\paren{p(x + y) - p(x - y) + ip(x + iy) - ip(x - iy)} > $ > *Proof*. Firstly, > $ > \begin{align*} > p(x + y) &= \angles{Tx, x} + \angles{Tx, y} + \angles{Ty, x} + \angles{Ty, y} \\ > p(x - y) &= \angles{Tx, x} - \angles{Tx, y} - \angles{Ty, x} + \angles{Ty, y} \\ > p(x + y) - p(x - y) &= 2(\angles{Tx, y} + \angles{Ty, x}) > \end{align*} > $ > On the other hand, > $ > \begin{align*} > p(x + iy) &= \angles{Tx, x} - i\angles{Tx, y} + i\angles{Ty, x} + \angles{Ty, y} \\ > p(x - iy) &= \angles{Tx, x} + i\angles{Tx, y} - i\angles{Ty, x} + \angles{Ty, y} \\ > p(x + iy) - p(x - iy) &= 2i(\angles{Ty, x} - \angles{Tx, y})\\ > ip(x + iy) - ip(x - iy) &= 2(\angles{Tx, y} - \angles{Ty, x}) > \end{align*} > $ > Combining both parts gives the desired identity.