Let \(H\) be a complex Hilbert space, \(T : H \to H\) be a bounded linear operator and let \(T^{*}\) denote the adjoint of \(T\). Which of the following statements are always TRUE?
P: \(\forall x, y \in H, \langle Tx, y \rangle = \langle x, T^{*}y \rangle\)
Q: \(\forall x, y \in H, \langle x, Ty \rangle = \langle T^{*}x, y \rangle\)
R: \(\forall x, y \in H, \langle x, Ty \rangle = \langle x, T^{*}y \rangle\)
S: \(\forall x, y \in H, \langle Tx, Ty \rangle = \langle T^{*}x, T^{*}y \rangle\)
GATE Mathematics - 2012,
Answer: A
The statements that always hold are P and Q.
P is the definition of the adjoint: \(\langle Tx, y \rangle = \langle x, T^{*}y \rangle\) for all \(x, y\). Q follows from it by conjugate symmetry:
\[ \begin{aligned} \langle x, Ty \rangle &= \overline{\langle Ty, x \rangle} \\ &= \overline{\langle y, T^{*}x \rangle} = \langle T^{*}x, y \rangle \end{aligned} \]
R would force \(\langle x, (T - T^{*})y \rangle = 0\) for all \(x\), that is \(T = T^{*}\), which holds only for self-adjoint operators. S says \(\|Tx\|\) and \(\|T^{*}x\|\) agree in the polarised sense, which is the condition \(T^{*}T = TT^{*}\) for normal operators, not a property of every bounded operator. So options with R or S are wrong.
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