Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Algebra / 2 / a

Let $D=T^{\ast }T-T{T}^{\ast }$. This operator is self-adjoint, andIf condition (iii) holds, this quadratic form vanishes. For a real inner product, polarization givesTherefore $D=0$, which is condition (i). This supplies the nontrivial reverse implication needed for the equivalence.

By the definition of the adjoint,If (i) holds, the expression is zero for all $x,y$, giving (ii). Conversely, if this expression vanishes for every $x$, nondegeneracy of the inner product gives $(T^{\ast }T-T{T}^{\ast })y=0$ for every $y$, so (ii) also implies (i).

Taking $y=x$ in (ii) gives (iii). Together with (iii)$\Rightarrow$(i) from part (i) and (i)$\Rightarrow$(ii) from part (ii), this completes