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

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

For a normal operator, (a) gives $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^{\ast }}$.

For real $\lambda$, the adjoint of $T-\lambda I$ is $T^{\ast }-\lambda I$. Expanding both products shows that their difference is $T^{\ast }T-T{T}^{\ast }=0$. Thus $T-\lambda I$ is normal, and applying the kernel of a linear map conclusion to it givesThis proves equality of each real eigenspace, hence the same eigenvalues and eigenvectors on the stated real vector space. If neither has real eigenvalues, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline{\lambda }$ for the adjoint; no claim of the same complex eigenvalues is needed here.

In fact the adjoint invariance follows from the orthogonal decomposition and the invariance under $T$, without using the normal operator property. For $x\in V_i$ and $y\in V_j$ with $j\ne i$,since $Ty\in V_j$. Hence $T^{\ast }x$ is orthogonal to every other summand and therefore lies in $V_i$. Moreover, for $x,y\in V_i$, the adjoint identity restricted to that subspace shows

Restricting $T^{\ast }T=T{T}^{\ast }$ to $V_i$, using (i), gives $T_i^{\ast }{T}_i=T_i{T}_i^{\ast }$, so the global normal operator property implies the normal operator property of every restriction. Conversely, if the restrictions are normal and the adjoint preserves each summand, then for $x={\sum }_i{x}_i$,Thus $\boxed{T\text{ is normal}⟺\text{every }{T}_i\text{ is normal}}$, with condition (i) automatic under the stated decomposition.