Order on bounded self adjoint operators

Assume Countable Choice and let $H$ be a nonzero complex Hilbert space. Write

$\mathcal{B}(H{)}_{\mathrm{s}\mathrm{a}}:=\{T\in \mathcal{B}(H):T^{\ast }=T\}$

for the set of bounded self-adjoint operators on $H$ (Self-adjoint, positive, unitary and normal operators). For $S,T\in \mathcal{B}(H{)}_{\mathrm{s}\mathrm{a}}$ define

$S\le T⟺⟨(T-S)x,x⟩\ge 0\text{for every }x\in H\mathrm{.}$

Equivalently $T\ge S$; the notation $T\ge 0$ is exactly the positivity of Self-adjoint, positive, unitary and normal operators. Operators are compared only when both are self-adjoint: the relation is not defined for a general pair in $\mathcal{B}(H)$, and no conjugate-linear or non-real quadratic form is admitted by the definition.

$\mathcal{B}(H{)}_{\mathrm{s}\mathrm{a}}$ is a real vector space and $\le$ is a partial order on it. Sums and real scalar multiples of self-adjoint operators are self-adjoint, because the adjoint is conjugate-linear and $T^{\ast \ast }=T$ (Hilbert-adjoint identities); the zero operator is self-adjoint. So the comparisons below are between elements of a real vector space.

• Reflexivity. $S-S=0$ and $⟨0x,x⟩=0$, so $S\le S$.
• Transitivity. If $S\le T$ and $T\le U$, then for every $x$ $⟨(U-S)x,x⟩=⟨(U-T)x,x⟩+⟨(T-S)x,x⟩\ge 0,$ by additivity of the pairing in its first argument, so $S\le U$.
• Antisymmetry. If $S\le T$ and $T\le S$, then $⟨(T-S)x,x⟩=0$ for every $x$. The sesquilinear form $B(x,y):=⟨(T-S)x,y⟩$ is linear in $x$ and conjugate-linear in $y$ and satisfies $B(z,z)=0$ for every $z$, so the four-term expansion $4B(x,y)=B(x+y,x+y)-B(x-y,x-y)+iB(x+iy,x+iy)-iB(x-iy,x-iy)$ vanishes for all $x,y$. Hence $⟨(T-S)x,y⟩=0$ for all $x,y$, and fixing $x$ and taking $y=(T-S)x$ gives $\mathrm{\parallel }(T-S)x{\mathrm{\parallel }}^2=0$, so $T-S=0$ and $S=T$. The expansion is the displayed consequence of additivity and conjugate-linearity alone, so antisymmetry consumes no completeness of $H$ (Hilbert space).

Two conventions. First, the order is a partial order on the real vector space of self-adjoint operators; it is not a total order, and the extrema of the spectrum in the later results are taken in $\mathbb{R}$, not by comparing operators. Second, $T\ge 0$ refers to the quadratic form of a self-adjoint operator; the counterexample on the companion page shows that nonnegativity of the spectrum alone does not define positivity for operators that are not self-adjoint (The operator norm as the least bound and as the unit-sphere or unit-ball supremum for the operator data used throughout).