Self-adjoint, positive, unitary and normal operators

Assume the Axiom of Countable Choice, and let $H$ be a real or complex Hilbert space with $T\in \mathcal{B}(H)$ a bounded linear operator and $T^{\ast }$ its Hilbert adjoint (The Hilbert-space adjoint of a bounded operator, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

• $T$ is self-adjoint when $T^{\ast }=T$;
• $T$ is positive when $⟨Tx,x⟩$ is a real number in $[0,+\mathrm{\infty })$ for every $x\in H$;
• $T$ is unitary when $T^{\ast }T=T{T}^{\ast }=I$, the identity operator;
• $T$ is normal when $T^{\ast }T=T{T}^{\ast }$.

The definitions are read over either scalar field with the same inner product. Two immediate consequences. Every self-adjoint operator is normal, because $T^{\ast }T=T^2=T{T}^{\ast }$ when $T^{\ast }=T$. Every unitary operator is normal, because its defining identity says exactly that $T^{\ast }T$ and $T{T}^{\ast }$ are both the identity. Positivity is a condition on the values of the quadratic form and therefore forces those values to be real, which for a complex Hilbert space does not follow from boundedness alone; a positive operator on a complex Hilbert space is in fact self-adjoint, but that is proved later and is not assumed here.

Consistency of the vocabulary. The identity operator is self-adjoint, positive and unitary, and the zero operator is self-adjoint and positive; on the one-dimensional Hilbert space $\mathbb{C}$ the operator $\lambda I$ is self-adjoint exactly when $\lambda$ is real, unitary exactly when $\mathrm{\mid }\lambda \mathrm{\mid }=1$, and normal for every scalar $\lambda$.