Positive-negative decomposition of a Hermitian operator

For a bounded Hermitian operator $A$, the continuous functional calculus defines $|A|=(A^2{)}^{1/2}$ andBoth parts are positive operators, commute, and satisfy $A_{+}{A}_{-}=0$, so their supports are orthogonal. They apply the scalar functions $\max (t,0)$ and $\max (-t,0)$ to the spectrum. The spectral theorem gives the same definitions even with continuous spectrum; a finite eigenvalue sum is only a special case.

If $A$ is additionally a trace-class operator, $\Vert A{\Vert }_1=\mathrm{Tr}{A}_{+}+\mathrm{Tr}{A}_{-}$. Consequently $\mathrm{Tr}A=0$ implies $\mathrm{Tr}{A}_{+}=\mathrm{Tr}{A}_{-}=\Vert A{\Vert }_1/2$. The trace-class assumption is essential: a general bounded Hermitian operator need not have a finite trace or trace norm. For $A=\mathrm{diag}(2,-1)$, the two parts are $\mathrm{diag}(2,0)$ and $\mathrm{diag}(0,1)$.

For a bounded Hermitian operator $X$, its negative part of a Hermitian operator isIt is a positive operator, despite the word negative: its eigenvalues in a finite-dimensional diagonalization are the magnitudes of the negative eigenvalues of $X$. For such a diagonalization, $X_{-}={\sum }_i\max (-{\lambda }_i,0)P_i$. The spectral-integral formula also covers a continuous spectrum without eigenvectors. Its support is orthogonal to the support of the positive part of a Hermitian operator, and $X=X_{+}-X_{-}$.

For a bounded Hermitian operator $X$, its positive part of a Hermitian operator isIt is a positive operator whose support is the positive spectral subspace. In a finite-dimensional spectral decomposition, $X_{+}={\sum }_i\max ({\lambda }_i,0)P_i$; the integral definition remains valid without an eigenbasis. Together with the negative part of a Hermitian operator, it obeys $X=X_{+}-X_{-}$, $|X|=X_{+}+X_{-}$ and $X_{+}{X}_{-}=0$.