Hermitian operator

A bounded operator is Hermitian, or self-adjoint, when $T=T^{\ast }$.

For a complex $n\times n$ matrix $B$, the real linear map $A\mapsto BA{B}^{\ast }$ on the $n^2$-dimensional real vector space of Hermitian matrices has determinant $|\det B{|}^{2n}$. A real basis of Hermitian matrices is also a complex basis of all matrices, so the matrix of this restriction equals the matrix of the complex-linear congruence map. Left and right multiplication separately have determinants $(\det B{)}^n$ and $(\overline{\det B}{)}^n$.

A bounded Hermitian operator $T$ on a Hilbert space is positive when $⟨Tx,x⟩\ge 0$ for every $x$. Its spectrum is contained in $[0,\infty )$.

A symmetric operator $L$ on a real Hilbert space is strictly positive definite when $⟨Lv,v⟩>0$ for every nonzero $v$ in its operator domain. Strict positivity proves uniqueness of a solution of $Lu=f$, but in infinite dimension it need not prove existence for every $f$. A uniformly positive definite symmetric operator additionally has a positive lower bound independent of $v$. Symmetry must be stated separately when positivity is defined only by a real quadratic expression: adding a skew-symmetric matrix changes the operator without changing that expression.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

The support of a positive operator is the orthogonal complement of its kernel. In finite dimension it is the span of eigenvectors with positive eigenvalues, equivalently the operator's image. Quantum relative entropy is finite only when the first state's support lies in the reference state's support.

Every bounded positive operator $T$ has a unique positive operator $T^{1/2}$ satisfying $(T^{1/2}{)}^2=T$. The spectral theorem for normal operators on a separable Hilbert space constructs it by applying the scalar square-root function to the spectrum.

The spectral theorem for normal operators gives a unique decompositionwhere $A_{+},A_{-}\ge 0$ and their supports are orthogonal. They are called the positive and negative parts of $A$. If $\mathrm{Tr}A=0$, then $\mathrm{Tr}{A}_{+}=\mathrm{Tr}{A}_{-}=\Vert A{\Vert }_1/2$.

The negative part of a Hermitian operator is $X_{-}={\sum }_i\max (-{\lambda }_i,0)P_i=(|X|-X)/2$. In this convention $X_{-}$ is itself a positive operator, and $X=X_{+}-X_{-}$. Its support is orthogonal to that of the positive part of a Hermitian operator.

For a Hermitian operator with spectral decomposition $X={\sum }_i{\lambda }_i{P}_i$, its positive part is $X_{+}={\sum }_i\max ({\lambda }_i,0)P_i=(|X|+X)/2$. It is a positive operator supported on the positive spectral subspace. Together with the negative part of a Hermitian operator it satisfies $X=X_{+}-X_{-}$ and $|X|=X_{+}+X_{-}$.

For Hermitian operators $A$ and $B$, one writes $A\le B$ when $B-A$ is positive semidefinite. This partial order is also called the semidefinite order.

For bounded positive self-adjoint operators, $0\le A\le B$ implies $A^{1/2}\le B^{1/2}$. Thus the nonnegative square-root function is operator monotone.