Positive operator ($T\ge 0$)

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

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.