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.

For bounded positive operators $S,T$, membership $x\in \ker (S+T)$ gives $0=⟨Sx,x⟩+⟨Tx,x⟩$. Both terms are nonnegative, so both vanish. The Cauchy-Schwarz inequality for a positive operator makes $Sx=Tx=0$. The reverse inclusion is immediate. Commutativity is not required.

For a bounded positive operator $T$, the form $B(x,y)=⟨Tx,y⟩$ is positive semidefinite and Hermitian. Expanding $B(x+ty,x+ty)\ge 0$ for complex $t$ and minimizing its quadratic polynomial gives the displayed inequality. If $B(y,y)=0$, varying $t$ along the negative direction of $B(x,y)$ forces $B(x,y)=0$, covering the degenerate case. In particular $⟨Tx,x⟩=0$ implies $Tx=0$, by testing against arbitrary $y$.

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 $A$ is the closed subspaceThe range identity follows from image-kernel orthogonality for an adjoint and $A=A^{\ast }$. Equivalently, it is the range of the spectral projection $E_A((0,\infty ))$. In finite dimension, it is the span of eigenvectors with positive eigenvalues, and the range is already closed.

Closure matters in infinite dimension. On the l2 sequence space, $A{e}_n=e_n/n$ has zero kernel and hence full support, but is not onto: $(1/n)$ is square-summable and its formal inverse image $(1,1,\dots )$ is not. In quantum relative entropy, containment of the first state's support in the reference state's support is necessary for finiteness; finite dimensions also make it sufficient, while infinite dimensions can have divergence despite that containment.

Every bounded positive operator $T$ on a Hilbert space has a unique bounded positive operator $T^{1/2}$ with $(T^{1/2}{)}^2=T$. The continuous functional calculus applies the continuous scalar function $\sqrt{t}$ on the compact spectrum of $T$, givingIt also gives $⟨Tx,x⟩=\Vert T^{1/2}x{\Vert }^2$ and $\ker T^{1/2}=\ker T$.

For uniqueness, any positive $S$ with $S^2=T$ commutes with $T$ and hence with uniform polynomial approximations to $T^{1/2}$. Thus $S$ commutes with $T^{1/2}$, and commuting positive operators with equal squares are equal applies. Positivity is essential: $I$ has both $I$ and $-I$ as self-adjoint square roots, but only the former is positive. For $T=\mathrm{diag}(0,4,9)$, the positive square root is $\mathrm{diag}(0,2,3)$.

For bounded positive operators satisfying the hypotheses, $(S-T)(S+T)=0$, so $S-T$ vanishes on the closure of the range of $S+T$. The kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Self-adjointness and the Hilbert projection theorem decompose the whole Hilbert space as that closed range plus that kernel. Thus $S-T=0$. This proves the commuting case of uniqueness of a positive square root without using a spectral representation.