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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 $y$.

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.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

If each local term is already minimized by the global ground state, subtract its minimum $a_Z$ to obtain a positive operator. Nonnegative normalized spectral filtering of Hamiltonian terms preserves positivity and leaves the ground eigenvalue $a_Z$ unchanged. Thus the same state minimizes every filtered term. This preserves frustration freeness but does not create it for an initially frustrated decomposition.

A decomposition $H={\sum }_Z{h}_Z$ has frustration freeness when a global ground state minimizes every term simultaneously. For finite-dimensional Hermitian terms this implies $E_0(H)={\sum }_Z{E}_0(h_Z)$. Shifting each term by its minimum produces positive operators with a common null vector. The property depends on the decomposition, not only the total operator.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

The input penalty of a history Hamiltonian tests prescribed ancilla qubits at clock time zero while leaving the quantum witness unrestricted. A zero ancilla qubit is tested by the bad-state projector $|1⟩⟨1|$; a plus ancilla qubit is tested by $|-⟩⟨-|$.

Let $Q={\sum }_{j=1}^m{\Pi }_j$, where each ${\Pi }_j$ tests a different ancilla qubit. These orthogonal projections commute because they act on separate tensor product factors. Thus $Q$ is a positive operator with integer eigenvalues between zero and $m$. Zero is essential: its kernel contains exactly the valid ancillary input subspace, tensored with arbitrary quantum witness states. Only the nonzero eigenvalues are positive integers.

The clock-conditioned term is $H_{\mathrm{in}}=Q\otimes |c_0⟩⟨c_0|$. For a normalized computational history state $|\eta ⟩=(T+1{)}^{-1/2}{\sum }_t|{\psi }_t⟩|c_t⟩$ with clock labels forming an orthonormal basis,Together with correct propagation, zero input penalty selects histories with valid initial data. The displayed expectation concerns this fixed history state; a bound on the full Hamiltonian's lowest eigenvalue also has to account for propagation penalties.

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.

Let $A,B$ be positive operators on a finite-dimensional Hilbert space, with every nonzero eigenvalue at least $\gamma >0$. Let $P,Q$ be the orthogonal projections onto $\ker A,\ker B$ and define $\cos \vartheta =\Vert PQ\Vert$. ThenWhen the kernels have no common nonzero vector, this gives a positive lower bound. The projection formula also covers a zero kernel without taking a supremum over an empty unit sphere.

Indeed, $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. For $p\in \mathrm{ran}P$, $q\in \mathrm{ran}Q$, $|⟨p,q⟩|\le \Vert PQ\Vert \Vert p\Vert \Vert q\Vert$. The map $(p,q)\mapsto p+q$ consequently has squared operator norm at most $1+\Vert PQ\Vert$, so $P+Q\le (1+\Vert PQ\Vert )I$. Combining these inequalities proves the result. With common kernel $N$, restrict to $N^{\perp }$ and use the kernel projections there to bound the positive spectral gap.

In infinite dimension, replace the eigenvalue condition by the actual operator inequalities $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. An assertion only about eigenvalues can be vacuous in the presence of continuous spectrum, and does not ensure this uniform gap.

The positive operator in the polar decomposition $S=J{\Delta }^{1/2}$ of the Tomita operator. In a matrix-algebra standard representation with cyclic vector $D>0$, it is $\Delta (X)=D^{2}X{D}^{-2}$. For a tracial vector it is the identity.

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 bounded positive operators $0\le A\le B$ in the Löwner order, one has $A^{1/2}\le B^{1/2}$ even if $A$ and $B$ do not commute. A useful proof usesFor $t>0$, put $s=t{r}^2$ to obtain $2\sqrt{t}{\int }_0^{\infty }(1+r^2{)}^{-1}dr=\pi \sqrt{t}$ before the factor $1/\pi$. The zero case is immediate. The continuous functional calculus therefore gives the corresponding operator integral.

For $s>0$, inverse order yields $(B+sI{)}^{-1}\le (A+sI{)}^{-1}$. This follows by conjugating $B+sI\ge A+sI>0$ by $(A+sI{)}^{-1/2}$ and inverting its spectrum. ConsequentlyIntegration with positive weight proves the result. The integral converges in operator norm: near zero the integrand is bounded by $s^{-1/2}$, and near infinity by $\Vert A\Vert s^{-3/2}$, or the corresponding bound for $B$.

The partial trace of a positive operator is positive by this sum of nonnegative expectations in an orthonormal basis of the traced factor. Its trace equals the original trace, using the product orthonormal basis. Basis independence follows from $\mathrm{Tr}[({\mathrm{Tr}}_B\rho )A]=\mathrm{Tr}[\rho (A\otimes I)]$, which uniquely determines the reduced operator. Thus reductions of density matrices remain density matrices. For a positive trace-class operator the same argument uses convergent nonnegative basis sums.

For a positive operator, $⟨Tx,x⟩\ge 0$. We show directly that every nonreal or negative real $\lambda$ lies in the resolvent set.

If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality and reality of $⟨Tx,x⟩$ giveThus $T-\lambda I$ is bounded below, hence injective with closed range: a convergent sequence of outputs has Cauchy preimages, whose limit maps to that output. Its adjoint is $T-\bar{\lambda }I$, which has trivial kernel by the same estimate. The identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$, proved by the same inner-product calculation as in (i), makes its range dense. Being closed and dense, it is all of $X$, and the lower bound makes the inverse bounded.

If $\lambda <0$ is real, positivity gives $⟨(T-\lambda I)x,x⟩\ge (-\lambda )\Vert x{\Vert }^2$, so the same argument applies with lower bound $-\lambda$. Here the operator is self-adjoint, so its range is dense because its own kernel is zero. Hence

For $x\in \ker (S+T)$,Both terms are nonnegative because $S,T$ are positive operators. Their sum is zero, so each vanishes. In particular,

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.

Write both the input and output integer labels in most-significant-bit-first order, $x={\sum }_{j=1}^n{x}_j{2}^{n-j}$ and $y={\sum }_{j=1}^n{y}_j{2}^{n-j}$. Factoring the Fourier phase over the bits of $y$ gives the product decomposition of a Fourier phase state:Here $0.x_l\cdots x_n$ denotes the binary fraction ${\sum }_{r=l}^n{x}_r/2^{r-l+1}$; the integer part of $x/2^j$ does not affect the phase.

Build a dyadic quantum Fourier transform circuit as follows. Process wires $j=1,\dots ,n$ in that order. First apply a Hadamard gate to wire $j$. Then, for each $l=j+1,\dots ,n$, apply the controlled phase gate $R_{l-j+1}$ with control wire $l$ and target wire $j$. On a computational-basis state, every unprocessed control is still the bit $x_l$, so wire $j$ becomesOnce a wire has been processed it is not touched by later stages. Reversing the order of the output wires gives exactly the Fourier product above. Agreement on every computational-basis state proves agreement as linear operators on all superpositions, even though the intermediate operation on a general input can create entanglement.

There are $n$ Hadamard gates and $n(n-1)/2$ controlled phase gates. Output reversal may be treated as wire relabelling. If it must instead be implemented physically while using only the requested gate family, use $\lfloor n/2\rfloor$ swap operators. Each swap is three controlled-NOT gates, andThus the swaps use only Hadamard gates and Controlled-Z gates, which are controlled $R_1$ gates. An exact physical network haselementary gates from the stated family. This implements the positive-exponent Fourier transform with the correct output order in quadratic size. Omitting the output reversal without relabelling the wires would produce the reversed-order transform instead.

For the perturbation bound, the operator norm in the source is equivalently $\Vert A\Vert ={\sup }_{\Vert \psi \Vert =1}\Vert A\psi \Vert$. It satisfies the triangle inequality, is unchanged by left or right multiplication by a unitary operator, and a unitary operator has norm one. The exact telescoping identity isIt follows by replacing one factor at a time; alternatively all intermediate products cancel in the sum. Taking the operator norm and using unitary invariance gives the quantum circuit gate-error telescoping boundThe original PDF assumes strict $\Vert U_j-V_j\Vert <ϵ$, so its strict conclusion is justified. If the hypothesis were weakened to a non-strict inequality, the conclusion would likewise be $\le mϵ$. The TeX aid misreads this hypothesis; the PDF resolves it.

Apply the non-strict version to each approximate unitary gate in the Fourier network. Extending a gate by the identity on other wires leaves its operator norm error unchanged. Including the physical swap decompositions, all gates are of the required type and can obey the stated accuracy. HenceWith exact output relabelling rather than physical swaps the numerator is just $n(n+1)/2$. The proof retains gate phases and makes no assumption of cancellation between coherent errors.

For every normalized input the output-state norm error is at most this same bound. If $M$ is any positive operator satisfying $0\le M\le I$, the change in its measurement in quantum mechanics outcome probability is at most $2\delta$, where $\delta =\Vert U_n-{\mathrm{QFT}}_{2^n}\Vert$: expand the difference of the two expectations into two terms, each bounded by $\delta$ using the Cauchy-Schwarz inequality. Thus inverse-polynomial gate accuracy gives an inverse-polynomial total error, rather than requiring exponential precision in the number of qubits. For the phase gates this corresponds to specifying angles with $O(\log n)$ bits of precision, since $|e^{i\theta }-e^{i{\theta }^{\prime }}|\le |\theta -{\theta }^{\prime }|$. This is a robustness statement for unitary gate approximations; it does not by itself correct quantum decoherence, correlated stochastic faults, or limitations of hardware connectivity.

For a positive operator $A$, the spectral theorem writes $A={\sum }_j{a}_j|j⟩⟨j|$ with $a_j\ge 0$. Hence its unique positive square root of an operator is $\sqrt{A}={\sum }_j\sqrt{a_j}|j⟩⟨j|$. Conjugating this spectral decomposition shows $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$ for a unitary operator $U$.

Let $\Delta ={\rho }_1-{\rho }_2$ and ${\rho }_i^{\prime }=U{\rho }_i{U}^{†}$. Then ${\Delta }^{\prime }=U\Delta U^{†}$ and $({\Delta }^{\prime }{)}^{†}{\Delta }^{\prime }=U{\Delta }^{†}\Delta U^{†}$. The square-root covariance and the cyclic property of the trace giveFor the quantum fidelity, covariance first gives $({\rho }_1^{\prime }{)}^{1/2}=U{\rho }_1^{1/2}{U}^{†}$. ThereforeApply the same square-root identity and the cyclic property of the trace to obtainThus unitary invariance of trace distance and fidelity givesThe same $U$ acts on both quantum states: these quantities depend on the relative distinguishability of the quantum states, not on a choice of basis.

A linear map $\Phi$ is a completely positive map if ${\mathrm{id}}_R\otimes \Phi$ sends every positive operator to a positive operator for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.

For the Kraus representation, let $\Omega \ge 0$ be any operator on the auxiliary system and the input system. ThenFor any vector $v$, its expectation is ${\sum }_k⟨(I_R\otimes A_k^{†})v,\Omega (I_R\otimes A_k^{†})v⟩\ge 0$. Every amplification is therefore positive, proving the Kraus-form map is completely positive. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic quantum channel, one additionally requires ${\sum }_k{A}_k^{†}{A}_k=I$; a physical outcome branch instead obeys ${\sum }_k{A}_k^{†}{A}_k\le I$.

Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^{†}{\rho }^{T}v$ is the complex conjugate of ${\bar{v}}^{†}\rho \bar{v}$, hence is real and nonnegative. It fails complete positivity already for a qubit. Apply partial transposition to the second subsystem of the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator isOn the antisymmetric vector $(|01⟩-|10⟩)/\sqrt{2}$, its eigenvalue is $-1/2$. Thus the partial transpose is not positive on this entangled input, andThe same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.

For a unital star-subalgebra $A\subseteq B(H)$, the Von Neumann double commutant theorem statesHere $A^{\prime }$ is the commutant of an operator algebra, consisting of operators commuting with every element of $A$. The double commutant is closed in either topology: multiplication by a fixed bounded operator is continuous in both. Therefore both closures of $A$ lie in $A^{\prime \prime }$.

For the converse, let $T\in A^{\prime \prime }$ and choose finitely many vectors ${\xi }_1,\dots ,{\xi }_n$. In $H^n$ putIt is invariant under the diagonal action of $A$ and its adjoints, so its orthogonal projection $Q$ commutes with that diagonal action. Consequently each matrix entry $Q_{ij}$ belongs to $A^{\prime }$. The diagonal operator with entry $T$ therefore commutes with $Q$. Since $I\in A$, the tuple $\xi =({\xi }_1,\dots ,{\xi }_n)$ belongs to $K$, and hence so does $(T{\xi }_1,\dots ,T{\xi }_n)$. By the definition of $K$, one element $a\in A$ approximates this entire tuple arbitrarily closely. This is exactly approximation in every strong-operator neighborhood of $T$. It proves $A^{\prime \prime }\subseteq {\overline{A}}^{\mathrm{SOT}}$ and thus both asserted equalities. In particular, $M=A^{\prime \prime }$ is a Von Neumann algebra.

The Kaplansky density theorem strengthens the result by preserving norm bounds:It also holds for the self-adjoint and positive unit balls; the full ball can in fact be approximated in the strong-star topology, meaning strong convergence of both operators and adjoints. We prove the bound rather than merely scaling the unbounded approximants furnished by the double commutant theorem.

First put $C={\overline{A}}^{\Vert \cdot \Vert }$, a unital C-star algebra. Its self-adjoint part is weak-operator dense in $M_{\mathrm{sa}}$: symmetrize any weak approximating net. For a convex subset, strong and weak operator closures agree. To see this explicitly, separation from a strong closure supplies a continuous real functional on finitely many image vectors, hence one of the form $\mathrm{Re}{\sum }_j⟨a{\xi }_j,{\eta }_j⟩$, which is also weak-operator continuous. The Hahn-Banach separation theorem would then separate the point from the weak closure as well. Thus every self-adjoint $x\in M$ is a strong limit of self-adjoint $a_i\in C$, initially without a common norm bound.

The needed functional-calculus continuity follows directly from resolvents. The resolvent identity givesThe first factor has norm at most one, so the right side tends strongly to zero. The same holds at $-i$. Both resolvent nets are uniformly bounded, so their products converge strongly. Their scalar functions generate a self-adjoint algebra separating points of $\mathbb{R}$ and vanishing at infinity. The Stone-Weierstrass theorem, applied to the one-point compactification, makes this algebra uniformly dense in $C_0(\mathbb{R})$. Uniform approximation and the functional-calculus norm bound therefore showThis proves strong continuity of functional calculus through resolvents without an unjustified uniform bound on the original $a_i$.

If $x=x^{\ast }$ and $\Vert x\Vert \le 1$, choose a real compactly supported continuous $f$ equal to the identity on $[-1,1]$ and bounded in absolute value by one. Then $f(a_i)\in C$ are self-adjoint contractions and converge strongly to $f(x)=x$. For a positive contraction $x$, choose instead $0\le f\le 1$ equal to the identity on $[0,1]$; this gives positive contractions. These prove the two restricted versions for $C$.

For an arbitrary contraction $x\in M$, apply the self-adjoint result in the matrix algebra $M_2(C)$ toThe strong closure of $M_2(C)$ is $M_2(M)$, since finitely many entries may be approximated simultaneously. The upper-right entries $c_i$ of the approximating self-adjoint contraction matrices belong to $C$, satisfy $\Vert c_i\Vert \le 1$, and converge strongly to $x$. The lower-left entries are $c_i^{\ast }$ and converge strongly to $x^{\ast }$. This proves the full strong-star assertion for $C$.

Finally transfer the bounds from $C$ to the possibly nonclosed algebra $A$. A contraction $c\in C$ has an approximant $b\in A$ with $\Vert b-c\Vert <\epsilon$, so $b/(1+\epsilon )$ is an $A$-contraction within $2\epsilon$ of $c$. Self-adjoint approximants are obtained by symmetrizing first. For a positive contraction, approximate $c^{1/2}$ by $b\in A$ and use the positive contraction $b^{\ast }b/(1+\epsilon {)}^2$, which tends in norm to $c$ as $\epsilon \downarrow 0$. Combining these norm approximations with each finite-vector strong approximation proves all versions for $A$. The reverse inclusions follow because the relevant norm bounds and the inequalities defining a positive operator are preserved by strong limits.