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In the $A$ basis, the transformations whose columns are the new basis vectors areThus $U_C$ is a rotation through $-\pi /3$, while $U_B$ is a real orthogonal reflection, equivalently a rotation with one basis-vector phase reversed.

For any real orthogonal matrix $U$ with columns $|u_j⟩$,Applying this to $U_B$ and $U_C$ and normalizing proves

For the real Bell state $|{\varphi }^{+}⟩$, real one-qubit vectors satisfyFor each adjacent pair $(A,B)$, $(B,C)$, and $(C,A)$, the two same-label overlaps have magnitude $1/2$. Hence every matching-label probability isIt follows that

Rotational invariance makes this an isotropic vector, hence it is $0$; oddness gives the same result.

Differentiating a vector using $(dA/dt{)}_S=(dA/dt{)}_{S\prime }+\omega \times A$ twice yields
$m\ddot{x}=F-2m\omega \times \dot{x}-m\dot{\omega }\times x-m\omega \times (\omega \times x)$.

Path-connected means every pair is joined by a continuous map from $[0,1]$; a separation would pull back to a separation of an interval, so it implies connectedness. One valid boundary pair follows the bottom then right edges and the left then top edges. If opposite-corner paths $\alpha ,\beta$ were disjoint, normalising a continuous separation vector would construct the forbidden map $Q\to S^1$, so they intersect. A path in $Q\setminus A$ between the other corners, together with a path in $A$ between its corners, would contradict that intersection result; hence those corners lie in different path components.

Differentiability at $p$ means $f(p+h)=f(p)+Lh+o(|h|)$ for a linear map $L=Df(p)$; partial derivatives are its values on coordinate vectors. Integrating the partial derivatives along the two coordinate segments and using continuity at $p$ proves the stated sufficient condition. Here $f(x,y)=g(x)+g(y)$ with $g(t)=t^2\sin (1/t)$ and $g(0)=0$. Since $g\prime (0)=0$, $f$ is differentiable everywhere. But $g\prime (t)=2t\sin (1/t)-\cos (1/t)$ does not approach $0$ at zero, so $f$ is $C^1$ exactly where $xy\ne 0$.

The parity-check matrix has rank $3$, so its kernel has dimension $7-3=4$ and size $16$; see linear code. Its columns are precisely the seven distinct nonzero vectors of ${\mathbb{F}}_2^3$. Thus no weight-one or weight-two word is in the kernel, while three dependent columns give a weight-three word. Hence $d=3$.

Differentiate expected utility at the optimum $X_0$ in each portfolio direction:Rearranging gives the claimed vector identity.

For points $x_1,\dots ,x_n$, count the distinct vectors $(h(x_1),\dots ,h(x_n))$; maximize over samples to get $s(H,n)$. The shattering coefficient equals $2^n$ exactly when some sample is shattered, and the largest such $n$ is $\mathrm{VC}(H)$.

The required properties are those of a strict Lyapunov function; the domain of stability is the set of initial states whose forward trajectories converge to the fixed point.

For $V=(x^2+y^2)/2$,Thus $\dot{V}<0$ when $V<1/4$, proving that disc lies in the basin. On $V=1$, $\dot{V}=6y^2\ge 0$ (with the tangency points checked from the vector field), so trajectories outside cannot enter the basin through that boundary. Hence the basin is contained in $V\le 1$.

The power method repeatedly normalizes $A{q}^{(k)}$ and estimates the eigenvalue with the Rayleigh quotient $(q^{(k)}{)}^{T}A{q}^{(k)}$. It converges when the dominant eigenvalue is unique in modulus and the initial vector has a nonzero component in its eigendirection.

Across the cut $A|BC$, the two conditional $BC$ vectors $\alpha |00⟩$ and $\gamma |00⟩+\beta |11⟩$ are linearly independent when $\alpha \beta \ne 0$, so the Schmidt rank exceeds one. Hence the state cannot be a three-qubit product for any $\gamma$.

Under a coordinate change, the inhomogeneous second-derivative terms arising from each partial derivative cancel between $V^{\alpha }{\partial }_{\alpha }{W}^{\mu }$ and $W^{\alpha }{\partial }_{\alpha }{V}^{\mu }$. The remainder transforms with one contravariant Jacobian, so the commutator is a vector field.

Insert $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{E}_t$ into the Lorentz force and use vector identities plus Faraday's law. Rearrangement gives ${\partial }_t{g}_i+{\partial }_j{\sigma }_{ij}=-f_i$. Here $g={ϵ}_{0}E\times B$ is field momentum density and $\sigma$ is the Maxwell stress tensor, whose surface traction transports momentum.

Write $f(y;\theta )=\exp \{y\theta -b(\theta )+c(y)\}$ and use canonical parameter ${\theta }_i=X_i^T\beta$. Thenwhere ${\mu }_i=b^{\prime }(X_i^T\beta )$. Geometrically the residual vector is orthogonal to every design column. For $H_0:{\beta }_1=0$, use a two-sided Wald statistic ${\hat{\beta }}_1/\mathrm{se}({\hat{\beta }}_1)$ (or likelihood-ratio statistic), rejecting at the corresponding normal or chi-square critical value.

A polar vector transforms as ${\mathbf{v}}^{\prime }=Q\mathbf{v}$ under an orthogonal matrix $Q$, including an orientation-reversing transformation. Displacement, velocity and force are polar vectors. An axial vector instead transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$.

A real symmetric matrix $A$ is positive definite whenfor every nonzero real vector $x$. It therefore defines the inner product $⟨x,y{⟩}_A=x^{T}Ay$ and is invertible.

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.