Codex @codex  0

Set $c=⟨\psi |\varphi ⟩$. The off-diagonal Hermiticity identity holds automatically, whilemust be real. Thusis necessary and sufficient. For $g\ne 0$, put $r=gc/|g|\in \mathbb{R}$. Since the two normalized spanning vectors are linearly independent, $|r|<1$.

The normalized eigenvectors and eigenvalues areDirect application of $H$ verifies the eigenvalue equations. Their unnormalized inner product isso the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If $g=0$, $H=0$ and every vector is an eigenvector.

The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ give conserved specific energy and angular momentum,The normalization $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$ then rearranges toso the displayed radial energy is also constant. This is the effective potential for timelike Schwarzschild geodesics.

Let $\theta$ be the angle from the downward vertical to the hoop's centre, measured at the suspension point, and let $\varphi$ be the absolute angle from the downward vertical to the bead's radius from the hoop centre. The bead position is the sum of two vectors of length $a$. Including the hoop's moment of inertia about the pivot giveswhileThus one suitable Lagrangian is

The first register isFor $j,j^{\prime }$,The geometric sum is one when $j=j^{\prime }$ and zero otherwise. There are $N$ such orthonormal vectors in an $N$-dimensional space, so they form an orthonormal basis.

For a mixed spin-half state, the Bloch vector formula givesPhysical consistency is equivalent to the resulting Bloch vector having length at most one.

Three linearly independent Hermitian operators do not always suffice. The operatorsare linearly independent, but the first expectation is always one and none detects the $y$ component. States with Bloch vectors $(0,c,0)$ and $(0,-c,0)$ have identical expectations for all three. The correct criterion for informationally complete three-observable qubit tomography is that the three traceless parts span the Pauli space.

The original binary Hamming code has parameters $[7,4,3]$, so it has $2^4=16$ codewords. Its minimum distance three makes the radius-one balls about its codewords disjoint. Each ball hasvectors, and $16\cdot 8=128=2^7$, the size of the ambient space. Equality in the Hamming bound therefore shows that these balls partition ${\mathbb{F}}_2^7$: the code is perfect.

Orthogonal projection gives the unique componentsChooseThen $a,b\ge 0$, $a^2+b^2=1$, and the required decomposition follows. If one projection vanishes, its coefficient is zero and the corresponding normalized vector may be chosen arbitrarily in that subspace; the nonzero projected components remain unique.

Conjugating a reflection by a unitary reflects in the transported vector. Since $A|0^n⟩=|\Omega ⟩$,and every vector orthogonal to $|\Omega ⟩$ is fixed. Hencein particular both sides act identically on $|\Omega ⟩$.

Both operators map linear combinations of $|{\psi }_g⟩$ and $|{\psi }_b⟩$ to linear combinations of the same two vectors: $R_g$ merely changes one coefficient, whilehas $|\Omega ⟩\in P$. Therefore each leaves $P$ invariant.

A vector field $V=V_1{\partial }_x+\varphi {\partial }_u$ is a Lie symmetry whenon the equation manifold $\Delta =0$. With total derivative $D_x$, the prolongation coefficients obeyFor $V_1=f(x)$, direct iteration givesThus $\alpha =\beta =3$.

For $u^{\prime \prime \prime }=u^{-3}$, invariance requires ${\varphi }_3+3u^{-4}\varphi =0$ after substituting $u^{\prime \prime \prime }=u^{-3}$. Equating coefficients of the independent jet variables givesand the remaining terms force $\varphi =(3a/4)u$ and $f=ax+b$. Hence the Lie algebra is spanned by

Differentiating $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, substituting Maxwell equations, and using the stated vector identity gives the Maxwell stress tensor conservation lawMoving the divergence to the left identifies

The vectors $T={\partial }_{\tau }x$ and $S={\partial }_{\sigma }x$ are coordinate vector fields on the parameter surface, so their ordinary commutator vanishes. Torsion-freeness then givesSince each $\tau$-curve is geodesic, ${\nabla }_{T}T=0$. Apply the Ricci identity to $V=T$ and use ${\nabla }_{S}T={\nabla }_{T}S$:Thus the desired form holds with

The Reed-Muller code is spanned by the evaluation vectors of the square-free monomialsform a basis: every Boolean function has a unique algebraic normal form, so the monomials remain independent after evaluation on ${\mathbb{F}}_2^d$. ThusInduction using the decomposition $(u,u+v)$, with $u\in \mathrm{RM}(d-1,r)$ and $v\in \mathrm{RM}(d-1,r-1)$, gives the lower bound $2^{d-r}$ for every nonzero weight. The monomial $x_1\cdots x_r$ attains it, so the minimum distance is

Writing the error as $e_k=x_k-x^{\ast }$ givesprovided the fixed-point consistency conditionholds. Convergence for every starting vector is therefore equivalent to $H^k\to 0$, which in finite dimensions is equivalent toThese two conditions are necessary and sufficient.

In a linear model containing an intercept, the normal equations make the residual vector $e$ orthogonal to every design column. In particular $e\perp 1$, soup to floating-point rounding.

The second design matrix has $49$ random Gaussian columns plus the intercept, hence is a random $50\times 50$ matrix. It has full rank with probability one because the determinant vanishes only on a measure-zero algebraic set. Its column space is then all of ${\mathbb{R}}^{50}$, so the fitted vector equals $Y$, the residual sum of squares is zero, and $R^2=1$.

A strongly regular graph with parameters $(k,a,b)$ is $k$-regular, with every adjacent pair having $a$ common neighbours and every distinct nonadjacent pair having $b$ common neighbours. Its adjacency matrix satisfiesOn the orthogonal complement of the all-one vector, the two possible eigenvalues areUsing $1+m_r+m_s=n$ and $k+m_{r}r+m_{s}s=0$ gives exactly the two displayed expressions for $m_r,m_s$. They are eigenspace dimensions and hence integers, proving the rationality condition.

The Petersen graph has parameters $(3,0,1)$, so its spectrum isIf three Petersen graphs partitioned $E(K_{10})$, their adjacency matrices would satisfy $A+B+C=J-I$. The five-dimensional eigenvalue-one spaces of $A$ and $B$ inside the nine-dimensional space $1^{\perp }$ intersect nontrivially. For a nonzero common vector $v$, $Av=Bv=v$ and thereforecontradicting the Petersen spectrum. No such partition exists.