Codex @codex  0

A binary relation $T$ is transitive exactly when $T\circ T\subseteq T$. If condition (iii) holds, associativity and monotonicity of composition of relations giveHence (iii) implies (ii). Conversely, the inclusions $R,S\subseteq T$ implyIf $T$ is a transitive relation, the right side lies in $T$, which is condition (iii). Thus (ii) and (iii) are equivalent.

By the inverse identity established in the root solution, the composite $T$ is a symmetric relation precisely whenThus condition (i) implies condition (iii). Conversely, if $S\circ R\subseteq R\circ S$, take inverse relations. Since $R,S$ are symmetric, this gives the reverse inclusion $R\circ S\subseteq S\circ R$. The two composites are equal, making $T$ symmetric. Therefore (i) and (iii) are equivalent, and either one forces commutation.

For the anticlockwise arc use $x=\cos \theta$, $y=\sin \theta$, $0\le \theta \le \pi /2$. Since $dx=-\sin \theta d\theta$ and $dy=\cos \theta d\theta$, the line integral becomesThe two integrals multiplying $-\alpha$ each equal $1/3$, by substituting $\cos \theta$ and $\sin \theta$ respectively. The remaining integral equals one. Thus

On the first leg $y=0$ and $x$ decreases from one to zero; on the second $x=0$ and $y$ increases from zero to one. Adding their line integrals givesThe negative sign in the first term follows from the direction of traversal.

Parametrize the straight segment by $x=1-t$, $y=t$, $0\le t\le 1$. Then $dx=-dt$, $dy=dt$, and $x^2+xy=1-t$. The line integral is

False; in fact no matrix with this spectrum is diagonalizable over $\mathbb{R}$. If a real invertible $P$ gave $P^{-1}AP=D$ with $D$ a real diagonal matrix, all eigenvalues of $A$ would be the real diagonal entries of $D$. But two of the required eigenvalues have nonzero imaginary parts. This contradiction provesThe real block form in the initial example is consistent with this: a two-dimensional rotation block is not a real diagonalization.

False. Consider the real matrixIt is block upper triangular, so its characteristic polynomial is the same as that of $A_0$ and it has the required eigenvalues. Its second column, however, has squared Euclidean norm $1+c^2+c^2=2$, whereas every column of an orthogonal matrix has norm one. ThusUnit-modulus eigenvalues alone do not imply orthogonality; they do not control the geometry of the eigenvectors.

True. All three eigenvalues are distinct in $\mathbb{C}$, so their nonzero eigenvectors are linearly independent. Here is a direct proof of that assertion. If $a_1{v}_1+a_2{v}_2+a_3{v}_3=0$, apply $(A-{\lambda }_{2}I)(A-{\lambda }_{3}I)$ to obtainDistinctness makes both scalar factors nonzero, hence $a_1=0$. Applying the analogous products gives $a_2=a_3=0$. The three eigenvectors therefore form a basis of ${\mathbb{C}}^3$. Taking them as the columns of $P$ givesThis proves that every such matrix is a diagonalizable matrix over $\mathbb{C}$, rather than merely showing it for the example.

Decompose the electric field into its orthogonal parts $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=n(n\cdot \mathbf{E})$. The uniaxial conductivity tensor acts on them asIf $\mathbf{J}=0$, its perpendicular and parallel components each vanish. Since $\alpha \ne 0$ and $\alpha +\gamma \ne 0$, this forces both field components to be zero. The contrapositive givesEquivalently, the tensor has nonzero eigenvalues $\alpha ,\alpha ,\alpha +\gamma$ and determinant ${\alpha }^2(\alpha +\gamma )\ne 0$.

Writing the uniaxial conductivity tensor in vector form gives the electric current densityChoose the plane through the origin perpendicular to $n$:For an electric field in this plane, the anisotropic term vanishes and $\mathbf{J}=\alpha \mathbf{E}$. Thus every nonzero such field is parallel or antiparallel to its current according to the sign of the nonzero $\alpha$. The zero field gives zero current. This establishes collinearity throughout the plane without assuming that the tensor is isotropic.

Under the moving-coordinate transformation, the forcing is $H(y)$. Choose a stationary solution $U(s,y)=w(y)$, which requires precisely $w^{\prime \prime }+H=0$. The left spatial limit of $Ay+B$ is zero only if $A=B=0$. Therefore one solution is the transported step forcing in an advection-diffusion equation profileIt has $u_t=(x-t{)}_{+}$ and $u_x=-(x-t{)}_{+}$, so their sum vanishes; almost everywhere $u_{xx}=-H(x-t)$, canceling the source. The first derivatives are continuous on the moving interface. For each fixed $t$, the profile is zero far to the left and tends to minus infinity to the right, as required. The PDE holds pointwise off that interface and weakly across it, consistently with the requested continuously differentiable regularity.

For $y<0$, the Heaviside function vanishes, so $w=Ay+B$. For $y>0$, $w^{\prime \prime }=-1$, giving $w=-y^2/2+Cy+D$. Continuity of $w$ and $w^{\prime }$ at zero requires $D=B$ and $C=A$. Consequently the general continuously differentiable profile isIts second derivative jumps from zero to minus one. The equation is satisfied classically away from zero and in the weak, or distributional, sense across it; no globally twice-differentiable function can have the prescribed step as its second derivative. There is no Dirac delta contribution because $w^{\prime }$ is continuous. The value assigned to the Heaviside function at a single point does not change this weak solution.

Apply the chain rule to $U(s,y)=u(s,y+s)$. Holding $y$ fixed while differentiating in $s$ gives $U_s=u_t+u_x$, while $U_y=u_x$ and $U_{yy}=u_{xx}$. Thus the advection-diffusion equation becomesThe moving coordinate follows the unit-speed advection, leaving the heat equation with a transformed source.

For the given Möbius transformation, the fixed point equation is $z(z+3)=3z+1$, hence

For the general Möbius transformation, its finite fixed points satisfyWith $c\ne 0$ and nonzero discriminant, the quadratic formula gives the two distinct fixed points of a Möbius transformation:The fixed-point equation for $\alpha$ also gives $b-d\alpha =-\alpha (a-c\alpha )$. ThereforeDividing yields the conjugation of a Möbius transformation with two distinct fixed points:Neither factor in this quotient vanishes, since $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The formula extends over the Riemann sphere, including the transformation's pole. For the special transformation and the ordering $\alpha =1,\beta =-1$, $k=1/2$.

For every integer $m$, De Moivre's theorem states $(\cos \theta +i\sin \theta {)}^m=\cos (m\theta )+i\sin (m\theta )$. Taking the real part for $m=5$ givesReplace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ and expand:The resulting polynomial is the fifth Chebyshev polynomial.

A trace-zero Hermitian matrix has the unique formSet $x_3=h$, $x_1=\mathrm{Re}w$ and $x_2=-\mathrm{Im}w$. Then it is exactly $R(x)$; these assignments also prove uniqueness. This is the real Pauli matrix representation $R(x)=x_1{\sigma }_1+x_2{\sigma }_2+x_3{\sigma }_3$.

For a unitary matrix, $U^{-1}=U^{†}$. Thusand invariance of the matrix trace under similarity transformations gives $\mathrm{tr}(U^{-1}AU)=0$. Therefore conjugation maps the space to itself.

Finally,The determinant is invariant under similarity transformations, so $\det R(y)=\det R(x)$ implies

Let $[v{]}_{\times }$ denote the cross product matrix, with $[v{]}_{\times }x=v\times x$. ThenThe two eigenvalue calculations use an orthonormal basis adapted to $u,v$.

Differentiating givesThereforeFor real coordinates take $q>0$ and a local branch of the angle. The transformation is locally canonical transformation because its determinant is one, but periodicity in $p$ prevents a globally one-to-one map if all real $p$ are retained.

The Poisson bracket is $\{Q,P\}=\alpha q^{\beta -2}$. For it to be identically one on an open region, the exponent must vanish and the coefficient must equal one:The transformation is defined away from $q=0$, and its inverse there is $q=1/P$, $p=Q{P}^2$.

Unmodified Hebbian learning reinforces correlated input and output but has no mechanism limiting synaptic strength. The second rule adds activity-dependent negative feedback and stabilizes the norm at an input-scale-independent value, provided the input keeps exciting the output. The third has a cubic weight penalty independent of the instantaneous output; its equilibrium norm instead depends on the strength of the input second moment.

For the third averaged rule, an equilibrium along eigenvector $i$ has radial linearized eigenvalue $-2{\lambda }_i/\tau$ and transverse eigenvalues $({\lambda }_j-{\lambda }_i)/\tau$. Thus a positive simple largest eigenvalue gives stable equilibria in its two opposite directions, whereas lower eigen-directions are unstable to larger components. With generic initial weights, it learns the dominant correlation direction, linking the rule to principal component analysis. These are idealized normalization and feature-extraction mechanisms, not claims that biological synapses implement these exact equations.