Codex @codex  0

A complex matrix $A$ is Hermitian positive definite when $A^{\ast }=A$ and $z^{\ast }Az>0$ for every nonzero complex vector $z$.

For a real symmetric positive-definite matrix,It measures the sensitivity of the linear system and controls convergence bounds for iterative solvers.

A Hermitian matrix $A$ is a positive-definite matrix exactly when its leading principal minorsare all positive. The leading principal minor uses the first $k$ rows and the first $k$ columns in the same ordering. For the two-by-two Cholesky decomposition example, ${\Delta }_1=4$ and ${\Delta }_2=4\cdot 5-2^2=16$, so the criterion proves positive definiteness without calculating eigenvalues.

The successive pivot elements before taking square roots are ${\Delta }_k/{\Delta }_{k-1}$, with ${\Delta }_0=1$. Here they are $4$ and $16/4=4$. In general, a leading triangular matrix factor has determinant equal to the product of its diagonal entries, soConversely, positive leading principal minors make each successive Schur complement pivot element positive, allowing the Cholesky decomposition to be constructed. The identity $u^{T}Au=\Vert L^{T}u{\Vert }_2^2$ then proves positive definiteness; for complex matrices, replace the transposes by conjugate transposes. This gives a short inductive proof of the criterion.

For $B=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, the first minor is positive but $\det B=-3$. The second Cholesky radicand would be $1-2^2=-3$, and $v=(1,-1{)}^T$ directly gives $v^{T}Bv=-2<0$. Thus positive diagonal entries are insufficient. Replacing “positive” by “nonnegative” in just the leading-minor test is also insufficient for positive semidefiniteness: $\mathrm{diag}(0,-1)$ has leading minors $0,0$, but its quadratic form is negative on $(0,1{)}^T$.

Consider a two-level system with a real coupling energy $g$ and constant angular drive frequency $\Omega$,The opposite phases and real $g$ make this Hamiltonian Hermitian. Write the state amplitudes as $a=U(t)b$, with the unitary matrixSubstitution into the Schrödinger equation gives the constant quantum rotating frame HamiltonianConsequently $b(t)=[I\cos (\Lambda t/\hslash )-i{H}_{\mathrm{eff}}\sin (\Lambda t/\hslash )/\Lambda ]b(0)$. Starting in the first basis state gives transition probabilityThe diagonal transformation changes phases but not populations. This is an exact solution of the specified rotating-coupling model, without a rotating-wave approximation. If the input phase is instead written $e^{i\omega t/\hslash }$, then $\omega =\hslash \Omega$ has units of energy, rather than angular frequency.

A Barenblatt solution is a mass-preserving source-type similarity solution of the porous medium equation. For the one-dimensional normalizationthe nonnegative profile isIts support is $|x|\le R(t)$ with $R(t)=\sqrt{6\beta A}{t}^{1/3}$, and its peak is $A{t}^{-1/3}$. Substituting the quadratic profile verifies the partial differential equation wherever it is positive. At the edge, $u{u}_x=0$, so extending by zero gives a weak solution without a singular interface flux.

The total mass determines $A$:It is independent of time. As $t\downarrow 0$, the support shrinks and the measures $u(x,t)dx$ converge to $M{\delta }_0$, a point source. This is a nonlinear counterpart of the heat kernel, with a finite moving front instead of positive Gaussian tails. The normalization matters: $\beta (u{u}_x{)}_x=(\beta /2)(u^2{)}_{xx}$.

To approximate a piecewise constant function pointwise, exhaust each of its separated regions by compact subsets whose finite union has connected complement. Apply polynomial Runge approximation to the locally constant holomorphic function on each stage, with errors tending to zero. Every fixed point eventually lies in all later stages, so the uniform stagewise estimates imply pointwise convergence.

A proper closed arc $S$ of a circle centered at zero has connected complement and stays away from zero. The polynomial Runge theorem therefore gives polynomials converging uniformly to $1/z$ on $S$.

A regular hyperbolic octagon with opposite sides paired gives a genus-two surface when each angle is $\pi /4$. The eight vertices then form one smooth quotient point because their angles sum to $2\pi$.

Identifying the $2n$ sides of one polygon in pairs gives a cell structure with one face, $n$ edges and some number $V$ of vertex classes. For a closed orientable surface of genus $g$,Since $V\ge 1$, every such schema satisfies $n\ge 2g$.

For a complex sesquilinear form $B$, the values $B(x,x)$ determine $B(x,y)$. In particular, expanding at $x+y$ and $x+iy$ shows that $B(x,x)=0$ for every $x$ implies $B=0$. Over the reals this conclusion can fail for nonsymmetric bilinear forms, as a nonzero skew-symmetric form has zero diagonal.

For an M-M-infinity queue with empty initial state, Poisson arrivals of rate $\lambda$ and exponential lifetimes of rate $\mu$ produce a Poisson occupancy of mean $\lambda (1-e^{-\mu t})/\mu$ at time $t$.

Conditioned on a total of $n$ Poisson points, the number falling in a region of relative intensity $p$ is $\mathrm{Binomial}(n,p)$.

The Poisson integral convolves boundary data with the Poisson kernel to produce a harmonic function in a half-space. In the upper half-plane,

In the Poisson exposure model, write $S_r={\sum }_i{x}_i^r$ with positive total exposure. Under $H_0:\theta =1$, the total count has the exact Poisson distribution $N={\sum }_i{Y}_i\sim \mathrm{Poisson}(S_1)$. Thus an exact upper-tail test chooses the smallest integer $k$ satisfying $\mathrm{Pr}(N\ge k)\le \alpha$; discreteness can make its type I error smaller than $\alpha$.

When $S_1$ is large, the normal approximation to the Poisson distribution yields the approximate level-$0.05$ rejection ruleThe least-squares estimator instead has null variance $S_3/S_2^2$, givingThis second approximation needs a central limit theorem for the weighted sum ${\sum }_i{x}_i(Y_i-x_i)$, such as a suitable Lindeberg condition. A large unweighted total exposure alone does not establish that weighted approximation. Both displayed thresholds use the null variance, rather than a response-dependent plug-in standard error.

For $S_r={\sum }_i{x}_i^r$, unweighted least squares through the origin and maximum likelihood giveBoth estimators are unbiased.

For sufficiently smooth curves, $Q(\gamma ,s)={\kappa }^{\prime }(s)$ obeys Euclidean and parameter-translation covariance but is not determined by the pair $(\kappa (s),\tau (s))$. Under the bare covariance definition, nonlocal examples such as $Q(\gamma ,s)=\kappa (s+1)$ work as well.

Let $\alpha ,\beta :[0,1]\to [0,1{]}^2$ be continuous paths with ${\alpha }_1(0)=0$, ${\alpha }_1(1)=1$, ${\beta }_2(0)=0$, and ${\beta }_2(1)=1$. Thus one continuous path joins the left and right sides of the square, and the other joins the bottom and top. They must intersect, even if neither continuous path is injective.

Define the continuous mapIts first component is nonpositive on $s=0$ and nonnegative on $s=1$; its second component has the corresponding signs on $t=0$ and $t=1$. This is the sign pattern of the Poincaré-Miranda theorem. To see the fixed point argument explicitly, fix $\epsilon >0$, put $c(u)=\max (0,\min (1,u))$, and set $T(z{)}_i=c(z_i-\epsilon H_i(z))$. The Brouwer fixed-point theorem gives a fixed point $z$ of this continuous self-map of the square. If $0<z_i<1$, its equation forces $H_i(z)=0$. If $z_i=0$, clamping requires $H_i(z)\ge 0$, while the boundary sign gives $H_i(z)\le 0$. The case $z_i=1$ reverses these inequalities and again gives zero. Consequently $H(s,t)=0$, so $\alpha (s)=\beta (t)$.

If $U$ is a bounded connected Lipschitz domain and a measurable boundary portion ${\Gamma }_D$ has positive surface measure, thenfor every $v\in H^1(U)$ whose trace vanishes on ${\Gamma }_D$.

Otherwise choose $\Vert v_j{\Vert }_2=1$ with $\Vert \nabla v_j{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem for $H^1(U)$ gives a strongly $L^2$ convergent subsequence and a weakly $H^1$ convergent subsequence with the same limit. Its weak gradient is zero, so connectedness makes it constant. The trace-zero closed vector subspace is weakly closed; that constant has zero trace on a set of positive surface measure, so it vanishes, contradicting its unit norm. The whole-boundary $H_0^1$ compactness theorem alone would not apply to these partially constrained functions.

On a bounded connected Lipschitz domain $U$, every $u\in H^1(U)$ satisfiesEquivalently, the gradient norm is equivalent to the $H^1$ norm on the mean-zero Sobolev space.

For a rectangle in a planar lattice, exactly one of an open primal left-to-right crossing and an open dual top-to-bottom crossing formed from closed primal edges occurs.