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If $\dim V=n$, then $\dim \mathrm{End}(V)=n^2$. The $n^2+1$ endomorphisms $I,T,\dots ,T^{n^2}$ are therefore linearly dependent, so some nonzero polynomial $p$ satisfies $p(T)=0$. The minimal polynomial is the monic polynomial of least degree annihilating $T$. Uniqueness follows because the difference of two such monic polynomials would be a nonzero annihilator of smaller degree.

A Jordan block $J_r(\lambda )$ has minimal polynomial $(x-\lambda {)}^r$, and the minimal polynomial of a direct sum is the least common multiple of those of its blocks. Thus an eight-dimensional example is

A sufficient statistic is a function of the sample whose conditional sample distribution, given that statistic, does not depend on the parameter. The factorization criterion for sufficiency states, for a dominated family, that $T$ is sufficient exactly when the joint density has form $g_{\theta }(T(x))h(x)$, where $h$ is independent of $\theta$.

Let $m=X_{(1)}={\min }_i{X}_i$ and $M=X_{(n)}={\max }_i{X}_i$. The joint uniform density isso $\boxed{T=(m,M)}$ is sufficient for $(a,b)$ by the factorization criterion for sufficiency. The uniform order statistics have expectationsFor example, the standardized minimum has survival probability $(1-t{)}^n$ on $[0,1]$, whose integral is $1/(n+1)$; the maximum follows by reflection. Solving these expectation equations gives the unbiased estimatorThe assumption $n\ge 2$ makes the denominator positive.

Choose the photon direction in the $x^{\prime }$–$y^{\prime }$ plane. Its four-momentum in the stellar rest frame isLet $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. The Lorentz transformation to $S$ givesUsing $p^0=h\nu /c$ and $p^x=(h\nu /c)\cos \theta$ yields the relativistic Doppler effect and relativistic aberration:

For the oppositely directed transverse photons, their rest-frame spatial momenta cancel and their total energy is $E=2h{\nu }^{\prime }$. Boosting this combined four-vector gives laboratory energy $\gamma E$ and momentumThe star loses rest mass $E/c^2$. At unchanged speed its momentum therefore decreases by $\gamma (E/c^2)v$, exactly the photon momentum. Its laboratory energy loss is likewise $\gamma E$. There is no rest-frame recoil from the symmetric emission, and constant speed is consistent with four-momentum conservation because the stellar mass decreases.

Away from $x=\xi$, the Dirichlet Green function solves the homogeneous equation. The endpoint conditions select $A{y}_1(x)$ on the left and $B{y}_2(x)$ on the right. Continuity and the unit derivative jump needed for a Dirac delta function giveWith Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2\ne 0$, solve to obtain $A=y_2(\xi )/W(\xi )$ and $B=y_1(\xi )/W(\xi )$. ThusThe first derivative term creates no extra delta because $G$ is continuous. The value at $x=\xi$ is the common limiting value.

For the specified constant-coefficient operator, takeTherefore its explicit Green function isFor the forcing step and the homogeneous endpoint conditions inherited from this Green-function problem, $y(x)=2{\int }_{x_0}^{1}G(x,\xi )d\xi$. When $x<x_0$, every integration point lies on the first branch, soWithout those endpoint conditions, an arbitrary homogeneous solution could also be added.

Use the Fourier transform convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{-ikx}dx$ and inverse factor $1/(2\pi )$. Laplace equation gives ${\hat{u}}_{yy}-k^2\hat{u}=0$. Selecting its decaying Fourier branch gives $\hat{u}(k,y)={\hat{u}}_0(k)e^{-|k|y}$, the standard bounded solution for bounded continuous boundary data. The inverse transform of the multiplier is the Poisson kernel for the upper half-plane:The convolution theorem therefore gives the Poisson integralFor suitably regular boundary data, this kernel is an approximate identity as $y\downarrow 0$, recovering $u_0$.

For $a>0$, contour integration gives ${\hat{u}}_0(k)=(\pi /a)e^{-a|k|}$. Multiplication by $e^{-y|k|}$ and inversion yieldThe stated form assumes $a>0$; if $a<0$, replace it by $|a|$ because the boundary function only depends on $a^2$. The case $a=0$ has singular boundary data and is outside this regular calculation.

The transform argument requires a growth condition. A zero-boundary harmonic function with pointwise vertical decay shows that the printed conditions alone do not ensure uniqueness:vanishes at $y=0$ and tends to zero as $y\to \infty$ for every fixed $x$. It can be added without changing those conditions, but its growth in $x$ excludes it from the bounded solution class. The displayed Poisson integral is the standard bounded solution for regular bounded boundary data; pointwise vertical decay by itself cannot justify a uniqueness claim.

If an open set meets $\overline{A}$, it meets $A$. Thus if two open sets meet the closure of an irreducible set, they both meet $A$, their intersection meets $A$ by irreducibility, and therefore it also meets $\overline{A}$. This proves irreducibility of the closure. A singleton is irreducible, so its closure is an irreducible closed subset.

In a Hausdorff space, two distinct points have disjoint open neighborhoods. An irreducible set cannot contain both, so every nonempty irreducible closed subset is a singleton, which is closed and has a unique generating point. Hence every Hausdorff space is a sober topological space. An infinite set with the cofinite topology is itself irreducible, because any two nonempty open sets meet, but every singleton is closed. Its whole space is therefore not the closure of a singleton, proving that it is not sober.

For the proposed sobrification, let $\hat{X}$ consist of nonempty irreducible closed subsets. The basic operations areThe last identity uses irreducibility of each member of $\hat{X}$. Hence these sets already form a topology. The map $U\mapsto \hat{U}$ is injective: $\overline{\{x\}}$ meets an open $U$ exactly when $x\in U$, so equality of the hats forces equality of the original opens. It is surjective onto the topology just constructed by definition.

Now let $C$ be a nonempty irreducible closed subset of $\hat{X}$. Its complement is $\hat{W}$ for a unique open $W\subseteq X$. Put $F=X\setminus W$. ThenIf opens $U,V$ meet $F$, choose points in those intersections. Their singleton closures lie in $C$, so $\hat{U},\hat{V}$ meet $C$. Irreducibility of $C$ forces $\widehat{U\cap V}$ to meet $C$, and hence $U\cap V$ meets $F$. Thus $F$ is nonempty, closed and irreducible, so $F$ is itself a point of $\hat{X}$.

The closure of this point in $\hat{X}$ is exactly $\{A:A\subseteq F\}$: inclusion makes every open meeting $A$ meet $F$, whereas a point of $A\setminus F$ is separated by the open $X\setminus F$. Hence $C=\overline{\{F\}}$. Any other point $G$ with the same closure satisfies $G\subseteq F$ and $F\subseteq G$, forcing $F=G$. This proves the constructed space is sober.

Write $L(f)={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)$ and $B(f)={\sup }_z|f(z)|$. The quantity in the question is the bounded Lipschitz norm $\Vert f{\Vert }_{\mathrm{BL}}=L(f)+B(f)$. Both terms are nonnegative and absolutely homogeneous, and each satisfies the triangle inequality. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a vector space. If the sum is zero, $B(f)=0$ forces $f=0$. Thus it is a norm.

For the sequence bounded by one, both $L(f_i)\le 1$ and $B(f_i)\le 1$. Given real $x$, choose a nearby rational $q$. ThenFirst make $|x-q|$ small, then use convergence at $q$ to make the last term small. The values at every real $x$ form a Cauchy sequence, so $f_i$ converges pointwise to a real function $f$.

To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple $x\ne y,z$:Pass to the pointwise limit and take the independent suprema over $(x,y)$ and $z$. This gives $\boxed{\Vert f{\Vert }_{\mathrm{BL}}\le 1}$, the pointwise closure of the bounded Lipschitz unit ball.

To obtain diagonal compactness for bounded Lipschitz functions, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in $[-1,1]$, so the Bolzano-Weierstrass theorem supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The diagonal subsequence argument gives one subsequence converging at every rational. Applying the preceding argument proves there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one.

Parametrize the graph by $X(u,v)=(u,v,f(u,v))$. The first fundamental form has coefficients $E=1+f_u^2$, $F=f_u{f}_v$, $G=1+f_v^2$. With unit normal $(-f_u,-f_v,1)/\sqrt{1+|\nabla f{|}^2}$, the second fundamental form has coefficients $e=f_{uu}/\sqrt{1+|\nabla f{|}^2}$, $f_{\mathrm{II}}=f_{uv}/\sqrt{1+|\nabla f{|}^2}$, and $g=f_{vv}/\sqrt{1+|\nabla f{|}^2}$. Consequently the Gaussian curvature of a graph surface is

A compact embedded surface has a highest point in the vertical direction. At that point its tangent plane is horizontal, so it is locally a graph with a local maximum of $f$. Its Hessian matrix is negative semidefinite, giving $\det D^{2}f\ge 0$ and hence $K\ge 0$. Thus the Gaussian curvature cannot be everywhere negative.

An embedded ring torus, with major radius $R$ larger than minor radius $r>0$, is a compact surface without boundary. Its standard parametrization is $((R+r\cos \theta )\cos \varphi ,(R+r\cos \theta )\sin \varphi ,r\sin \theta )$, and the Gaussian curvature of a ring torus isSince the denominator is positive, curvature is positive on the outer side and negative on the inner side, so this example changes sign.

Prove the submodule theorem for free modules over a principal ideal domain by induction on the rank of $R^n$. For a submodule $N$, project onto its first coordinate. The image is an ideal $aR$. If $a=0$, then $N$ is a submodule of $R^{n-1}$ and induction applies. Otherwise choose $v\in N$ projecting to $a$. Every $w\in N$ has first coordinate $ra$ and hence $w-rv$ lies in the projection kernel. The domain property shows $Rv$ meets that kernel only at zero. Thusand the kernel is a free module by induction. This proves that $N$ is free, of rank at most $n$.

To prove free modules are projective, choose a basis $(e_{\lambda })$ of a free module $P$. For the given surjection $f$, choose $m_{\lambda }\in M$ with $f(m_{\lambda })=g(e_{\lambda })$ and extend $e_{\lambda }\mapsto m_{\lambda }$ linearly. Finite support of basis expansions makes the resulting module homomorphism well defined, and $f\circ h=g$.

Finally, take a finite generating set of a projective module $P$ to obtain a surjection $\pi :R^n\to P$. Projectivity applied to ${\mathrm{id}}_P$ gives a section $s:P\to R^n$ with $\pi s={\mathrm{id}}_P$. Thus $s$ is injective and identifies $P$ with a submodule of $R^n$. The proved submodule theorem for free modules over a principal ideal domain shows that every finitely generated projective module over a principal ideal domain is free.

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.

Write $u_k$ for the quantity denoted ${\xi }_k$ in the question, to distinguish it from the random-walk increments. Absorption at zero at time zero gives $u_0=1$, and starting at $N$ gives $u_N=0$. For $1\le k\le N-1$, condition on the first step and use independence of the remaining increments. That step contributes one factor of $s$, so the discounted gambler's ruin hitting probability satisfiesThe absorption time is finite almost surely: in each block of $N$ steps there is a probability bounded below by a positive constant of reaching a boundary by successive steps in one direction.

Trying $u_k=r^k$ gives $ps{r}^2-r+qs=0$, with rootsThey are positive and distinct because $4pq\le 1$ and $0<s<1$. Hence the general solution is $A{r}_{+}^k+B{r}_{-}^k$. Enforcing both boundary values givesThis includes the unbiased walk without a special limiting case, since the discount keeps the roots distinct. For uniqueness, a difference of two solutions has zero boundary values and its largest absolute value $M$ satisfies $M\le sM$ if attained in the interior. Since $s<1$, $M=0$.

Treat the four fixed row totals as four independent samples from binomial distributions. Under independence, each has the same unknown probability $p$ of a third-child boy. The pooled maximum-likelihood estimator is $\hat{p}=100/200=1/2$, so each expected cell count is $25$. The Pearson chi-squared test of homogeneity statistic isThere are four independently varying row probabilities under the alternative, and one estimated common probability under the null; the degrees of freedom are therefore $4-1=3$. Equivalently this is the $(4-1)(2-1)$ contingency-table count. The fact that the pooled estimate happens to equal one half does not make it a specified null probability and does not change the degrees of freedom to four.

Since $10.08>7.8147$, reject independence at the 5% level using the stated chi-squared calibration. All expected counts are sufficiently large for the usual asymptotic test.

Hermiticity of the Hamiltonian operator gives $⟨{\psi }_m,H{\psi }_n⟩=⟨H{\psi }_m,{\psi }_n⟩$, soThe distinct eigenvalues force orthogonality for $m\ne n$, and the given normalization gives one for $m=n$. Therefore $⟨{\psi }_m,{\psi }_n⟩={\delta }_{mn}$.

Under the time-independent Schrödinger equation, each energy component acquires its own phase:The sum converges in the state Hilbert space because the squared coefficients are summable. Orthogonality makes its squared norm ${\sum }_{n\ge 0}{2}^{-(n+1)}=1$ at every time.

The Born rule for energy measurement givesBoth depend only on squared coefficient magnitudes, so both are time-independent. The expectation follows either by summing energy times measurement probability or by inserting the orthogonal expansion in the quadratic form of $H$.

Normalization alone does not guarantee a finite mean for an arbitrary spectrum. The displayed energy series is finite when $\sum |E_n|2^{-(n+1)}<\infty$. Since the spectrum is bounded below by $E_0$, a divergent positive tail gives an extended expectation $+\infty$, still independent of time; for example an abstract diagonal Hamiltonian with $E_n=4^n$ has this property. No growth assumption on $E_n$ is printed, so finiteness should not be asserted without this qualification.

With $\Re \alpha <0$, the causal exponential is integrable and its Fourier transform isFor the pendulum system, a transfer function is the Fourier transform of its causal impulse response, equivalently the multiplier relating the transformed forcing to the transformed zero-initial-data response. Transforming derivatives givesPartial fractions therefore giveThe exponential transform identifies the causal response as $R(t)=\frac{1}{2}{e}^{-t}\sin 2t$ for $t>0$ and zero for $t<0$. It solves the homogeneous equation for $t>0$, with $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, the jump required by a unit impulse. The forcing is understood as switched on at zero; its transform is $1/(i\omega +4)$, and convolution with $R$ gives, if the time response is wanted,This has both stated initial values equal to zero.