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A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently its only subsets that are both open and closed are the empty set and the whole space. For a subspace topology, open means relatively open within that subspace. A path-connected space is connected, since a separation would pull back along a path to a separation of an interval.

The determinant isEquivalently it is the alternating multilinear function of the columns with value one on the identity matrix. To prove multiplicativity, write column $j$ of $AB$ as ${\sum }_k{b}_{kj}{A}_k$, where $A_k$ is column $k$ of $A$. Multilinear expansion gives a sum over choices $k_1,\dots ,k_n$; terms with repeated choices vanish by alternation. The remaining permutations contributeIf $C$ is invertible, $CB=C(BC)C^{-1}$, so multiplicativity makes $\det (CB{C}^{-1}-\lambda I)=\det (BC-\lambda I)$ and $p_{CB}=p_{BC}$. For arbitrary $A$, put $C=A+tI$. Its determinant is a nonzero polynomial in $t$, so it is invertible except for finitely many values. Thus for all other $t$,For each fixed $\lambda$ both sides are polynomials in $t$, and equality at infinitely many $t$ makes them identical. Setting $t=0$ proves the Characteristic polynomials of AB and BA identity $\boxed{p_{BA}=p_{AB}}$ without an invertibility assumption on $A$.

In the convention $p_A(\lambda )=\det (A-\lambda I)$, the coefficient of ${\lambda }^{n-1}$ is $(-1{)}^{n-1}\mathrm{tr}A$: in the determinant expansion such a term chooses one diagonal entry of $A$ and $n-1$ factors $-\lambda$; nonidentity permutations have at least two nonfixed positions and contribute lower degree. Therefore $\boxed{p_A=p_B⟹\mathrm{tr}A=\mathrm{tr}B}$.

For $n\ge 1$, an eigenvalue is a number $\lambda \in \mathbb{C}$ for which there is a nonzero vector $v$ satisfying $Av=\lambda v$. Equivalently, $\ker (A-\lambda I)\ne \{0\}$, or $\det (A-\lambda I)=0$. The characteristic polynomial $\det (tI-A)$ has degree $n$ and leading coefficient one. By the fundamental theorem of algebra it has a complex root $\lambda$. At that root $A-\lambda I$ is a singular matrix, so it has a nonzero kernel vector, proving that $A$ has an eigenvalue.

Distinguish the full velocity potentials ${\Phi }_i=U_{i}x+{\phi }_i$ from their perturbations ${\phi }_i$. The linear boundary conditions use the perturbations. This distinction reconciles the notation in the printed conditions with the full potentials written in the mode ansatz. All equations below neglect products of perturbations. Assume positive densities and no surface tension.

A topology on $X$ is a family $\mathcal{T}$ of subsets containing $\varnothing$ and $X$, closed under arbitrary unions and finite intersections. Its members are the open sets. A Hausdorff space separates each pair of distinct points by disjoint open neighbourhoods; a compact space is one for which every open cover has a finite subcover.

Let $\Omega$ be the Earth's angular velocity magnitude. Use the local east, north and upward unit vectors ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. At latitude $\lambda$, the Earth's angular velocity and the train's ground-relative velocity areThus the east component of the Coriolis force isFor a north–south track, the train has no eastward acceleration relative to Earth. Its path curvature lies in the local north–vertical plane. The centrifugal acceleration and Newtonian gravity also lie in that plane, and the Earth's constant angular velocity gives no Euler acceleration. Hence the sideways force from the track must cancel the eastward Coriolis force:Its magnitude is $2m\Omega V\sin \lambda$, and its direction is west in the Northern hemisphere. The eastward tendency is the apparent Coriolis force; the requested track force has the opposite direction.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

Write a Möbius transformation as $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, interpreted on the Riemann sphere. Let $T_u(z)=z+u$, $D_{\lambda }(z)=\lambda z$ with $\lambda \ne 0$, and $I(z)=1/z$.

If $c=0$, then $a,d\ne 0$ and $M(z)=(a/d)z+b/d=T_{b/d}\circ D_{a/d}(z)$. If $c\ne 0$, algebraic division givesConsequentlyThe scaling coefficient is nonzero because $ad-bc\ne 0$. Thus translations, nonzero complex scalings and reciprocal inversion generate every Möbius map. The formulas extend over their poles and infinity by the sphere interpretation.

For a normal subgroup $H⊴G$, define multiplication of left cosets byTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product is $g{h}_{1}k{h}_2=gk(k^{-1}{h}_{1}k)h_2$, and normality puts $(k^{-1}{h}_{1}k)h_2$ in $H$. Thus the product coset is independent of the representatives. Associativity follows from multiplication in $G$, the identity is $H$, and the inverse of $gH$ is $g^{-1}H$. These operations form the quotient group $G/H$.

For the unheaded finite-index conclusion, let $G$ act on the $n$ left cosets of its proper subgroup $H$ by $g\cdot (xH)=gxH$. This is well-defined even if $H$ is not normal: replacing $x$ by $xh$ leaves $gxH$ unchanged. The maps are bijections with inverses supplied by $g^{-1}$, and their compositions give a group homomorphism $\phi :G\to S_n$. Its kernel of a group homomorphism $N$ is normal, by the calculation in part (ii).

The action is not trivial, because some $g\notin H$ sends the coset $H$ to $gH\ne H$; hence $N\ne G$. If $N$ were trivial, $\phi$ would be injective, implying $|G|\le |S_n|=n!$. Under the given strict inequality this is impossible. Thus $N$ is a nontrivial proper normal subgroup, andMore precisely, $N={\bigcap }_{x\in G}xH{x}^{-1}$ is the subgroup core: $g$ fixes every coset if and only if $x^{-1}gx\in H$ for every $x$. This identifies the normal subgroup produced by the coset action.

Use the right-handed rotation matrix which sends the positive $x$ direction to the positive $y$ direction:The Cartesian second-rank tensor transformation law givesHere isotropic tensor means invariant under every proper rotation, as is needed for the rank-three assertion involving the Levi-Civita symbol. A passive coordinate rotation through the same angle uses $R^{\mathsf{T}}$ instead; the invariance conclusions are identical.

Setting $S^{\prime }=S$ forces $S_{11}=S_{22}$, $S_{12}=-S_{21}$, and $S_{13}=S_{23}=S_{31}=S_{32}=0$. No symmetry assumption on $S$ has been made: it now has the formInvariance under a right-handed quarter-turn about the $x$ axis next gives $S_{22}=S_{33}$ and $S_{12}=-S_{13}$, hence $a=c$ and $b=0$. Conversely $R(\lambda I)R^{\mathsf{T}}=\lambda I$ for every orthogonal matrix. The general isotropic second-rank tensor is therefore $S_{ij}=\lambda {\delta }_{ij}$.

An isotropic vector must be fixed by the half-turn about the $z$ axis, so its first two components vanish. The half-turn about the $x$ axis then forces its third component to vanish. Thus the only isotropic vector is zero. The general isotropic third-rank tensor under proper rotations is $A_{ijk}=\lambda {ϵ}_{ijk}$. If invariance under the whole orthogonal group were required of an ordinary rank-three tensor, inversion $R=-I$ would send $A$ to $-A$, forcing $A=0$ instead.

Measure temperature relative to ambient by ${\theta }_i=T_i-T_{\infty }$. Newton's law of cooling models ordinary cooling by ${\theta }_i^{\prime }=-a{\theta }_i$, since heat loss is proportional to the excess temperature. The additional negative forcing represents the cooling from the milk, with the temperature drop or rate normalized to one in the chosen units.

The Dirac delta function has unit integral and is supported at the instant of addition. It therefore imposes a sudden unit drop for cup 1. The difference $H(t-1)-H(t-2)$ of Heaviside step functions is one during the addition interval and zero outside it, giving a unit cooling rate for cup 2. The second temperature is continuous; only its derivative changes at the switching times. The value assigned to a Heaviside step function exactly at its jump does not change the integrated solution. This is the impulse versus finite-duration cooling comparison.

Taking the conjugate transpose of $U=A+iB$ and using that both parts are Hermitian matrices gives $U^{†}=A-iB$. Addition and subtraction recover the Hermitian parts of a unitary matrix:These depend only on $U$, so any such decomposition has the same $A,B$ and is unique. Conversely the displayed expressions are Hermitian matrices and sum to $U$ in the form $A+iB$, establishing existence as well as uniqueness. Their commutation and quadratic relation are proved in the two labelled parts.

First establish the interior convergence lemma for a power series. If $\sum a_n{w}^n$ converges at some $w\ne 0$, its terms are bounded: $|a_n{w}^n|\le M$ for a finite $M$. For every $|z|<|w|$,Comparison with a convergent geometric series proves absolute convergence at $z$.

Now letThe set contains zero, since at $w=0$ only the constant term remains. If $|z|<R$, the definition of supremum gives a point of convergence $w$ with $|w|>|z|$. The lemma proves absolute convergence at $z$. If $|z|>R$, convergence would place $|z|$ in the set defining $R$, a contradiction. This proves the existence of a radius of convergence, including $R=0$ and $R=\infty$:The proof leaves behavior on $|z|=R$ open; it need not be uniform around that circle. The two examples illustrate opposite possibilities.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.