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Let $r=\dim \mathrm{im}T$. Since the kernel of a linear map and its image agree, the rank-nullity theorem gives $\dim V=r+r$. Thus $\dim V=2r$ is even. Every $Tv$ lies in $\ker T$, so $T^2=0$. The minimal polynomial therefore divides $X^2$. It cannot be $1$, and it cannot be $X$ since $T\ne 0$; hence the minimal polynomial is $X^2$.

Use the universe $\Omega =\{B\subseteq A:|B|=k\}$ and define the exact-occupancy events $E_i=\{B\in \Omega :|B\cap A_i|=3\}$. The desired collection is ${\bigcup }_{i=1}^k{E}_i$, so apply inclusion-exclusion to these events.

Fix $r$ distinct block indices. For $B$ to belong to the intersection of their events, it must contain exactly three elements in each selected block. There are ${\left(\genfrac{}{}{0ex}{}{m}{3}\right)}^r$ ways to make those choices. None of the other elements in these selected blocks may be included; the remaining $k-3r$ elements must come from the $n-rm$ elements outside the selected blocks. ThusThere are $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$ choices of block indices, and the intersection is empty when $3r>k$. Substitution into inclusion-exclusion givesHere binomial coefficients with a lower argument exceeding a nonnegative upper argument are zero. In particular, when $m<3$ every event is empty, and when $k<3$ the sum is empty; both give the correct answer zero. The subtraction of entire selected blocks, not merely the $3r$ chosen elements, is what enforces exact rather than at-least-three occupancy. This is inclusion-exclusion for exact block occupancy with the number of blocks and subset size both equal to $k$.

False. On the additive topological group $\mathbb{R}$, put $d(x,y)=|x-y|$ and $d^{\prime }(x,y)=\sqrt{|x-y|}$. Both are invariant under left and right translations and induce the usual topology. The triangle inequality for $d^{\prime }$ follows from $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$ for $a,b\ge 0$. Butcan be arbitrarily close to zero and arbitrarily large. Consequently no $K>0$ can satisfy both proposed inequalities. Compatibility of the two invariant metrics is topological equivalence, which need not be bilipschitz equivalence.

True. Given a compatible left-invariant group metric $d$, definePullback by the bijection $x\mapsto x^{-1}$ preserves all metric axioms, and because inversion is a homeomorphism, $d_R$ induces the same topology. For right translation by $a$, left invariance givesThus $d_R$ is the required compatible right-invariant group metric. No metrizability theorem is needed.

False. Give the two-element group $C_2$ the indiscrete topology and take $G=\mathbb{R}\times C_2$ with the product topology. The indiscrete factor is a topological group, since every map into an indiscrete space is continuous, and the product group operations are continuous. Points $(0,e)$ and $(0,a)$, $a\ne e$, have identical neighbourhoods, so $G$ is not a Hausdorff space. However, its continuous projection onto $\mathbb{R}$ is surjective. If $G$ were compact, its image $\mathbb{R}$ would be compact, which it is not.

Let $G_1=\varphi (\mathbf{r}|{\mathbf{r}}_1)$ and $G_2=\varphi (\mathbf{r}|{\mathbf{r}}_2)$. Green's second identity givesThe common Robin boundary condition makes the surface integrand vanish. Where $\beta \ne 0$, substitute ${\partial }_n{G}_i=-\alpha G_i/\beta$; a Dirichlet portion also contributes zero. The delta sources therefore give the Helmholtz Green-function reciprocityThis is a bilinear identity, with no complex conjugation. Assume the boundary-value problem has a unique Green kernel; in an exterior problem use the same outgoing radiation condition for both kernels.

For the weighted operator, assume smooth positive $f$ and set $\mathbf{b}=\nabla \log f$. ThenIts unweighted transpose, obtained by integration by parts, isThe associated boundary identity isTo cancel it when $v$ satisfies the given Robin condition, the required adjoint kernel satisfiesUsing these two delta sources in the boundary identity proves $\psi ({\mathbf{r}}_2|{\mathbf{r}}_1)=\tilde{\psi }({\mathbf{r}}_1|{\mathbf{r}}_2)$, the adjoint reciprocity for a weighted Helmholtz operator. As a check, $L^{\mathrm{T}}(fv)=fLv$, soThis is equivalent to weighted acoustic Green-function reciprocity, $f({\mathbf{r}}_2)\psi ({\mathbf{r}}_2|{\mathbf{r}}_1)=f({\mathbf{r}}_1)\psi ({\mathbf{r}}_1|{\mathbf{r}}_2)$. Unit delta normalization is essential here. A Hermitian adjoint would instead conjugate coefficients and change the radiation convention; the requested equality uses the transpose.

Let $z_f(t)$ be the filling-box first front, with $z_f(0)=H$. Below it the original fluid has not yet been through the plume. In the filling box model, the turbulent plume adjusts rapidly and occupies a negligible fraction of the room's area. A point source with negligible source volume flux supplies $Q(z)=C{F}_s^{1/3}{z}^{5/3}$. All its volume crossing the front grows the processed region, so volume conservation givesPut $a_p=\pi C{F}_s^{1/3}/A_c$. Integrating $d(z_f^{-2/3})/dt=2a_p/3$ yieldsThe first front approaches the floor only as $t\to \infty$. A nonzero source volume flux would require an equal exhaust to keep the room volume fixed; it is treated in the following parts.

Since $2p-q=1$, the leading turbulent viscosity is linear:Using $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$ and $1/p=2-\sigma$ gives

Let $p$ be the best uniform approximation to the exponential function from the degree-at-most-$n-1$ polynomial space. Suppose that $E_{n-1}(e^x)=E_n(e^x)$. Then the same $p$ is best in the larger space. Its error $g(x)=e^x-p(x)$ is nonzero, since the exponential is not a polynomial. The Chebyshev alternation theorem supplies $n+2$ ordered points where $g$ has alternating signs. By the intermediate value theorem, $g$ has at least one zero in each of the $n+1$ intervening intervals, hence at least $n+1$ distinct zeros.

Applying Rolle's theorem successively $n$ times forces a zero of $g^{(n)}$. But $\mathrm{deg}p\le n-1$, soeverywhere. This is a contradiction. Together with the automatic non-increase of best errors, it provesThe argument is an instance of strict polynomial error decrease from nonvanishing derivatives.

The qualitative prediction of galactic chemical evolution is useful: later stellar populations tend to inherit more enriched gas, and old stellar populations can preserve low stellar metallicity. The closed-box metallicity distribution, however, predicts many more metal-poor long-lived dwarfs in the solar neighbourhood than are observed. This is the G-dwarf problem, also apparent with other long-lived tracers. A single well-mixed box cannot describe the different age and abundance distributions of the galactic disk, galactic bulge, and stellar halo of the Milky Way. Nor need its exponentially declining star formation rate match the history of ongoing disk star formation.

Prolonged relatively metal-poor galactic gas inflow supplies fuel while diluting the enrichment, allowing many stars to form after the earliest enrichment stage; pre-enrichment of the initial reservoir also removes a pristine tail. Galactic outflows lower the effective yield, but a constant well-mixed outflow alone still gives an exponential metallicity distribution function, and is not automatically a cure for the G-dwarf problem. Spatially varying star formation, radial transport and migration, and selection of local survivors complicate a comparison with one global box. A varying initial mass function or stellar yield changes the enrichment law. Finally, finite stellar lifetimes and Type Ia supernovae make iron enrichment delayed, so an observed iron-abundance distribution is not interchangeable with the instantaneous total-metal distribution. Gas exchange, spatially distinct populations, and delayed enrichment are essential qualifications to the closed-box prediction.

Divide metal enrichment by gas depletion in the closed-box model of galactic chemical evolution:The logarithm appears on integrating from $M_0$ and $Z=0$:With the gas fraction of a galaxy $\mu =M_g/M_0$, this isIt gives $Z=0$ at $\mu =1$ and increasing gas-phase metallicity as gas is consumed, independently of the time dependence of the star formation rate. For this closed initially star-free system, $M_0=M_g+M_s$, so this initial-reservoir fraction equals the present gas-to-total baryonic fraction. That equality fails in an open system.

A newly formed stellar population inherits the current gas-phase metallicity. Eliminate time between locked-mass growth and enrichment:For $Y>0$, integration from pristine gas gives $M_g=M_0{e}^{-Z/Y}$ and hence the closed-box metallicity distributionTo obtain numbers rather than mass, specify a fixed initial mass function and count a long-lived tracer population. Let $\eta$ be its number of surviving stars per unit locked stellar mass. At the final gas-phase metallicity $Z_f$,with zero density outside that interval. The cumulative count is $N(<Z)=\eta M_0(1-e^{-Z/Y})$ within the interval, and the normalized metallicity distribution function isThese equations do not use $\psi (t)$: any positive star formation rate gives the same distribution under the same chemical assumptions. If the tracer number is instead normalized per gross formed mass, replace $\eta$ by that normalization divided by $1-F$. Short-lived stars require an age-dependent survival factor and generally do not follow this number distribution. In a logarithmic metallicity distribution, $dN/d{\log }_{10}Z=(\ln 10)ZdN/dZ$; its interior peak is at $Z=Y$, whereas the distribution per linear interval decreases from $Z=0$.

Let $Z$ be the well-mixed gas-phase metallicity, measured as a metal mass fraction. Removing gas into stars takes metals at rate $Z\psi$; prompt return restores existing metals at rate $ZF\psi$, together with newly synthesized metals at rate $Y(1-F)\psi$. Thus the closed-box model of galactic chemical evolution has metal budgetExpand the left side and use ${\dot{M}}_g=-(1-F)\psi$. The terms proportional to existing gas-phase metallicity cancel, leavingSince $\psi =k{M}_g$, initially pristine gas obeysThe stellar yield here is per permanently locked mass, so inserting another factor $1-F$ into $Y$ would double-count recycling. The approximation treats metals as a trace constituent and ignores finite stellar-return delays. Apply it while $Y\lambda t\ll 1$; the formal unlimited linear growth is not a physical prediction for almost entirely metallic gas.

Write the gross star formation rate as $\psi =k{M}_g$, with $k>0$ constant. Use the instantaneous recycling approximation: a fraction $F$ returns promptly, while a fraction $1-F$ stays in long-lived stars and remnants. Assume $0\le F<1$, instantaneous mixing, and no galactic gas inflow or galactic outflow. If $M_s$ denotes the permanently locked stellar mass, mass conservation givesSet $\lambda =(1-F)k$. Integration, with $M_g(0)=M_0$, gives the exponential closed-box gas consumption lawThe locked mass is $M_s=M_0(1-e^{-\lambda t})$, so $M_g+M_s=M_0$ at every time. The decay time is $[(1-F)k{]}^{-1}$, longer than the gross gas-processing time $k^{-1}$ because returned gas can form subsequent generations. The formal $F=1$ limit retains all gas and locks no mass; a finite stellar yield defined per locked mass then needs separate treatment.

Here the conserved total angular momentum is ${\mathbf{J}}_t=(0,0,1)$, while the black-hole spin magnitude remains $\sqrt{3}$. The hole again aligns with the positive $z$ direction in the persistent-torque regime. ThusSince $1-\sqrt{3}<0$, the disk points in the negative $z$ direction and has magnitude $\sqrt{3}-1$. This realizes counteralignment of a black hole and accretion disk even though the hole itself is aligned with the conserved total vector. The black hole and disk do not always finish with aligned spins.

Choose an auxiliary null vector $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The screen-space projector and its induced positive metric areThey annihilate $U,N$, obey $P^2=P$, and have trace two in four spacetime dimensions. Since both $U$ and $N$ are parallel along $U$, ${\nabla }_{U}P=0$. Equivalently, the screen represents the two-dimensional quotient $U^{\perp }/\mathrm{span}(U)$.

Define the optical matrix of a null congruence byFor a transverse deviation, its screen projection $\zeta =PZ$ satisfies ${\nabla }_U{\zeta }^a={\hat{B}}^a{}_b{\zeta }^b$. Decompose the matrix asThese are the null expansion, null shear and null twist. Null expansion gives the fractional change of a small transverse area, $\theta =d\log (dA)/d\lambda$. Null shear is symmetric trace-free distortion; null twist is the antisymmetric rotation part. This null expansion convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the null geodesic congruence.