This assertion is false in general. In dimension two take $A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$, $B=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $C=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. Then $ABC=A$, whereas $ACB=0$. Their characteristic polynomials are $\lambda (\lambda -1)$ and ${\lambda }^2$, respectively, so $\boxed{p_{ABC}\ne p_{ACB}}$.

Write $f=u+iv$ and $z=x+iy$. The Cauchy-Riemann equations give $u_x=v_y$ and $u_y=-v_x$. Thus the real Jacobian determinant isThis identity also holds at points where $f^{\prime }(z)=0$; there the Jacobian is zero and local angle preservation is not guaranteed.

For a regular smooth curve $z=z_j(t)$ passing through $z_0$, the complex tangent of its image is $d{w}_j/dt=f^{\prime }(z_0)z_j^{\prime }(t)$. When $f^{\prime }(z_0)\ne 0$, both tangents are multiplied by the same nonzero complex number. Multiplication rescales lengths by $|f^{\prime }(z_0)|$ and rotates every direction by $\arg f^{\prime }(z_0)$, so the difference of tangent arguments is unchanged. Hence the analytic function defines a conformal map at $z_0$: oriented angles, and therefore ordinary angles between the curves, are preserved. The nonzero derivative condition also ensures that the image tangents are nonzero.

False in general. Take $A=\mathrm{diag}(1,6,6)$ and $B=\mathrm{diag}(2,2,9)$. Both have trace $13$ and determinant $36$, but their eigenvalue multisets are different. For $n>3$, append equal identity blocks to preserve this counterexample. In dimensions one and two the assertion is true: in dimension two the characteristic polynomial is $t^2-(\mathrm{tr}A)t+\det A$, whose roots, with multiplicity, are determined by those two quantities.

False. Every entry of $A=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ is strictly positive, yet $A(1,-1{)}^T=-(1,-1{)}^T$. Its eigenvalues are $3$ and $-1$, so one has negative real part. Positivity of entries does not imply the positive-definite matrix property.

False. The matrix $A=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has real eigenvalues $1$ and $2$, because its characteristic polynomial is $(t-1)(t-2)$, but $A^{\ast }=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)\ne A$. Real spectrum alone does not imply that it is a Hermitian matrix.

True. If $Av=\lambda v$ with $v\ne 0$ and $A=A^{\ast }$, thenSince $v^{\ast }v>0$, $\lambda =\bar{\lambda }$. Thus every eigenvalue of a Hermitian matrix is real.

The unrestricted maximum-likelihood estimators remain $(\bar{X},\bar{Y})$. Subject to the null constraint, minimizing $({\mu }_X-\bar{X}{)}^2$ on $[A,\infty )$ gives ${\hat{\mu }}_X=\max (A,\bar{X})$, while ${\hat{\mu }}_Y=0$. ConsequentlyWe must calibrate this likelihood-ratio test over the entire composite null. Under that null let $W=\sqrt{n}(\bar{X}-A)$ and $Z=\sqrt{n}\bar{Y}$. Then $W\sim N(\delta ,1)$ with $\delta =\sqrt{n}({\mu }_X-A)\ge 0$, while $Z\sim N(0,1)$ independently, and $T=Z^2+(-W{)}_{+}^2$. Couple all null distributions by $W=W_0+\delta$, with a fixed standard Gaussian $W_0$. Increasing $\delta$ decreases $(-W_0-\delta {)}_{+}^2$ pointwise, so the largest rejection probability occurs at $\delta =0$, namely ${\mu }_X=A$.

At this boundary the sign of $W$ is independent of $W^2$, with each sign having probability one half. On $W\ge 0$, $T=Z^2\sim {\chi }_1^2$; on $W<0$, $T=Z^2+W^2\sim {\chi }_2^2$. Thus the least favourable null hypothesis law is the exact chi-bar-squared distribution $\frac{1}{2}{\chi }_1^2+\frac{1}{2}{\chi }_2^2$. Choose $c_{\alpha }>0$ byThe tail is a continuous function and strictly decreasing from one to zero, so this threshold is unique for $0<\alpha <1$. The null rejection probability is at most $\alpha$ and equals $\alpha$ at ${\mu }_X=A$. This proves the required size, rather than using an interior chi-squared approximation.

To compare, put $c_1=z_{1-\alpha /2}^2$, the threshold in (i). Since ${\chi }_2^2$ is stochastically larger than ${\chi }_1^2$ (add an independent squared standard Gaussian), the mixture threshold obeys $c_{\alpha }>c_1$. When $\bar{X}\ge A$, test (ii) therefore requires stronger evidence from $\bar{Y}$ than test (i). When $\bar{X}<A$, test (ii) also responds to violation of the mean constraint and can reject even with $\bar{Y}=0$. Neither critical region contains the other. In the coordinates $(W,Z)$, the acceptance region in (i) is the strip $|Z|\le \sqrt{c_1}$; in (ii) it is the right half-strip $W\ge 0$, $|Z|\le \sqrt{c_{\alpha }}$ together with the left half-disk $W<0$, $W^2+Z^2\le c_{\alpha }$. The extra constraint changes both the statistic and its correct null calibration.

Write $\bar{X}=n^{-1}{\sum }_i{X}_i$ and $\bar{Y}=n^{-1}{\sum }_i{Y}_i$. Completing the square in each sample, the log likelihood is, up to a data-dependent constant,The maximum-likelihood estimators follow because it is strictly maximized over unrestricted means at $(\bar{X},\bar{Y})$. Under the null it is maximized at $(\bar{X},0)$. Thus the likelihood ratio is $\Lambda =\exp (-n{\bar{Y}}^2/2)$, and the likelihood-ratio test rejects for large $T=-2\log \Lambda =n{\bar{Y}}^2$.

Under every null value of ${\mu }_X$, $Z=\sqrt{n}\bar{Y}$ has the standard Gaussian distribution; hence $T$ has the chi-squared distribution with one degree of freedom. Let $z_p={\Phi }^{-1}(p)$, where $\Phi$ is the standard Gaussian cumulative distribution function. The exact size-$\alpha$ test isThe null rejection probability is $2[1-\Phi (z_{1-\alpha /2})]=\alpha$, independently of the nuisance mean ${\mu }_X$. No randomization is required because the statistic has a continuous distribution.

Incompressibility makes each perturbation velocity potential a harmonic function. A horizontal mode $e^{i(kx+\ell y)+st}$ has $q=\sqrt{k^2+{\ell }^2}>0$. Decay away from the interface selects ${\phi }_1=A_1{e}^{qz}{e}^{i(kx+\ell y)+st}$ below and ${\phi }_2=A_2{e}^{-qz}{e}^{i(kx+\ell y)+st}$ above. The kinematic boundary condition givesInsert these into the dynamic boundary condition for an inviscid interface and cancel a nonzero mode amplitude to obtainWriting ${\rho }_T={\rho }_1+{\rho }_2$ and $\bar{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/{\rho }_T$, completion of the square givesConsequentlyThe imaginary advection term does not change growth rates. A mode grows exponentially precisely when the radicand is positive, equivalentlyIf the upper fluid is heavier, every nonzero wavevector is unstable even without shear: this is Rayleigh-Taylor instability. If the lower fluid is heavier, gravity stabilizes sufficiently long waves, but any nonzero velocity difference destabilizes sufficiently short waves with $\ell =0$: this is Kelvin-Helmholtz instability. For equal densities, shear destabilizes modes with $k\ne 0$. A negative radicand gives purely imaginary frequencies and oscillatory modes; equality is the marginal threshold for exponential growth.

Without surface tension, the exact dynamic boundary condition for an inviscid interface is equality of pressure, $p_1=p_2$, at $z=\zeta$. The Unsteady Bernoulli equation in each irrotational fluid isChoose the potential gauge so that $C_i=p_0/{\rho }_i+U_i^2/2$, its unperturbed value. Expanding the kinetic term as $U_i^2/2+U_i{\partial }_x{\phi }_i$ and evaluating gravity at the displaced interface givesto first order. Equality of these pressures gives

The interface is a material surface, so its defining function $F=z-\zeta (x,y,t)$ obeys $DF/Dt=0$ on each side. Since ${\mathbf{u}}_i=\nabla {\Phi }_i$, the exact kinematic boundary condition isSubstitute ${\Phi }_i=U_{i}x+{\phi }_i$. Products such as ${\phi }_{ix}{\zeta }_x$ are second order, as is the correction from evaluating ${\phi }_{iz}$ at $z=\zeta$ instead of zero. Linearization therefore gives

For the charge-generated electric field, use the electric potentialThe vector potential is time independent, so its usual additional contribution $-{\partial }_t\mathbf{A}$ is zero. The continuity equation now givesThus the second term in part (iii) is exactly the displacement current contribution ${\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$, andThis is the Ampère-Maxwell equation. The constructed field also has zero curl, consistent with the time-independent magnetic field and Faraday's law. A static addition to the electric field does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.

Use the identity $\nabla \times \mathbf{B}=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$. The Laplacian identity for the Coulomb kernel gives, component by component,Also $\nabla (s^{-1})=-{\nabla }^{\prime }(s^{-1})$, so integration by parts yieldsHere the surface integral at infinity vanishes under the assumed localization of the current. As usual, this assumption means sufficient decay for these integrals and the boundary limit, rather than merely pointwise convergence to zero. Combining the two terms gives

Let $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$ and introduce the magnetic vector potential $\mathbf{A}(\mathbf{r})={\mu }_0(4\pi {)}^{-1}\int \mathbf{J}({\mathbf{r}}^{\prime })/sd{V}^{\prime }$. Differentiation is with respect to $\mathbf{r}$, so $\mathbf{J}({\mathbf{r}}^{\prime })$ is constant under it. HenceTaking the curl through the integral reproduces the Biot-Savart law, proving $\boxed{\mathbf{B}=\nabla \times \mathbf{A}}$ and the requested representation.

Charge conservation gives the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. Since $\mathbf{J}$ is independent of time, so is ${\partial }_t\rho =-\nabla \cdot \mathbf{J}$. Integrating at each fixed position gives

Use the same indented upper-semicircle contour and branch of the complex logarithm, now for $F(z)=z^{1/2}\mathrm{Log}z/(1+z^2)$. Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. On the negative axis, $z^{1/2}=i|z{|}^{1/2}$ and $\mathrm{Log}z=\log |z|+i\pi$. Consequently the two straight contributions tend toThe outer arc is $O(R^{-1/2}\log R)$ and the small arc is $O({\epsilon }^{3/2}(|\log \epsilon |+1))$, so neither contributes in the limit. At the enclosed pole $i$,The residue theorem gives $(1+i)J-\pi I(1/2)=i{\pi }^2{e}^{i\pi /4}/2$. Part (a) gives $I(1/2)=\pi /\sqrt{2}$. Substituting and using $e^{i\pi /4}=(1+i)/\sqrt{2}$, the right side after moving the $I$ term is ${\pi }^2(1+i)/(2\sqrt{2})$. Hence

Put $a=1/n$, so $0<a<1$, and write $I(a)={\int }_0^{\infty }{x}^a/(1+x^2)dx$. Choose the branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, whose cut is the downward imaginary axis. Define $z^a=\exp (a\mathrm{Log}z)$ and integrate $z^a/(1+z^2)$ along the positively oriented upper semicircle, indented above zero. The straight pieces run from $-R$ to $-\epsilon$ and from $\epsilon$ to $R$; the small arc runs clockwise from argument $\pi$ to zero.

On the negative real axis $z^a=|z{|}^a{e}^{i\pi a}$, so the two straight pieces tend to $(1+e^{i\pi a})I(a)$. The outer arc is $O(R^{a-1})$ and the inner arc is $O({\epsilon }^{a+1})$, both tending to zero. The only enclosed pole is $i$, with residue $e^{i\pi a/2}/(2i)$. The residue theorem therefore givesUsing $1+e^{i\pi a}=2e^{i\pi a/2}\cos (\pi a/2)$ yieldsThe same bounds show convergence of the real integral at both zero and infinity.

A Hausdorff, noncompact topology. The condition says precisely that each point of $U$ lies in a residue class $n+k\mathbb{Z}$ contained in $U$. These classes form a basis: the intersection of two classes containing a common point contains the class through that point with modulus equal to the least common multiple of their moduli. Thus arbitrary unions obey the condition, and so do finite intersections. Also $\varnothing$ and $\mathbb{Z}$ obey it. This is the profinite topology on $\mathbb{Z}$.

If $a\ne b$, choose $k>|a-b|$. Then $a+k\mathbb{Z}$ and $b+k\mathbb{Z}$ are disjoint open neighbourhoods, proving the Hausdorff property. To test compactness, defineEvery $F_n$ is nonempty and both open and closed: its complement is a finite union of other residue classes. Since $a_{n+1}-a_n=4^n$, $F_{n+1}\subseteq F_n$. If $x$ belonged to their intersection, then $3x+1$ would be divisible by $4^n$ for every $n$, forcing $3x+1=0$, impossible for an integer. Therefore $U_n=\mathbb{Z}\setminus F_n$ is an increasing open cover of $\mathbb{Z}$. Any finite subcollection has union $U_N$ for its largest index and misses the nonempty $F_N$. This proves that the profinite topology on the integers is not compact.

Not a topology. For every integer $k\ge 2$, the set $U_k=\mathbb{Z}\setminus k\mathbb{Z}$ is periodic with period $k$, so belongs to the family. Their union is $\mathbb{Z}\setminus \{0\}$: for a nonzero $n$, choose $k>|n|$, which cannot divide $n$. But this union has no positive integer period. If it had period $k$, membership of $-k$ would force membership of zero. Again arbitrary unions fail the topology axiom.