Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 1

The minimal polynomial $m_A$ is the monic polynomial of smallest degree satisfying $m_A(A)=0$. An annihilating polynomial exists, for example because $I,A,\dots ,A^{n^2}$ are linearly dependent in the $n^2$-dimensional space of matrices. A nonzero constant cannot annihilate $A$, so the minimal degree is positive.

If $p$ and $q$ are two such monic polynomials of minimal degree $d$, then $(p-q)(A)=0$ and $\mathrm{deg}(p-q)<d$ unless $p=q$. Minimality therefore forces $p=q$. Polynomial division also shows that $m_A$ divides every annihilating polynomial: the remainder after division would otherwise be an annihilator of smaller degree. This proves uniqueness rather than merely choosing one of several minimal-degree polynomials.

If $A$ is real, coefficientwise complex conjugation of $m_A(A)=0$ gives $\overline{m_A}(A)=0$. The conjugated polynomial is monic of the same degree, hence uniqueness gives $\overline{m_A}=m_A$. The minimal polynomial of a real matrix has real coefficients.

For $n\ge 3$, use the block diagonal matrixThe two-dimensional Jordan block is annihilated by $(t-1{)}^2$ but not by $t-1$, while the scalar block requires the factor $t+1$. A polynomial annihilates the direct sum exactly when it annihilates every block. Thus its minimal polynomial is the least common multiple $(t-1{)}^2(t+1)$. For $n=3$ the final identity block is omitted.

In the upper half-plane model $H=\{(x,y):y>0\}$ the hyperbolic metric isThe length of a differentiable curve is $\int \sqrt{{\dot{x}}^2+{\dot{y}}^2}dt/y$. The area element is the square root of the metric determinant times $dxdy$, so the hyperbolic area of a measurable region $S$ is ${\int }_S{y}^{-2}dxdy$.

The Gauss-Bonnet theorem for a geodesic hyperbolic triangle of angles $\alpha ,\beta ,\gamma$ in curvature $-1$ gives area $\pi -\alpha -\beta -\gamma$; an ideal vertex has angle zero. The horizontal edge in the present region is not a geodesic, so it is simpler to use the area element directly:The open boundary does not affect the integral because its area is zero.

Write the holomorphic function as $f=u+iv$. Analyticity gives smooth real and imaginary parts satisfying the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$. Differentiating these identities showsThus both parts are harmonic functions. The product rule then givesSince $f^{\prime }=u_x+i{v}_x$, this proves the Laplacian identity for the squared modulus of a holomorphic function,

Interpret $m$ as the rest mass of each initial ball and assume that no energy or momentum escapes the collision. Let $\gamma =(1-v^2/c^2{)}^{-1/2}$ be its Lorentz factor. The total initial energy and momentum areTheir conservation determines the final four-momentum. For a single lump, $P={\gamma }^{\prime }{m}^{\prime }{v}^{\prime }$ and $E={\gamma }^{\prime }{m}^{\prime }{c}^2$, henceThe final invariant mass follows from $m^{\prime 2}{c}^4=E^2-c^2{P}^2$:This is the rest mass of a coalescing relativistic collision. For $v>0$, $m^{\prime }>2m$: some kinetic energy becomes internal energy in the final rest frame. Setting $m^{\prime }=2m$ would discard that energy and contradict conservation. At small speeds $v^{\prime }\sim v/2$, while $v=0$ gives $m^{\prime }=2m$.

The displayed evolution law uses the usual assumptions of incompressible flow, constant density and conservative body force. With $\mathbf{\omega }=\nabla \times \mathbf{u}$, the Euler equation isUse $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ and take a curl. Gradients have zero curl, andHere $\nabla \cdot \mathbf{\omega }=0$ because it is a curl, and $\nabla \cdot \mathbf{u}=0$ by incompressibility. The vorticity equation is thereforeThe material derivative combines local change with advection by fluid particles. The right-hand side is vortex stretching and tilting: spatial variation of velocity along a vortex line changes its strength and direction. Along each particle path this is a homogeneous linear equation for vorticity. For a smooth velocity field its solution with zero initial vorticity remains zero by uniqueness. Thus an initially irrotational flow remains irrotational while these assumptions and smoothness hold.

For a plane flow $\mathbf{u}=(u(x,y,t),v(x,y,t),0)$, the vorticity is $\mathbf{\omega }=(0,0,\zeta )$ and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}=\zeta {\partial }_z\mathbf{u}=0$. Hence $D\zeta /Dt=0$. Every particle retains its initial value, so an initially uniform value givesInviscid motion alone is insufficient for the printed simplified law: compressible variable-density motion generally adds $-\mathbf{\omega }\nabla \cdot \mathbf{u}+\nabla \rho \times \nabla p/{\rho }^2$. Those terms vanish under the assumptions used above.

Use columnwise partial pivoting: select the largest absolute entry in the current column and interchange rows. In the first column the pivot is $4$, so interchange rows 1 and 2. The multipliers are $1/2$ and $-1/2$, and elimination givesThe second pivot is $5/2$, so interchange the two remaining rows. Their previously stored first-column multipliers must also be interchanged. The final multiplier is $(1/2)/(5/2)=1/5$, giving the LU decompositionThe permutation matrix orders the original rows as $2,3,1$. Multiplication verifies the displayed identity.

If “column pivoting” instead denotes interchanging columns rather than searching within a column, that convention gives $AQ=\tilde{L}\tilde{U}$ withThis second identity uses a different pivot convention; it does not change the row-pivoted answer above.

At the first step of Gaussian elimination, the first column of a nonsingular matrix cannot be zero. Choose a nonzero pivot there using a row interchange and eliminate below it. The resulting matrix has block formElementary elimination and row permutations preserve nonsingularity, so $p\det B\ne 0$ and the trailing matrix $B$ is nonsingular. Its first column again has a nonzero entry. Repeating this argument inductively provides a nonzero pivot at every step and produces an upper-triangular factor. The recorded elimination multipliers, with previous entries permuted when necessary, form the unit lower-triangular factor. Every nonsingular square matrix admits $PA=LU$ under columnwise partial pivoting.

For the column-interchange convention from part (a), use the same induction with a nonzero entry in the first row of each nonsingular trailing matrix. Column interchanges put it in the pivot position, and elimination gives $AQ=LU$. These existence statements are in exact arithmetic; they do not assert that all choices have equally good numerical stability.

For the independent normal distribution observations, the log likelihood isIts derivative is ${\sum }_i(X_i-\theta )$ and its second derivative is $-n<0$, so the unique maximum-likelihood estimate isFor the prior $\theta \sim N(\mu ,{\tau }^{-1})$ with $\tau >0$, multiplying prior density by likelihood and completing the square givesThus normal-normal conjugacy gives the posterior distributionFor quadratic loss, the conditional expected loss of action $a$ is the posterior variance plus $(a-\mathbb{E}[\theta \mid X]{)}^2$. Its unique minimizer is the posterior mean, so the Bayes estimator isThis weights the sample and prior means by their respective precisions.

For maximization of $f(x)$ subject to $g_j(x)\le 0$ and $h_k(x)=0$, the Lagrangian sufficiency theorem says: if a feasible $x^{\ast }$ globally maximizes $L(x)=f(x)-{\sum }_j{\lambda }_j{g}_j(x)+{\sum }_k{\nu }_k{h}_k(x)$, where ${\lambda }_j\ge 0$, and ${\lambda }_j{g}_j(x^{\ast })=0$, then $x^{\ast }$ globally maximizes $f$ over the feasible set. Indeed, at every feasible $x$,This requires a global maximum of the optimization Lagrangian; a stationary point alone is not sufficient without additional concavity or another justification.

Put $q=p/(p-1)$ and $C=({\sum }_i|a_i{|}^q{)}^{1/q}$. If every $a_i=0$, the maximum is zero and every feasible vector is optimal. Otherwise setSince $p(q-1)=q$, ${\sum }_i|x_i^{\ast }{|}^p=1$. For $g(x)={\sum }_i|x_i{|}^p-1$, the optimization Lagrangian is $L(x)={\sum }_i{a}_i{x}_i-\lambda g(x)$. Its gradient vanishes at $x^{\ast }$ becauseThe function ${\sum }_i|x_i{|}^p$ is a strictly convex function for $p>1$, so $L$ is a strictly concave function. Its supporting-hyperplane inequality proves that this stationary point is its global maximum. Complementary slackness and feasibility hold, so the sufficiency theorem applies. Substitution givesThe maximizing vector is unique when $a\ne 0$. It is the norming vector for Holder inequality, and the value is the corresponding dual norm.

For a linear map $\theta :U\to V$, its rank of a linear map is $r(\theta )=\dim \mathrm{im}\theta$, and its nullity is $n(\theta )=\dim \ker \theta$. Choose a basis $e_1,\dots ,e_s$ of its kernel and extend it to a basis $e_1,\dots ,e_d$ of $U$. Then $\theta e_{s+1},\dots ,\theta e_d$ span its image. They are linearly independent: a linear combination mapped to zero would belong to the kernel, contradicting independence of the extended basis unless all its coefficients vanish. Hence $r(\theta )=d-s$, provingFor endomorphisms of $U$, define $(\theta +\varphi )(u)=\theta (u)+\varphi (u)$ and $(\theta \varphi )(u)=\theta (\varphi (u))$. The image of the sum lies in the sum of the two images. ThusTo prove the product inequality, restrict $\varphi$ to $K=\ker (\theta \varphi )$. This map has kernel $\ker \varphi$ and image $\mathrm{im}\varphi \cap \ker \theta$: every vector in the intersection has a preimage in $K$. Applying the rank-nullity theorem to the restricted map givesNow put $d=\dim U$ and $s=r(\theta )+r(\varphi )$. If both equalities hold, $r(\theta +\varphi )=s\le d$, while $n(\theta \varphi )=2d-s\le d$ forces $s\ge d$. Therefore $s=d$, the sum has full rank, and the product has nullity $d$. The sum is an isomorphism and the product is zero.

Conversely, if $\theta \varphi =0$, then $\mathrm{im}\varphi \subseteq \ker \theta$, giving $s\le d$. If the sum is also an isomorphism, its rank is $d\le s$. Thus $s=d$, and the two inequalities become equalities. This proves the equality in the rank-sum and nullity-product inequalities in both directions.

Use the following standard facts: $A_5$ has order $60$ and is a nonabelian simple group; the coset action on the cosets of an index-$j$ subgroup is a transitive homomorphism into $S_j$. If $H<A_5$ has index $j>1$, that action is nontrivial. Its kernel is a normal subgroup, hence is trivial by simplicity. It would therefore embed $A_5$ in $S_j$. For $j=2,3,4$ this is impossible since $j!<60$. There are no subgroups of indices 2, 3 or 4.

An index-5 subgroup has order $12$ by Lagrange's theorem. Consider its natural action on the five letters. It cannot be transitive, since an orbit of size 5 would divide its order by the orbit-stabilizer theorem. If it had no fixed letter, its orbits would have sizes 2 and 3. It would then be a subgroup of the permutations preserving that partition, namely $S_2\times S_3$. Exactly half of those 12 permutations are even, so $(S_2\times S_3)\cap A_5$ has order 6, too small to contain $H$.

Thus $H$ fixes a letter $i$. The stabilizer of $i$ in $A_5$ is the copy of $A_4$ consisting of even permutations of the other four letters; it has order 12. Containment and equality of orders force $H$ to equal that stabilizer. Conversely each such stabilizer has index 5. ConsequentlyThese five subgroups are distinct: for $i\ne j$, a three-cycle on letters other than $i$ can move $j$, so it belongs to the stabilizer of $i$ but not that of $j$.

Let $S_N(x)={\sum }_{n=1}^N[a_n\cos (nx)+b_n\sin (nx)]$. Under the first coefficient bounds,This proves the uniform Cauchy condition on the whole real line and hence uniform convergence to $f$. It is the Weierstrass M-test applied with the summable majorant $2c{n}^{-1-\epsilon }$. Every partial sum is continuous. To see continuity of the limit explicitly, at a chosen $x_0$ first choose $N$ with small uniform error; then make $|S_N(x)-S_N(x_0)|$ small by continuity. The inequalityproves continuity. Also $S_N(x+2\pi )=S_N(x)$ for every $N$, so taking limits gives $f(x+2\pi )=f(x)$.

Under the stronger coefficient bounds, the derivative series has summable uniform majorant $2c{n}^{-1-\epsilon }$. Thus $S_N^{\prime }$ converges uniformly to a continuous function $g$ given by that series. The original series also converges uniformly. On any finite interval, the fundamental theorem of calculus gives $S_N(x)-S_N(0)={\int }_0^x{S}_N^{\prime }(t)dt$. Uniform convergence permits passage to the limit inside this finite integral, since the error is at most $|x|\Vert S_N^{\prime }-g{\Vert }_{\infty }$. Therefore $f(x)-f(0)={\int }_0^{x}g(t)dt$, and continuity of $g$ gives $f^{\prime }=g$. This justifies Termwise differentiation of a Fourier series rather than assuming it:

The product topology on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. Equivalently, these rectangles form a basis.

The empty set is the empty union, and $X\times Y$ is itself an allowed rectangle. An arbitrary union of sets of this form is still a union of allowed rectangles. Finally,which is an allowed rectangle because the factor topologies are closed under finite intersections. Distributing intersection over unions shows that the intersection of any two product-open sets is product-open; induction handles all finite intersections. These verify every topology axiom. Rectangles alone are generally not a topology, which is why their arbitrary unions are included.

Parametrize the logarithmic spiral byThe map is continuous into the ambient plane, hence into $X$ with its subspace topology, and it is onto by the definition of the spiral. It is injective because its radius $|h(t)|=e^t$ determines $t$ uniquely, even when the angular coordinate winds through repeated turns.

Every point of $X$ is nonzero. The inverse is thereforethe restriction of a continuous function on the punctured plane. Hence $h$ is a homeomorphism and the spiral is homeomorphic to the real line. Its accumulation at the origin causes no failure of this argument: the origin is not included in $X$.

Let $F(z)=e^{az}/(1+e^z)$ and integrate counterclockwise around the rectangle with vertices $-R,R,R+2\pi i,-R+2\pi i$. It contains a single simple pole at $z=\pi i$, with residueOn the top edge $F(x+2\pi i)=e^{2\pi ia}F(x)$, and its orientation reverses that of the bottom edge. On the right edge $|F|\le e^{aR}/(e^R-1)$, so its integral tends to zero for $a<1$. On the left $|F|\le e^{-aR}/(1-e^{-R})$, which tends to zero for $a>0$. The residue theorem therefore givesSince $1-e^{2\pi ia}=-2i{e}^{\pi ia}\sin (\pi a)$,The restriction is also necessary for ordinary convergence: at $-\infty$ the integrand behaves as $e^{ax}$, requiring $a>0$, while at $+\infty$ it behaves as $e^{(a-1)x}$, requiring $a<1$. At either endpoint one tail approaches a nonzero constant; outside this range one tail grows. There is no cancellation because the real integrand is positive.

In orthonormal Cartesian coordinates, a second-rank tensor has components transforming under a change of axes $Q$ as $M_{ij}^{\prime }=Q_{ik}{Q}_{j\ell }{M}_{k\ell }$, where $Q{Q}^T=I$. Its contraction is invariant:Thus $M_{ii}$ is a scalar. This is tensor contraction with the Euclidean metric. In general nonorthonormal coordinates the invariant expression is $g^{ij}{M}_{ij}$, not an unweighted sum of diagonal covariant components.

For the planar plate, every mass element has $z=0$. The inertia tensor is consequentlyIt follows directly that ${\mathbf{e}}_z$ is an eigenvector andThe real symmetric matrix has three real eigenvalues. Its trace is $2{\int }_D(x^2+y^2)\rho dS=2M_{\perp }$, so $M_1+M_2+M_{\perp }=2M_{\perp }$. Hence the perpendicular axis theorem gives $M_{\perp }=M_1+M_2$.

For a disc of constant areal density ${\rho }_0$, symmetry gives ${\int }_{D}xydS=0$, and polar integration gives ${\int }_D{x}^{2}dS={\int }_D{y}^{2}dS=\pi a^4/4$. Thereforewhere the scalar $M$ in the last expression denotes the total mass rather than the tensor.

No: ordering the attractive potentials does not impose the proposed ordering of reflection probabilities. Take $k=\pi /(4a)$ and let $M$ denote the particle mass. Choose $V_2=0$ everywhere andThese satisfy every stated ordering and support condition. The incoming energy is ${\hslash }^2{k}^2/(2M)$; the interior wavenumber in the finite square well is $q=\sqrt{k^2-2M{V}_1/{\hslash }^2}=3k$.

Here is an explicit matching calculation. For a general well of width $L=2a$, take exterior waves $e^{ik(x+a)}+r{e}^{-ik(x+a)}$ on the left and $t{e}^{ik(x-a)}$ on the right. Shifting the incident phase in this way does not change the reflection probability. Continuity of the wavefunction and its first derivative gives the propagation relationInverting it and adding the two equations after dividing the derivative equation by $ik$ yieldsThe incident and reflected exterior waves have the same speed, so their flux ratio isFor the chosen parameters $qL=3\pi /2$ and $(q^2-k^2{)}^2/(4k^2{q}^2)=16/9$. ThusThe zero potential has no reflected component. This proves the failure with explicit probabilities. More generally the same expression vanishes at $qL\in \pi \mathbb{Z}$, explaining resonant transmission through a square well and why increasing attractive well depth does not monotonically control reflection.

Take zero electric potential at infinity and put $K=(4\pi {\epsilon }_0{)}^{-1}$. By spherical symmetry and Gauss's law, the electric field is radial, with magnitude determined by the total enclosed electric charge. Thus, away from the idealized shell surfaces,Integrating $E_r=-d\Phi /dr$ from infinity, or adding the three spherical-shell potentials, givesThe expressions agree at each interface, so the potential is continuous everywhere, including the origin. The electric field has different interior and exterior limits at each infinitely thin charged surface, with jump equal to surface charge density divided by ${\epsilon }_0$. An ideal sheet has no single classical field value on the sheet itself; the displayed adjacent limits specify the field there. In conducting material of nonzero thickness the internal field is zero.

The electrostatic energy is the integral of ${\epsilon }_0|\mathbf{E}{|}^2/2$. There is no contribution from $r<a$, and spherical integration givesThe dependence on $b$ cancels because the magnitude of enclosed electric charge is $|q|$ on both sides of the middle shell, even though the field direction changes.

Define source strength $m$ here as the volume flux through a small sphere around the source. For incompressible potential flow, $\mathbf{u}=\nabla \varphi$ and the governing conditions areEquivalently the Laplace equation holds away from the source, with local behavior $\varphi \sim -m/(4\pi |\mathbf{x}-{\mathbf{x}}_s|)$. The boundary condition expresses impermeability. The method of images uses an equal source below the plane:The image singularity lies outside the fluid, the two vertical derivatives cancel on the wall, and the source has the specified outward flux.

Let $s=\sqrt{x^2+y^2}$. On the wall $u_z=0$ and the tangential radial velocity is $u_s=ms/[2\pi (s^2+a^2{)}^{3/2}]$. With ambient pressure $p_{\infty }$, the steady Bernoulli equation in this irrotational flow givesThe fluid pressure is reduced everywhere on the wall except at its axis and in the infinite-distance limit. The finite hydrodynamic force is the excess over the uniform ambient-pressure force. Integrating its magnitude over the plane yieldsPutting $u=s^2$ gives the last integral as $\frac{1}{2}{\int }_0^{\infty }u(u+a^2{)}^{-3}du=1/(4a^2)$. HenceRelative to ambient pressure on the other side, the boundary is attracted toward the source. This is the hydrodynamic attraction of a plane wall to a point source.

There is an alternative source-strength convention: if $m$ denotes the coefficient in $u_r=m/r^2$ rather than total volume flux, replace the flux above by $4\pi m$. Then $\varphi =-m(1/r_{+}+1/r_{-})$, $p-p_{\infty }=-2\rho m^2{s}^2/(s^2+a^2{)}^3$, and $F=\pi \rho m^2/a^2$. The physical result is identical once the normalization is fixed.

A sufficient statistic $T(X)$ for $\theta$ is a function of the observations whose conditional distribution of $X$ given $T(X)$ does not depend on $\theta$, whenever that conditioning event is possible. It retains all parameter dependence of the likelihood.

In the discrete case the Fisher-Neyman factorization theorem provides a direct criterion: $f(x\mid \theta )=g(T(x),\theta )h(x)$, where $h$ is independent of the parameter. To justify sufficiency, for a value $t$ with positive probability, sum this expression over its fiber and divide:The parameter-dependent factor cancels. Conversely, parameter-independent conditional probabilities $h_t(x)$ give the factorization $f(x\mid \theta )={\mathbb{P}}_{\theta }(T=t)h_t(x)$ with $t=T(x)$. Thus the criterion is both necessary and sufficient, with conditional laws only required on fibers of positive probability.

For the shifted exponential sample the likelihood isIt factors through $T=X_{(1)}={\min }_i{X}_i$, so the sample minimum is a sufficient statistic. Writing $Y_i=X_i-\theta$, independence gives for $t\ge 0$Therefore $T-\theta$ is exponential with rate $n\lambda$. Its mean is $(n\lambda {)}^{-1}$ and its variance is $(n\lambda {)}^{-2}$. The unbiased endpoint estimator for a shifted exponential sample is consequentlyAlthough the parameter is nonnegative, this unbiased estimator can be negative. Truncating it at zero would change its expectation and lose the requested property of an unbiased estimator.

For the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, a stopping time $T$ takes values in $\{0,1,\dots \}\cup \{\infty \}$ and satisfies $\{T\le n\}\in {\mathcal{F}}_n$ for each $n$. Whether it has occurred by time $n$ can be decided from the observations through that time.

The Strong Markov property for a time-homogeneous Markov chain says that, on $\{T<\infty \}$, conditional on ${\mathcal{F}}_T$, the process $(X_{T+r}{)}_{r\ge 0}$ has the law of a fresh chain started at $X_T$. In particular its future conditional law depends on the past only through $X_T$. For example $\mathbb{P}(X_{T+r}=j\mid {\mathcal{F}}_T)=(P^r{)}_{X_{T}j}$ on that event, and the corresponding statement holds for every finite future path.

Start at $i$ and let $T_0=0$, $T_{r+1}=\inf \{n>T_r:X_n=i\}$, with $\inf \varnothing =\infty$. These are stopping times. Put $f={\mathbb{P}}_i(T_1<\infty )$. Each finite return restarts the chain at $i$, so the Strong Markov property gives inductivelyThe events decrease as $r$ increases; infinitely many visits occur exactly when every return time is finite. Continuity of probability therefore givesNo irreducibility assumption is needed for this zero-one conclusion at the starting state.

Let $N={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$, counting the initial visit. Summing nonnegative random variables and then using the tail-sum identity givesIf $f<1$, this is $1/(1-f)<\infty$. Thus divergence of the given sum forces $f=1$, and infinitely many visits then occur with probability one. This proves the recurrence criterion by return probabilities directly from the regeneration at return times; divergent probability sums alone would not suffice for arbitrary dependent events.