Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 3

Put $t=\mathrm{tr}A=a+d$ and $\delta =\det A=ad-bc$. Direct matrix multiplication gives the two-dimensional Cayley-Hamilton theorem identity $A^2-tA+\delta I=0$. If $A^2=0$, determinant multiplicativity gives ${\delta }^2=0$, hence $\delta =0$. The identity then says $tA=0$. If $A=0$, its trace is zero; otherwise a nonzero entry forces $t=0$. Thus $d=-a$ and $bc=ad=-a^2$.

Conversely, those two scalar relations give $t=\delta =0$, so the same Cayley-Hamilton theorem identity implies $A^2=0$. ThereforeThis is the square-zero criterion for a two-by-two matrix. If $A^3=0$, determinant multiplicativity first gives ${\delta }^3=0$, hence $\delta =0$. Now $A^2=tA$ and $A^3=t^{2}A$. Either $A=0$, or a nonzero entry gives $t^2=0$ and hence $t=0$; in both cases $A^2=0$. The converse follows by multiplying $A^2=0$ by $A$. Consequently $A^3=0⟺A^2=0$. A nonzero nilpotent matrix of this size therefore has nilpotency index two.

A Möbius transformation is $M_A(z)=(az+b)/(cz+d)$ with $\Delta =ad-bc\ne 0$. On the Riemann sphere, a zero denominator represents infinity, and $M_A(\infty )=a/c$ when $c\ne 0$; when $c=0$, infinity is fixed. The numerator and denominator cannot vanish together because $\Delta \ne 0$.

The matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on homogeneous coordinates $[z:1]$, so composition satisfies $M_A\circ M_B=M_{AB}$. The product is invertible since $\det (AB)=\det A\det B\ne 0$, proving closure. Composition is associative because these are maps of the sphere. The identity is $z\mapsto z$, andis again a Möbius transformation. Thus the transformations form a group. Nonzero scalar multiples of a matrix give the same transformation, but the identity and inverse assertions concern the transformations themselves.

Write $T_q(z)=z+q$, $S_k(z)=kz$ with $k\ne 0$, and $H(z)=1/z$. If $c=0$, then $a,d\ne 0$ and $M_A=T_{b/d}\circ S_{a/d}$. If $c\ne 0$, division givesHence the Möbius translation-scaling-inversion factorization isThe scaling coefficient is nonzero. At $z=-d/c$ the reciprocal sends the translated zero to infinity; at infinity it sends infinity to zero. Thus the factorization holds on the entire Riemann sphere, including both exceptional points, and proves the stated generating property.

For a differentiable scalar function and a differentiable curve, the multivariable chain rule isLet $L=x{\partial }_x-2y{\partial }_y$. Choose $u=\alpha (x)y$ to be an invariant of the method of characteristics. Then $Lu=y\{x{\alpha }^{\prime }(x)-2\alpha (x)\}$, so $x{\alpha }^{\prime }=2\alpha$ and a convenient choice is $\alpha (x)=x^2$. For $v=y/x$, the chain rule gives $Lv=-3v$. Thus, writing the unknown in the new coordinates,Integrating with $u$ fixed yieldswhere $F$ is an arbitrary differentiable function on the relevant interval. This gives all local solutions on each patch where the specified change of variables is invertible: its Jacobian determinant is $\partial (u,v)/\partial (x,y)=3y$.

The axes need care because $v$ is undefined at $x=0$ and the Jacobian vanishes at $y=0$. Reparameterizing $F(u)=u^{2}G(u)$ gives the simpler chart expressionIt is the general solution on either component $x>0$ or $x<0$, including $y=0$ when $G$ is differentiable there. Indeed $(x,u=x^{2}y)$ is a nonsingular coordinate chart for $x\ne 0$, and the weighted Euler first-order equation reduces to $x{\partial }_{x}f{|}_u=6f$.

For a patch crossing $x=0$ with $y\ne 0$, the characteristic invariant $w=x\sqrt{|y|}$ instead gives $f=|y{|}^{-3}K(w)$, with an arbitrary differentiable function on each sign component of $y$. These chart descriptions must agree on overlaps. On the axes the original equation gives $f(x,0)=C_{\pm }{x}^6$ and $f(0,y)=D_{\pm }|y{|}^{-3}$ on their separate components; a solution defined smoothly through the origin must have $f(0,0)=0$ and compatible smooth limits, in particular $D_{\pm }=0$. No boundary data select a particular arbitrary function.

Consider the globally smooth potential $\Phi (x,y,z)=x{y}^2\sin (xz)$. Direct differentiation givesThus the differential one-form is the exact differential $d\Phi$, proving exactness rather than only checking a necessary condition on its partial derivatives. Along any piecewise differentiable path, the chain rule gives $udx+vdy+wdz=d\Phi$. The fundamental theorem of calculus therefore makes the line integral depend only on its endpoints:

An eigenvalue of $C$ is a scalar $\lambda$ for which a nonzero vector $v$ satisfies $Cv=\lambda v$. Equivalently, $\det (\lambda I-C)=0$, because the homogeneous linear system has a nontrivial solution exactly when its coefficient matrix is singular. This characteristic polynomial is the monic quadraticA nonzero quadratic has at most two distinct roots, proving the eigenvalue bound. Eigenvalues and diagonalization are taken over $\mathbb{C}$, also for a matrix with real entries.

For the matrix commutator, the cyclic trace identity follows directly from its entries:Hence $\mathrm{tr}[A,B]=0$. Factoring the characteristic polynomial as $(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)$ and comparing the coefficient of $\lambda$ gives $\mathrm{tr}C={\lambda }_1+{\lambda }_2$, with the eigenvalues counted with algebraic multiplicity. If both eigenvalues are zero, trace and determinant are zero. The Cayley-Hamilton theorem then gives $C^2=0$; no diagonalizability assumption is needed.

Now put $M=[A,B]$. Since its trace is zero, its characteristic polynomial is ${\lambda }^2+\det M$ and the scalar square of a traceless two-by-two matrix identity givesThis proves the scalar-square assertion for both real and complex entries. If $\det M=0$, it gives $M^2=0$. If $\det M\ne 0$, the two eigenvalues are the distinct nonzero numbers $\pm \sqrt{-\det M}$. Their eigenvectors are linearly independent: a nonzero vector cannot satisfy two different eigenvalue equations. They form a basis, making $M$ a diagonalizable matrix. Thus either the commutator is diagonalizable over $\mathbb{C}$ or its square is zero. These alternatives need not be disjoint, since the zero matrix has both properties. Over $\mathbb{R}$ the diagonalization assertion would require real eigenvalues; complex diagonalization is essential to the unrestricted statement.

A left group action of $G$ on a set $X$ is a map $(g,x)\mapsto g\cdot x$ satisfying $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The group orbit of $x$ is $G\cdot x=\{g\cdot x:g\in G\}$, and its stabilizer subgroup is $G_x=\{g\in G:g\cdot x=x\}$. The stabilizer subgroup contains the identity, is closed under products, and contains $g^{-1}$ whenever it contains $g$, so it is indeed a subgroup.

The orbit-stabilizer theorem for a finite group saysTo prove it, map a left coset $g{G}_x$ to $g\cdot x$. This is well-defined because every element of $G_x$ fixes $x$. Conversely, $g\cdot x=h\cdot x$ precisely when $h^{-1}g\in G_x$, which is precisely equality of the two cosets. Thus this map is a bijection between the cosets and the group orbit. Each coset has $|G_x|$ elements, by the bijection $k\mapsto gk$, and the cosets partition $G$, giving the formula.

Apply this to the rotational symmetry group of a cube acting on its six faces. The action is transitive: rotations through right angles about coordinate axes can send the top face to each of the other faces. A rotation fixing a face must fix its outward normal, so it rotates about the axis through that face's center and the opposite face's center. Preservation of its square boundary permits exactly four angles modulo $2\pi$: $0,\pi /2,\pi ,3\pi /2$. All four preserve the cube, so this face stabilizer subgroup has order four. The group is finite because a rotation permutes the six face normals and is determined by their images. The orbit-stabilizer theorem givesThese are orientation-preserving rotations; reflections are not being counted.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$; in particular $|H|$ divides $|G|$. This follows by partitioning $G$ into cosets, each in bijection with $H$.

If $|G|=p$ is prime, take any $g\ne e$. Its cyclic subgroup has order dividing $p$, but its order is not one. Thus it has order $p$ and equals $G$. HenceThis describes a single group isomorphism type for each prime.

Now let $G=⟨x⟩$ have order $n$, and let $H$ be any subgroup. Choose the least positive integer $d$ with $x^d\in H$; this exists because $x^n=e\in H$, even when $H$ is trivial. Divide $n=qd+r$ with $0\le r<d$. Since $x^r=x^n(x^d{)}^{-q}\in H$, minimality forces $r=0$, so $d$ divides $n$. Likewise, dividing an exponent $k=qd+r$ for any $x^k\in H$ shows $r=0$. Therefore $H=⟨x^d⟩$.

Conversely, for each divisor $d$ of $n$, $⟨x^d⟩$ is a subgroup of order $n/d$. The least positive exponent property makes $d$ unique. This proves the complete subgroups and quotients of a cyclic group classification:The choices $d=n$ and $d=1$ include the trivial subgroup and the whole group.

Compose permutations from right to left. The indicated cycle lengths immediately give $X^3=P^2=Q^2=e$. Under conjugation, $X^{-1}(ab)X=(X^{-1}a{X}^{-1}b)$. Since $X^{-1}$ sends $1,2,3$ to $3,1,2$ and fixes $4$,Direct composition also gives $PQ=QP=(14)(23)$. Thus $V=\{e,P,Q,PQ\}$ is a Klein four-group. Conjugation by $X$ cycles its three nonidentity elements, so $X$ normalizes $V$.

The cosets $V,VX,V{X}^2$ have four elements each and are distinct. If two coincided, a nontrivial power of $X$ would belong to $V$, impossible because it has order three whereas every nonidentity element of $V$ has order two. Hence the subgroup generated by $X,P,Q$ contains at least twelve elements. These generators are even permutations, so the subgroup lies in the alternating group $A_4$, whose order is $4!/2=12$; multiplying by one fixed transposition pairs even and odd permutations to justify this count. ThereforeThe calculation also describes $A_4$ as the semidirect product $V⋊⟨X⟩$, with the order-three factor permuting the three involutions.

The direct product of groups uses componentwise multiplication:Associativity follows in each factor. The maps $g\mapsto (g,e_H)$ and $h\mapsto (e_G,h)$ are injective group homomorphisms onto subgroups, giving the required copies of $G$ and $H$.

Use the PDF's convention that the subscript in $D_n$ is the group order. Write the hexagon group as $D_{12}=⟨r,s\mid r^6=s^2=e,srs=r^{-1}⟩$ and the triangle group as $D_6=⟨a,b\mid a^3=b^2=e,bab=a^{-1}⟩$. Let $z$ generate $C_2$. In $D_{12}$, $r^2$ has order three and $s$ conjugates it to its inverse, so $⟨r^2,s⟩$ is a copy of $D_6$. The element $r^3$ has order two and is central, since $s{r}^{3}s=r^{-3}=r^3$.

DefineThe generator relations are preserved and $r^3$ commutes with the other images, so $\Phi$ is a group homomorphism from the direct product. The six residues $2i+3k$ are distinct modulo six: reduction modulo two first determines $k$, then reduction modulo three determines $i$. Together with $j$, they give all twelve distinct normal forms $r^m{s}^j$ in $D_{12}$. Thus $\Phi$ is bijective andThis is the dihedral splitting when the half-rotation order is odd with half-rotation order three.

Finally, $D_{12}$ contains an element of order six, namely $r$. The even cycle types on four symbols are the identity, a 3-cycle, and two disjoint transpositions; their orders are respectively one, three and two. Therefore $A_4$ has no element of order six. Since a group isomorphism preserves element orders,

At a critical point $x_0$ of a twice continuously differentiable real function, the Taylor expansion isThe Hessian matrix is real symmetric, so an orthogonal eigenbasis diagonalizes its quadratic form. If every eigenvalue is positive, that form is bounded below by a positive multiple of $|h{|}^2$, which dominates the remainder and gives a strict local minimum. If every eigenvalue is negative, the same argument gives a strict local maximum. If both signs occur, displacement along the corresponding eigenvectors produces values above and below $f(x_0)$, so the point is a saddle point of a scalar function. These exhaust a nonsingular Hessian matrix.

A singular Hessian matrix does not always make the classification impossible: nonzero eigenvalues of both signs still give a saddle. If the Hessian matrix is positive or negative semidefinite, higher-order terms in its null directions must be examined. For example, $x^2+y^4$ and $x^2-y^4$ have the same Hessian matrix $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle. The function $x^2$ has a non-strict minimum along a whole line.

For the given polynomial,A stationary point has $y=3x^2/2$ and then $x(2\alpha -x^2)=0$. Hence the origin is always critical, and two additional points exist exactly for $\alpha >0$:The Hessian matrix isAt the origin it is $\mathrm{diag}(2\alpha ,2)$, giving a strict local minimum for $\alpha >0$ and a saddle for $\alpha <0$. At either extra critical point it has determinant $64\alpha -36(2\alpha )=-8\alpha <0$, so both extra points are saddles.

At $\alpha =0$, complete the square:Along $x=0$ it is positive away from zero, whereas along $y=3x^2/2$ it is negative away from zero. Thus the origin remains a degenerate saddle despite its nonnegative Hessian matrix. This is the parabolic quartic critical-point classification. The complete result isFor $\alpha >0$ the local minimum is not global, since along the same parabola $f=\alpha x^2-x^4/4\to -\infty$.

For a one-to-one continuously differentiable change of variables with nonzero Jacobian determinant, the change of variables formula rule iswhere $E$ is the transformed region and the inverse map covers $D$ once. The absolute value is required even if the coordinate map reverses orientation.

Here $x,y$ are positive. For $u=y/x$, $v=xy$, the inverse is $x=\sqrt{v/u}$ and $y=\sqrt{uv}$. On the first portion of $D$, the bounds give $v\ge 1$, $u\le 4$ and $v\le u$ because $x\le 1$. Together these give $1\le v\le u\le 4$; conversely these inequalities imply $1/2\le x\le 1$ and the original bounds. On the second portion, they give $u\ge 1$, $v\le 4$ and $v\ge u$, hence $1\le u\le v\le 4$, with the converse following similarly. The two triangles combine into the rectangle $E=\{1\le u\le 4,1\le v\le 4\}$, sharing only the diagonal corresponding to $x=1$.

The Jacobian isAlso $x{y}^3=u{v}^2$ and $x^2+y^2=v(1+u^2)/u$. The transformed integral consequently factors:Therefore

The divergence theorem states that a continuously differentiable vector field on a bounded region $V$ satisfieswith outward unit normal $\mathbf{n}$ and a sufficiently regular boundary; piecewise smooth boundaries are also allowed. For $\mathbf{u}=\mathbf{c}\Omega$ with constant $\mathbf{c}$, the left side is $\mathbf{c}\cdot {\int }_V\nabla \Omega dV$ and the right side is $\mathbf{c}\cdot {\int }_{\partial V}\Omega \mathbf{n}dS$. Equality for every constant vector gives the gradient volume-to-boundary identityFor the cone and its cap, the volume has $0\le z\le 1$ and $0\le r\le \sqrt{3}z$ in cylindrical coordinates. Its volume is ${\int }_0^1\pi (\sqrt{3}z{)}^{2}dz=\pi$. Since $\nabla \Omega =-{\mathbf{e}}_z$,The cap at $z=1$ has area $3\pi$, outward normal ${\mathbf{e}}_z$ and $\Omega =a-1$, so its contribution to the surface integral is $3\pi (a-1){\mathbf{e}}_z$.

Parameterize the side by $\mathbf{R}(r,\theta )=(r\cos \theta ,r\sin \theta ,r/\sqrt{3})$, $0\le r\le \sqrt{3}$, $0\le \theta <2\pi$. The outward vector area element ispointing radially outward and downward from the region above the cone. The radial part integrates to zero around the circle. The vertical part of the side integral isAdding cap and side givesThe apex and circular rim have zero surface area. Alternatively, removing a tiny apex neighborhood and taking its radius to zero justifies applying the smooth-boundary theorem there; the additional flux vanishes because the scalar field is bounded.

Let ${\varphi }_1,{\varphi }_2$ be two sufficiently regular solutions with the same source and boundary data, and set $w={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}w=0$ and $\alpha {\partial }_{n}w+w=0$. Green's first identity givesThe left side is nonnegative, so it vanishes. Thus $w$ is constant on each connected component. Its normal derivative is zero, and the boundary condition then forces that constant to be zero. This proves at most one solution. The Poisson uniqueness with a nonnegative Robin normal coefficient proof never divides by $\alpha$, so it includes the points or boundary portions where $\alpha =0$.

For the exponential-sine modes, direct differentiation givesTheir sum is zero, so both modes are harmonic functions for every real $\ell$.

For the square, the horizontal Dirichlet data suggest separation of variables in the form $\varphi (x,y)=F(x)\sin (ky)$, where the positive integer $k$ makes the sine vanish at both horizontal edges. The Laplace equation gives $F^{\prime \prime }-k^{2}F=0$, hence $F=A\cosh (kx)+B\sinh (kx)$. At $x=0$, the outward normal points in the negative $x$ direction. The Robin boundary condition is therefore $F(0)-F^{\prime }(0)=0$, or $A=kB$. The right edge requires $F(\pi )=1$, fixing the remaining coefficient. The resulting separated Laplace mode with a Robin edge isThe denominator is positive for $k>0$. The function satisfies the Laplace equation and all four boundary conditions directly; for example its left-edge value and $x$ derivative agree, giving the correct outward-normal Robin sign. It extends regularly to the corners.

On the left edge the uniqueness coefficient is $\alpha =1$; on the other three edges it is $\alpha =0$ with the prescribed Dirichlet value. The same Green's first identity argument applies to the square's piecewise smooth boundary, whose corners have zero boundary measure. Thus this is the unique regular solution, or equivalently the unique finite-energy solution with these boundary traces. Unbounded corner singularities are outside that boundary-value class.