Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 3

Let $c=1/\sqrt{2}$. A concrete real matrix isIts first coordinate has eigenvalue $-1$. The lower block has characteristic polynomial $(\lambda -c{)}^2+c^2$, whose roots are $c\pm ic=(1\pm i)/\sqrt{2}$. Thus this example has all three required eigenvalues. It is also an orthogonal matrix, but orthogonality is not forced by its spectrum, as the second part shows.

True. All three eigenvalues are distinct in $\mathbb{C}$, so their nonzero eigenvectors are linearly independent. Here is a direct proof of that assertion. If $a_1{v}_1+a_2{v}_2+a_3{v}_3=0$, apply $(A-{\lambda }_{2}I)(A-{\lambda }_{3}I)$ to obtainDistinctness makes both scalar factors nonzero, hence $a_1=0$. Applying the analogous products gives $a_2=a_3=0$. The three eigenvectors therefore form a basis of ${\mathbb{C}}^3$. Taking them as the columns of $P$ givesThis proves that every such matrix is a diagonalizable matrix over $\mathbb{C}$, rather than merely showing it for the example.

False. Consider the real matrixIt is block upper triangular, so its characteristic polynomial is the same as that of $A_0$ and it has the required eigenvalues. Its second column, however, has squared Euclidean norm $1+c^2+c^2=2$, whereas every column of an orthogonal matrix has norm one. ThusUnit-modulus eigenvalues alone do not imply orthogonality; they do not control the geometry of the eigenvectors.

False; in fact no matrix with this spectrum is diagonalizable over $\mathbb{R}$. If a real invertible $P$ gave $P^{-1}AP=D$ with $D$ a real diagonal matrix, all eigenvalues of $A$ would be the real diagonal entries of $D$. But two of the required eigenvalues have nonzero imaginary parts. This contradiction provesThe real block form in the initial example is consistent with this: a two-dimensional rotation block is not a real diagonalization.

Both subgroups contain the identity element, so $H\cap K$ is nonempty. If $x,y\in H\cap K$, then $x{y}^{-1}$ belongs to both $H$ and $K$, hence to their intersection. This gives closure under inverses by setting $x=e$, and closure under products by using the inverse of $y$. Therefore $H\cap K$ is a subgroup of $G$, and also of both $H$ and $K$.

Lagrange's theorem for finite groups states that the order of a subgroup divides the order of its finite ambient group; more precisely $|L|=[L:M]|M|$ for $M\le L$. Applying it twice givesSince $p$ and $q$ are coprime, their only positive common divisor is one. Thus

Take the unit normal to point outwards. The divergence of the field isThe divergence theorem turns the closed surface integral into a volume integral over $[0,1{]}^3$:For a direct check of the surface integral, on $x=1$ the outward component is $F_x=2y$, contributing ${\int }_0^1{\int }_0^{1}2ydydz=1$. On $x=0$ it vanishes. On both $y=0$ and $y=1$, $F_y=xy(y-1{)}^2=0$. On $z=1$, $F_z=0$, whereas on $z=0$ the field component is $F_z=-1$ and the outward normal is $-\hat{\mathbf{z}}$, giving a contribution $1$. The six face contributions are therefore $1,0,0,0,0,1$, again totaling 2. The positive bottom-face contribution is a consequence of its outward orientation.

Use suffix notation, summing repeated indices, and write ${\partial }_i=\partial /\partial x_i$. For a twice continuously differentiable vector field, mixed derivatives commute. Thus the divergence of its curl satisfiesbecause the derivatives are symmetric in $i,j$, while the Levi-Civita symbol is antisymmetric in those indices. HenceFor the second identity, use the contraction of two Levi-Civita symbols:Rearranging each component proves the vorticity cross-product identityThe second identity needs first derivatives; the first additionally uses equality of the mixed second derivatives.

For the centralizer of a subset, the identity commutes with every $h\in A$. If $g,k$ commute with each member of $A$, then for every such $h$,Also $gh=hg$ implies $g^{-1}h=h{g}^{-1}$ after multiplication by $g^{-1}$ on both sides. Thus identity, products and inverses lie in $C(A)$, proving that it is a subgroup.

For the conjugation action, identity acts trivially, andThese are the defining laws of a left group action. Its orbits of a group action are the conjugacy classes. The stabilizer of $h$ is exactly its centralizer, since $gh{g}^{-1}=h$ if and only if $gh=hg$. The orbit-stabilizer theorem therefore givesA class has one element exactly when that element commutes with all of $G$, so the singleton classes are precisely the elements of $C(G)$, the centre of a group. Partitioning $G$ into all its conjugacy classes now proves the class equationFor $|G|=p^r$, Lagrange's theorem for finite groups makes every class size a power of $p$. A nonsingleton class has size at least $p$ and is therefore divisible by $p$. Reducing the class equation modulo $p$ yields $|C(G)|\equiv 0(\mathrm{mod}p)$, because $r>0$. The identity lies in $C(G)$, so its order is positive and divisible by $p$. ConsequentlyThis proves the nontrivial center of a finite p-group directly from the conjugation classes.

A group homomorphism sends identity to identity. For $\theta (g),\theta (h)$ in its image,which is still in the image; hence the image of a group homomorphism is a subgroup of $H$. Let $N=\ker \theta$. It is a subgroup by the same identity, and for $k\in N$ and $g\in G$,Thus $gN{g}^{-1}\subseteq N$, and replacing $g$ by $g^{-1}$ gives equality, proving that $N$ is a normal subgroup.

Define a map from the quotient group byIf $gN=hN$, then $h^{-1}g\in N$ and $\theta (g)=\theta (h)$, so the map is well defined. Conversely equality of the images implies $h^{-1}g\in N$, giving injectivity. It is onto by the definition of image, and $\Phi ((gN)(hN))=\theta (gh)=\theta (g)\theta (h)$. This proves the first isomorphism theorem for groups in this case:An orthogonal matrix $Q$ has $Q^{\mathsf{T}}Q=I$, so $(\det Q{)}^2=1$. The determinant is therefore a group homomorphism $O(3)\to \{1,-1\}$, with image both signs because $\mathrm{diag}(-1,1,1)$ has determinant $-1$. Its kernel is the special orthogonal group $SO(3)$. Hence $SO(3)$ is normal in $O(3)$, andFor the last request, the determinant-twist projection in odd dimension gives an explicit example:Indeed $\det \pi (Q)=(\det Q{)}^4=1$, and scalar factors commute, so $\pi (QR)=\pi (Q)\pi (R)$. If $\pi (Q)=I$, then $Q=(\det Q)I$, making $Q=I$ or $Q=-I$. Both lie in the kernel, which is therefore exactly $\{I,-I\}$, of order two. The map also fixes every member of $SO(3)$, so it is onto.

Since $\sigma$ is a group homomorphism, $\sigma (0)=I$, $\sigma (x+y)=\sigma (x)\sigma (y)$, and $\sigma (-x)=\sigma (x{)}^{-1}$. The product is closed on $\mathbb{R}\times {\mathbb{R}}^2$. Its two associative bracketings giveand coincide by the homomorphism property. The identity element is $(0,0)$, and the two-sided inverse isFor example, multiplying on the right gives $v-\sigma (x)\sigma (-x)v=0$, and multiplying on the left gives $-\sigma (-x)v+\sigma (-x)v=0$. Thus this law defines a group, specifically a semidirect product.

If it is an abelian group, comparing $(x,0)\ast (0,w)$ and $(0,w)\ast (x,0)$ gives $\sigma (x)w=w$ for every $w$. Hence $\sigma (x)=I$ for every $x$. Conversely this trivial action makes the product ordinary vector addition. Therefore the unique action giving a commutative group isFor the final condition, compare $(0,a{e}_1)\ast (x,w)$ and $(x,w)\ast (0,a{e}_1)$. They are $(x,a{e}_1+w)$ and $(x,w+\sigma (x)a{e}_1)$ respectively. They coincide for every $a,x,w$ exactly when $\sigma (x)e_1=e_1$ for every $x$. The first column of $\sigma (x)$ is then $(1,0{)}^{\mathsf{T}}$; since it lies in the special linear group, its determinant is one and its other diagonal entry is also one. ThusMatrix multiplication gives $r(x+y)=r(x)+r(y)$. Conversely any such additive function gives a homomorphism into $SL(2,\mathbb{R})$ fixing $e_1$ and therefore makes these translations central. This proves the full characterization of the central translations in a linear semidirect product:No continuity or measurability is assumed, so it would be unjustified to restrict the answer to $r(x)=cx$.

Let $T(z)=(az+b)/(cz+d)$, with $ad-bc\ne 0$. If $c=0$, then $d\ne 0$ anda composition of a nonzero scaling and a translation. If $c\ne 0$, rearrange the fraction asThus every Möbius transformation is a composition, in order, of translation by $d/c$, reciprocal inversion, nonzero scaling by $(bc-ad)/c^2$, and translation by $a/c$. All formulas act on the Riemann sphere, including zero and infinity. The scaling parameter must be nonzero for that factor to be a Möbius transformation.

For three distinct finite points, define the cross-ratio normalizationIt sends $z_1,z_2,z_3$ to $0,1,\infty$ respectively. If $z_3=\infty$, use $(\zeta -z_1)/(z_2-z_1)$; if $z_1=\infty$, use $(z_2-z_3)/(\zeta -z_3)$; if $z_2=\infty$, use $(\zeta -z_1)/(\zeta -z_3)$. These are the corresponding limits of the same formula and cover all possible triples. Form the analogous normalization $F_w$ of the target triple. Thenhas all three specified images, proving existence. For uniqueness, the quotient of any two such maps fixes the three source points. Conjugating by $F_z$ gives a Möbius transformation fixing $0,1,\infty$. Fixing zero and infinity forces it to be $\lambda \zeta$, and fixing one forces $\lambda =1$. The quotient is identity, so the two maps are equal.

The Möbius transformations permuting three points are therefore in one-to-one correspondence with permutations of the triple. For the normalized set the six maps and their ordered images of $(0,1,\infty )$ areThey realize all six permutations. The action map to the symmetric group is a homomorphism, is injective by the three-point uniqueness proof, and is onto by this list. Hence

Parametrize the straight segment by $x=1-t$, $y=t$, $0\le t\le 1$. Then $dx=-dt$, $dy=dt$, and $x^2+xy=1-t$. The line integral is

On the first leg $y=0$ and $x$ decreases from one to zero; on the second $x=0$ and $y$ increases from zero to one. Adding their line integrals givesThe negative sign in the first term follows from the direction of traversal.

For the anticlockwise arc use $x=\cos \theta$, $y=\sin \theta$, $0\le \theta \le \pi /2$. Since $dx=-\sin \theta d\theta$ and $dy=\cos \theta d\theta$, the line integral becomesThe two integrals multiplying $-\alpha$ each equal $1/3$, by substituting $\cos \theta$ and $\sin \theta$ respectively. The remaining integral equals one. Thus

When $\alpha =2\beta$, all three values reduce to $-\beta /3$. To prove path independence for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ givesIndeed ${\Phi }_x=2\beta (x^2+xy)=\alpha (x^2+xy)$ and ${\Phi }_y=\beta (x^2+y^2)$. The integrand is the exact differential $d\Phi$ throughout ${\mathbb{R}}^2$. The fundamental theorem for line integrals therefore gives, for every such path from $A$ to $B$,

For a smooth vector field near an oriented piecewise smooth surface $S$, Stokes theorem stateswhere the boundary orientation is induced by the surface unit normal through the right-hand rule. For a constant vector $\mathbf{k}$, the curl of $\varphi \mathbf{k}$ is $\nabla \varphi \times \mathbf{k}$. Apply Stokes theorem and use the scalar triple product:Since this holds for every constant $\mathbf{k}$, equality of all vector components proves the vector-valued Stokes identityFor the specified field, the two axis segments contribute zero because $x^2{y}^4$ vanishes on them. On the arc the direct line integral isIts two terms giveandThereforeOne can also check the orientation and sign by Stokes theorem. For the upward-oriented quarter disk, $(\nabla \times \mathbf{A}{)}_z=2x{y}^4-4x^2{y}^3$. Polar integration givesconsistent with the positively oriented boundary in the question.

Let $u={\varphi }_1-{\varphi }_2$. It is a harmonic function in the bounded region and vanishes on its entire boundary. For smooth solutions and boundary, the divergence theorem applied to $u\nabla u$ gives the energy identityThe integrand is nonnegative, so $\nabla u=0$ in $R$. Thus $u$ is constant on each connected component, and its zero Dirichlet boundary condition makes every constant zero. HenceThis proves the Uniqueness of the Dirichlet problem rather than assuming it.

For the separated expression, write $\varphi =F(x)G(y)$. The hyperbolic factor obeys $F^{\prime \prime }={\lambda }^{2}F$ and the trigonometric factor obeys $G^{\prime \prime }=-{\lambda }^{2}G$, sofor all the constants, including $\lambda =0$. In the half-strip take $a>0$ so the region has positive width. For every nonzero integer $n$, the decaying harmonic sine mode in a half-strip isIts second derivatives cancel, it vanishes at $y=0,a$, and it has the prescribed sine trace at $x=0$. The absolute value in the exponential ensures decay for negative as well as positive $n$. For $n=0$, the zero function is the solution. Thus all integer cases are covered, and the decay is uniform across the strip.

A Cartesian isotropic tensor is unchanged by every proper rotation matrix. For a rank-two tensor this means $B^{\prime }=RB{R}^{\mathsf{T}}=B$, or $B_{ij}^{\prime }=R_{ik}{R}_{jl}{B}_{kl}=B_{ij}$. No symmetry of $B$ is assumed. For a quarter-turn about the $z$ axis,Comparing entries yields $B_{11}=B_{22}$, $B_{12}=-B_{21}$, $B_{13}=-B_{23}=-B_{13}$, and $B_{31}=-B_{32}=-B_{31}$. In particular,The remaining matrix has the form $\left(\begin{array}{ccc}b& c& 0\\ -c& b& 0\\ 0& 0& d\end{array}\right)$. A quarter-turn about the $x$ axis givesInvariance now forces $c=0$ and $d=b$, proving that an isotropic second-rank tensor isa scalar multiple of the Kronecker delta.

For the final unheaded conductivity identity, let $P=n{n}^{\mathsf{T}}$, the orthogonal projection onto the unit vector $n$. Then $P^2=P$ and the uniaxial conductivity tensor is $\sigma =\alpha I+\gamma P$. The antisymmetric tensor $D$ sends a vector $v$ to $v\times n$. Applying it twice and using the vector triple product givesConsequentlyThis equals $-\sigma =-\alpha I-\gamma P$ exactly when $(\alpha +\gamma )P=0$. Since $P$ has rank one and is nonzero, the requested parameter isAt this value conductivity vanishes along $n$; it is precisely the singular case excluded in part (ii), so there is no conflict with that part's conclusion.

Writing the uniaxial conductivity tensor in vector form gives the electric current densityChoose the plane through the origin perpendicular to $n$:For an electric field in this plane, the anisotropic term vanishes and $\mathbf{J}=\alpha \mathbf{E}$. Thus every nonzero such field is parallel or antiparallel to its current according to the sign of the nonzero $\alpha$. The zero field gives zero current. This establishes collinearity throughout the plane without assuming that the tensor is isotropic.

Decompose the electric field into its orthogonal parts $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=n(n\cdot \mathbf{E})$. The uniaxial conductivity tensor acts on them asIf $\mathbf{J}=0$, its perpendicular and parallel components each vanish. Since $\alpha \ne 0$ and $\alpha +\gamma \ne 0$, this forces both field components to be zero. The contrapositive givesEquivalently, the tensor has nonzero eigenvalues $\alpha ,\alpha ,\alpha +\gamma$ and determinant ${\alpha }^2(\alpha +\gamma )\ne 0$.