Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3

Since $AB=0$, the image of a linear map represented by $B$ lies in the kernel of $A$. The matrix $B$ is nonzero, so this image of a linear map contains a nonzero vector; consequently $\dim \ker A\ge 1$. On the other hand, $A\ne 0$ gives $\mathrm{rank}A\ge 1$. The rank-nullity theorem in dimension two therefore givesThus the image of $A$ is a line through the origin, and $A$ maps onto that line.

Reversing the order of the matrix multiplication need not give zero. For example,gives $AB=0$ but $BA=B\ne 0$. This also shows why information about one composition of linear maps does not establish the corresponding property of the reversed composition.

Write $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Its rank is one, so its determinant is zero. Choose its adjugate matrix:Direct matrix multiplication gives $AC=CA=(ad-bc)I=0$. Since the entries of $C$ are the entries of the nonzero matrix $A$, rearranged with possible sign changes, $C\ne 0$. This establishes both required annihilation properties.

A group $G$ is a cyclic group if some element $g\in G$ generates it:For a finite cyclic group, the order of a group element of a generator equals the number of elements of the group.

An example of a finite abelian group that is not a cyclic group is the Klein four-group, realized aswith componentwise addition modulo two. Addition commutes, so this is an abelian group of four elements. Each of its three nonzero elements satisfies $g+g=0$, so the cyclic subgroup it generates has just two elements; the identity generates only itself. No element generates all four elements.

Yes: every finite group of rotations about the origin in the plane is cyclic. A planar rotation is determined by an angle modulo $2\pi$, and composition adds those angles. If $G$ consists only of the identity, it is a cyclic group. Otherwise, finiteness allows us to choose the smallest positive representative angle $\alpha \in (0,2\pi )$ occurring in $G$.

Take any planar rotation in $G$ with representative angle $\theta \in [0,2\pi )$. Write $\theta =q\alpha +r$, where $q$ is an integer and $0\le r<\alpha$. The composition $R_{\theta }{R}_{\alpha }^{-q}$ belongs to $G$ and has angle $r$. Minimality of $\alpha$ forces $r=0$, so every element of $G$ is a power of $R_{\alpha }$.

Similarly, write $2\pi =m\alpha +r$ with $0\le r<\alpha$. If $r>0$, the planar rotation $R_{\alpha }^{-m}$ has representative angle $r$, a contradiction. Thus $\alpha =2\pi /m$ andThis proves finite groups of planar rotations are cyclic by an elementary angle argument.

For an exact differential $Md\theta +Nd\varphi =dF$, equality of mixed partial derivatives requires $M_{\varphi }=N_{\theta }$. Here the two partial derivatives areThey are not identically equal, so the differential form is not a closed differential form and hence is not an exact differential. Equality along special lines such as $\theta =0$ does not supply a potential on a two-dimensional open neighborhood, because every such neighborhood contains points where the two partial derivatives differ.

To construct a potential for the exact differential, first integrate its $dx$ coefficient with respect to $x$:Matching the $dy$ coefficient gives $g_y=-2z^2$, hence $g=-2y{z}^2+h(z)$. The remaining partial derivative is thenMatching the $dz$ coefficient forces $h^{\prime }=0$. Therefore a potential, valid throughout ${\mathbb{R}}^3$, isDifferentiating this expression recovers all three coefficients, proving that the differential form really is an exact differential.

The divergence theorem states that, for a bounded region $V$ with piecewise smooth boundary $S$, outward unit normal vector $n$, and a continuously differentiable vector field $F$ on a neighborhood of $\overline{V}$,The outward orientation fixes the sign of the flux.

On the sphere of radius $R$, the outward unit normal vector is $\mathbf{r}/R$. Thus $r^n\mathbf{r}\cdot n=R^{n+1}$. Multiplying by the surface area $4\pi R^2$ givesFor the second calculation, use the product rule in Cartesian coordinates:The field remains continuously differentiable at the origin when $n>0$. Indeed, for $r>0$ its Jacobian matrix entries are $r^n{\delta }_{ij}+n{r}^{n-2}{x}_i{x}_j$, all tending to zero as $r\to 0$; its derivative at zero is also zero because $|r^n\mathbf{r}|/r=r^n\to 0$. The divergence theorem therefore applies to the entire ball. In spherical coordinates, it yieldsin agreement with the direct surface integral.

Using Einstein summation convention and the contraction of two Levi-Civita symbols,This proves the vector triple product identity directly from the Levi-Civita symbol and Kronecker delta.

Define the scalar triple product by $[a,b,c]=a\cdot (b\times c)$. Its expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by cyclically relabeling the three indices: a three-cycle has positive sign. ThusIn particular, let $\Delta =[e_1,e_2,e_3]$. Since the three vectors form a basis, their determinant $\Delta$ is nonzero, so their reciprocal basis vectors are well defined. Cyclic invariance and vanishing of a scalar triple product with repeated vectors giveTo calculate the new scalar triple product explicitly, apply the already proved vector triple product identity with $A=e_3\times e_1$:Consequentlyso the reciprocal vectors also form a basis. Taking the reciprocal basis of that basis now givesThe same cyclic calculation gives ${\hat{\hat{e}}}_2=e_2$ and ${\hat{\hat{e}}}_3=e_3$, proving reciprocal basis involution.

Finally, write $V=V_1{e}_1+V_2{e}_2+V_3{e}_3$. Taking a dot product with ${\hat{e}}_j$ isolates $V_j$. Applying the same argument to the reciprocal basis, whose reciprocal is the original basis, gives the two sets of coordinates:No orthonormality assumption is needed; the reciprocal basis is what makes these coordinate-extraction formulas work.

Set $w=(x-y)/2$. The sum condition then gives $x=a+w$ and $y=a-w$, and their dot product isTherefore $|w{|}^2=|a{|}^2-c$. Since $c<|a{|}^2$, this is a strictly positive squared radius. The complete solution isEvery choice of unit vector $n$ satisfies both conditions, and every solution has this form. The endpoints lie on the sphere with center $a$ and radius $R$, and their midpoint is $a$. They are therefore opposite ends of one diameter of that sphere.

Translate one vertex of the tetrahedron to the origin and denote the other vertex vectors by $a,b,c$. Translation preserves edge vectors and their dot products. We may label the two known perpendicular pairs of opposite edges of a tetrahedron as the pairs represented by $a,c-b$ and $b,a-c$. Their orthogonality givesHence all three pairwise dot products have the same value $s$. The third pair satisfiesso the third pair of opposite edges is also perpendicular.

Expanding squared lengths using the dot product givesThus the sums of squared opposite-edge lengths are equal. These two conclusions are the basic identities for orthogonal opposite edges of a tetrahedron.

For the linear map $\alpha$ represented by $A$, a solution to the system of linear equations exists precisely when $b\in \mathrm{im}\alpha$. If $x_0$ is one solution, thenThus the affine solution space of a linear equation completely describes the possibilities: no solutions when $b\notin \mathrm{im}\alpha$; one solution when $b\in \mathrm{im}\alpha$ and $\ker \alpha =\{0\}$; and infinitely many solutions when $b\in \mathrm{im}\alpha$ and $\ker \alpha \ne \{0\}$. In the last case, a nonzero $w\in \ker \alpha$ supplies the distinct solutions $x_0+tw$ for every real $t$. In dimension three, the rank-nullity theorem also says that a trivial kernel is equivalent to an invertible matrix $A$.

For the matrix equation, $AX=B$ means that $\beta =\alpha \circ \chi$, where $\chi$ is represented by $X$. Necessarily $\mathrm{im}\beta \subseteq \mathrm{im}\alpha$. Conversely, if that containment holds, choose an $\alpha$-preimage of each $\beta (e_j)$ for the three standard basis vectors, and use those preimages as the columns of $X$. Their linear extension gives $AX=B$. This proves the image criterion for a matrix factorization:Whenever it is solvable, every solution is $X_0+N$ with $AN=0$. It is unique exactly when $\ker \alpha =\{0\}$. Indeed, a nonzero $w\in \ker \alpha$ produces a nonzero matrix $N$ with first column $w$ and other columns zero, so uniqueness fails for any compatible $B$ if the kernel is nontrivial.

For the printed numerical matrix $A$, its determinant is $-2$, so the solution is unique. To calculate it, solve $A(x,y,z{)}^T=(b_1,b_2,b_3{)}^T$. The last equation gives $z=b_3-3y$; the second then gives $x=b_2-b_3+y$. Substitution into the first yieldsApplying these formulas to the three columns of the PDF's $B$ givesMultiplication gives $AX=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 3& 1& 2\end{array}\right)$, confirming the result against the printed matrix.

The three known vertices have pairwise squared distances $8$, so the regular tetrahedron must have every squared edge length equal to $8$. Comparing the squared distance from $d$ to $b$ with that from $d$ to $a$ givesEquality forces $x=-1$. Similarly,so $y=-1$. The resulting point has squared distance $8$ from each of the other three vertices. Therefore

Take the half-turn about the $z$-axis, represented by the rotation matrixIt is an orthogonal matrix with determinant $1$, and it fixes the $z$-axis while reversing both perpendicular coordinates. Direct substitution gives $Ma=b$, $Mb=a$, $Mc=d$ and $Md=c$. A linear map that permutes the four vertices also preserves their convex hull, which is the solid tetrahedron. Hence this rotation matrix leaves $T$ invariant. Its axis joins the midpoints of the two opposite edges $ab$ and $cd$.

The chain rule along the specified straight line givesHere equality of mixed partial derivatives follows from smoothness. Apply Taylor theorem in the single variable $t$:Writing $h_1=x_1-a=\alpha t$ and $h_2=x_2-b=\beta t$ produces the linear and quadratic termswith all displayed partial derivatives evaluated at $(a,b)$. For a remainder valid uniformly over directions, apply the Taylor formula with integral remainder to $s\mapsto f((a,b)+sh)$:Continuity of the Hessian matrix makes the difference between the integral and $\frac{1}{2}{h}^{T}H(a,b)h$ be $o(|h{|}^2)$. The printed ellipsis is therefore justified as a finite Taylor expansion with a remainder. Smoothness alone does not assert convergence of an infinite Taylor series.

At a stationary point, the linear term vanishes. The symmetric Hessian matrix satisfiesIf $H_{11}>0$ and $\det H>0$, this is a positive-definite quadratic form. Its minimum over the unit circle is a positive number $c$, so $h^{T}Hh\ge c|h{|}^2$. The Taylor remainder is smaller than $c|h{|}^2/4$ for sufficiently small $h$, proving $f((a,b)+h)>f(a,b)$ for nonzero nearby $h$. Thus the stationary point is a strict local minimum.

For the given polynomial, the gradient equations reduce toThe stationary points are $(0,0),(1,-1),(-1,1)$, andAt both points with $x_1=\pm 1$, we have $H_{11}=10>0$ and $\det H=16>0$. ConsequentlyAt the origin the determinant of the Hessian matrix is $-8$, giving a saddle point. An independent check is the identity $f=(x_1^2-1{)}^2+(x_1+x_2{)}^2-1$, which proves that the two displayed local minima are also the only global minima.

Choose coordinates that measure the curved boundary and the slope of a ray:On the positive interior, the inverse change of variables isThe two rays become $v=a$ and $v=1$, while the curve becomes $u=1$. The interior of $S$ therefore maps to the rectangle $0<u<1$, $a<v<1$. For $a=0$, the lower edge $v=0$ describes a limiting boundary, since the original region is unbounded; the origin also has no uniquely defined ratio $y/x$. For $a=1$ the region has zero area. These endpoint cases do not invalidate the change of variables on the positive interior.

For the three-dimensional integral, retain $z$ as the third coordinate. The transformed region isThe planar Jacobian determinant isThe integrand cancels this Jacobian determinant especially simply:HenceOne can first use the change of variables formula with $z\ge \epsilon >0$ and then pass to the limit. The nonnegative integrand allows this by the monotone convergence theorem, so the unbounded limiting slice at $z=0$ causes no omitted contribution.

In Einstein summation convention, the curl of a cross product has componentsThe second line uses the contraction of two Levi-Civita symbols; the last uses the product rule. These components prove the curl of a cross product identity, including both divergence and directional-derivative terms.

Now apply that identity to $m\times v$. On the orientable surface, $m=n$, $m\cdot v=0$, and $|m|=1$. Differentiating $m\cdot m=1$ gives $m_i{\partial }_j{m}_i=0$. Taking a dot product with $m$ in the curl of a cross product formula therefore gives, on $S$,Only tangency on the surface is needed; no claim is made that the extension $v$ is tangent away from it. The right side is the surface divergence of the tangent vector field.

With the boundary oriented according to Stokes theorem, cyclic invariance of the scalar triple product giveswhere $u$ is the outward conormal of a surface boundary. Thus Stokes theorem provesThis is the surface divergence theorem for a tangent field.

For the disc in the $z=0$ plane, take $n=(0,0,1)$ and $v=\mathbf{r}=(x,y,z)$. The field is tangent on the disc, and ${\partial }_j{v}_i={\delta }_{ij}$. The contraction is $3-|n{|}^2=2$, so the surface integral is $2\pi R^2$. On the positively oriented circular boundary, $u=(\cos \theta ,\sin \theta ,0)$ and $v=Ru$, so the boundary flux is $R(2\pi R)=2\pi R^2$. Both sides agree.

In Cartesian coordinates, put $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$. Away from the origin the partial derivatives of $\psi =1/r$ areSumming the three second partial derivatives yieldsThus $1/r$ is a harmonic function on the punctured space and solves Laplace equation there. The origin is excluded: the singularity cannot be treated as an ordinary harmonic point.

Use the product rule for the divergence of the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two dot products cancel. Apply the divergence theorem with outward unit normal vector $n$ to obtain Green's second identity:Smoothness of the functions and boundary supplies the regularity needed for the divergence theorem.

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.