Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its centre is $2\sqrt{\alpha \beta }$ on the real axis and its radius is $\alpha +\beta$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

Put $q=e^{i\theta }$. Since $q\ne 1$, the finite geometric series giveswhere the second equality multiplies numerator and denominator by $1-\overline{q}$. Taking imaginary parts gives the finite trigonometric sumTaking real parts of the same finite geometric series givesThe denominator is nonzero by the restriction on $\theta$. Equivalently, these finite trigonometric sums are $\sin (N\theta /2)\sin ((N+1)\theta /2)/\sin (\theta /2)$ and $\sin (N\theta /2)\cos ((N+1)\theta /2)/\sin (\theta /2)$, respectively.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

For $0<x<\pi /2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\le x\le \tan x$. HenceThe cosine is a continuous function at zero, so the squeeze theorem yields the right-hand limit of a function $1$. The quotient $\sin x/x$ is even, so the left-hand limit of a function is the same. Thus the limit is $1$.

For $x$ sufficiently close to zero, $1+x>0$. Apply the mean value theorem to the logarithm between $1$ and $1+x$: there is a point ${\xi }_x$ between these endpoints withAs $x\to 0$ from either side, ${\xi }_x\to 1$, so this quotient tends to $1$. The exponential function is continuous, and thereforeThe hypotheses of the mean value theorem hold because $\log t$ is continuous and differentiable for $t>0$, with derivative $1/t$.

For $x>0$, the exponent $x/(1+x)$ lies in $(0,1)$, and $0\le {\cos }^{4}x\le 1$. ConsequentlyThe last inequality uses the positive-term power series $e^x={\sum }_{k\ge 0}{x}^k/k!\ge x^2/2$ for $x>0$. The final bound tends to zero, so the squeeze theorem gives limit $0$.

A complex power series with radius of convergence $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the complex plane.

If $p$ is the zero polynomial, every coefficient vanishes, so $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,The ratio test therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the polynomial-coefficient power series has $R=1$ for every nonzero polynomial. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.

The Cauchy-Schwarz inequality in ${\mathbb{R}}^n$ isHere the Euclidean norm is induced by the dot product. If $b=0$, the result is immediate. If $b\ne 0$, the squared Euclidean norm of the component of $a$ perpendicular to $b$ is nonnegative:Rearrangement proves the Cauchy-Schwarz inequality. Equality holds precisely when $a$ and $b$ are linearly dependent, including the cases of a zero vector.

Expanding the squared Euclidean norm and using the Cauchy-Schwarz inequality givesTaking nonnegative square roots gives the triangle inequality. Applying the triangle inequality once more gives

Subtract the two constraints to obtain $x\cdot (a-b)=A-B$. The Cauchy-Schwarz inequality gives $|A-B|\le \Vert x\Vert \Vert a-b\Vert$. Since $a\ne b$, division and squaring give the distance bound from two hyperplane constraintsIf $a=b$ and $A\ne B$, the intersection is empty. If $a=b\ne 0$ and $A=B$, the constraints describe the same hyperplane. Its points satisfy $\Vert x{\Vert }^2\ge A^2/\Vert a{\Vert }^2$, with equality at $x=Aa/\Vert a{\Vert }^2$. The fraction involving $a-b$ is undefined in this case and cannot be used.

If $a=b=0$, both equations are satisfied by every $x$ when $A=B=0$, and otherwise there is no solution. These degenerate cases distinguish an inconsistent constraint from a repeated one.

Apply the triangle inequality to the path from $x_1$ through $y_2$ to $y_1$:This is one direction of the reverse triangle inequality. A second application of the triangle inequality, this time through $x_2$, givesThe resulting four-point triangle inequality follows without any assumptions on coincidences among the four points.

Write a vector in the domain as $(u,v,w,t{)}^T$. The first and third equations defining the kernel of a linear map imply $v=2\alpha u-2t$. Substitution into the second equation then givesFor $\alpha \ne 1$, this forces $t=(\alpha +1)u$, $v=-2u$ and $w=3u$. HenceThe kernel of a linear map has dimension one, so the rank-nullity theorem gives rank three and proves the assertion about the image of a linear map. Alternatively, the minor formed from columns $2,3,4$ has determinant $2(\alpha -1)$, including a nonzero value at $\alpha =-1$.

For $\alpha =1$, the parameters $u,t$ are free and $v=2u-2t$, $w=-3u+3t$. ThusThe third row is the first row minus the second. The image of a linear map is therefore contained in $y_1-y_2-y_3=0$. The first and third columns, $(1,2,-1{)}^T$ and $(1,0,1{)}^T$, are linearly independent, so the matrix rank is two and this containment is equality:This is the sole kernel jump in a parameter-dependent linear map; there is no further exceptional case at $\alpha =-1$.

Let $e_1,\dots ,e_n$ be the standard basis and define $a_j=f(e_j)$. By linearity,Thus every linear functional has this form. Its representing vector is unique because evaluating $a\cdot x=b\cdot x$ at each $e_j$ gives $a_j=b_j$.

For the prescribed kernel of a linear map, the third listed vector is the first minus the second, so it adds no independent constraint. The first two requireTaking $a_1=s$ and $a_4=t$ gives $a_2=2s-2t$, $a_3=-3s+3t$. Consequently all possible representing vectors areConversely, every such vector is perpendicular to each of the three prescribed vectors, so its linear functional vanishes on them. This also includes the zero linear functional.

The characteristic polynomials, in the convention $\det (\lambda I-T)$, areSolving the eigenvalue equations gives the complete eigenspacesEvery nonzero vector in a listed eigenspace is an eigenvector, and these are all the eigenvectors. For $A$, the repeated eigenvalue $2$ has a two-dimensional eigenspace; for $B$, all three eigenvalues are simple. The three displayed generating vectors form a basis because their column matrixhas determinant $2$. Thus both matrices are diagonalizable, with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this basis.

Suppose $C=S{\Lambda }_C{S}^{-1}$ and $D=S{\Lambda }_D{S}^{-1}$ for an invertible $S$ and diagonal matrices ${\Lambda }_C,{\Lambda }_D$. Since diagonal matrices commute,Therefore simultaneous diagonalization implies commutation. This argument works over either $\mathbb{R}$ or $\mathbb{C}$ and explains why simultaneous diagonalization produces commuting operators.

Let $Dx=\lambda x$ with $x\ne 0$. Commutation givesThus $Cx$ belongs to the eigenspace $E_{\lambda }(D)$, with the zero vector allowed. Since $D$ has $n$ distinct eigenvalues, each eigenspace is one-dimensional. Hence $Cx=\mu x$ for some scalar $\mu$. This is the fact that commuting maps preserve eigenspaces.

The standard linear independence theorem for eigenvectors belonging to distinct eigenvalues supplies a basis $x_1,\dots ,x_n$ of eigenvectors of $D$. The preceding argument makes every one an eigenvector of $C$ as well. Their column matrix therefore simultaneously diagonalizes $C$ and $D$. This is simple spectrum gives simultaneous diagonalization.

For real matrices, this conclusion always holds over $\mathbb{C}$; a real common basis follows when the distinct eigenvalues of $D$ are real. Real matrix entries alone do not ensure a real eigenbasis: a quarter-turn of ${\mathbb{R}}^2$, with $C=D$, has eigenvalues $\pm i$. The matrices in this question have real spectra, so their common basis is real.

Direct multiplication givesTake $D=B$ and $C=A$ in the preceding result: $B$ has the three distinct eigenvalues $2,4,6$ and the matrices commute. The common eigenbasis from part (i) givesIndeed $AS=S\mathrm{diag}(1,2,2)$ and $BS=S\mathrm{diag}(2,4,6)$, and $\det S=2\ne 0$. These equalities directly verify the simultaneous diagonalization.

For the standard complex inner product, $\Vert y{\Vert }^2=y^{†}y$. Using the conjugate transpose and the defining equality for a normal matrix,Both norms are nonnegative, so $\Vert Ax\Vert =\Vert A^{†}x\Vert$ for every $x$.

Set $M=A-\lambda I$. Its conjugate transpose is $M^{†}=A^{†}-\overline{\lambda }I$, soSince $A$ is a normal matrix, the two products coincide. Therefore $A-\lambda I$ is normal for every complex $\lambda$.

If $Ax=\lambda x$, then $M=A-\lambda I$ is a normal matrix by part (ii), and $Mx=0$. Part (i), applied to $M$, yields $\Vert M^{†}x\Vert =\Vert Mx\Vert =0$. ThusThe nonzero vector $x$ is consequently an eigenvector of the conjugate transpose with eigenvalue $\overline{\lambda }$. This adjoint eigenvector identity for a normal matrix will also determine the special spectral restrictions below.

The adjoint eigenvector identity for a normal matrix gives $A^{†}{x}_{\lambda }=\overline{\lambda }{x}_{\lambda }$. HenceAs $\mu \ne \lambda$, $x_{\lambda }^{†}{x}_{\mu }=0$. Thus eigenvectors in distinct eigenspaces of a normal matrix are orthogonal in the complex inner product.

Start with the assumed basis of eigenvectors, grouping them by eigenvalue. In each eigenspace, apply the Gram-Schmidt process, which replaces a linearly independent list by an orthonormal list with the same span. Every new vector stays within that eigenspace, since an eigenspace is a vector subspace.

By part (iv), different eigenspaces are orthogonal. The combined lists therefore form an orthonormal basis of the entire space, still consisting of eigenvectors. Let $U$ have these vectors as columns. Then $U$ is a unitary matrix, soThis gives the required unitary diagonalization of a normal matrix, including repeated eigenvalues.

For a Hermitian matrix, $A^{†}=A$, so $A^{†}A=A^2=A{A}^{†}$ and $A$ is a normal matrix. If $Ax=\lambda x$ with $x\ne 0$, part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Since also $A^{†}x=Ax=\lambda x$, we obtain $\lambda =\overline{\lambda }$. Thus every eigenvalue is real.

For a skew-Hermitian matrix, $A^{†}=-A$, so both $A^{†}A$ and $A{A}^{†}$ equal $-A^2$. It is therefore a normal matrix. On an eigenvector with eigenvalue $\lambda$, part (a)(iii) givesThus $\overline{\lambda }=-\lambda$, and every eigenvalue lies in $i\mathbb{R}$, including the possible eigenvalue zero.

For a unitary matrix, $A^{†}=A^{-1}$, so both defining products for a normal matrix are $I$. The matrix is invertible, hence no eigenvalue is zero. If $Ax=\lambda x$, then $A^{-1}x={\lambda }^{-1}x$, while part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Therefore $\overline{\lambda }={\lambda }^{-1}$, andThus all eigenvalues have unit modulus.

Since $a_j=S_j-S_{j-1}$, shifting an index yields the summation by parts identityThe sum on the last line is empty when $m=n$, so the identity also covers that case.

The series with a bounded monotone multiplier converges. Here is a proof that also permits conditional convergence. Choose $M>0$ with $|b_j|\le M$. The Cauchy sequence criterion for the convergent series $\sum a_j$ ensures that, for any $\eta >0$ and sufficiently large $m$, all the partial tails $T_j={\sum }_{k=m}^j{a}_k$ satisfy $|T_j|<\eta$. Apply summation by parts to these partial tails:Since the multiplier is a monotone sequence, ${\sum }_{j=m}^{n-1}|b_j-b_{j+1}|=|b_m-b_n|$. ConsequentlyTaking $\eta =\epsilon /(3M)$ proves that the weighted partial sums form a Cauchy sequence, and hence converge. This is a special case of bounded-variation multipliers of a convergent series.

The series $\sum n^{1/n}{a}_n$ also converges. The multiplier need only be monotone eventually, since a finite initial segment cannot affect convergence. For $h(x)=\log x/x$,Thus $n^{1/n}=\exp (h(n))$ decreases for $n\ge 3$; it is bounded and tends to $1$, because $\log n/n\to 0$. Apply the proved multiplier result to the tail beginning at $3$. The fact that the first few values are not monotone causes no difficulty.

We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.

Away from integer multiples of $\pi$, the function is a composition of continuous functions. At zero, the bound $|g(x)|\le \sqrt{|x|}$ gives $g(x)\to 0=g(0)$ by the squeeze theorem. This is oscillation multiplied by a vanishing amplitude.

Fix $n\ne 0$ and set $t_k=\pi /2+2\pi k$ and $x_k=n\pi +(-1{)}^n\arcsin (1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, andSo $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its set of continuity points is

On $[0,\pi ]$, $g(x)\le \sqrt{x}<\sqrt{\pi }$ for $x<\pi$, while $g(\pi )=0$. With $x_k=\pi -\arcsin (1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to \sqrt{\pi }$. Consequently the supremum is $\sqrt{\pi }$ and is not attained. This illustrates that supremum can fail to be attained at an oscillatory endpoint.

On $[\pi ,3\pi /2]$, set $x_0=\pi +\arcsin (2/(3\pi ))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi /2$, so $g(x_0)=\sqrt{x_0}>\sqrt{\pi }$. Choose $0<\delta <x_0-\pi$. On $[\pi ,\pi +\delta ]$,On $[\pi +\delta ,3\pi /2]$, $g$ is a continuous function, so the extreme value theorem gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore $g$ does attain its supremum on $[\pi ,3\pi /2]$. This is the interior-value criterion for an attained maximum.

The intermediate value theorem says that if $f:[a,b]\to \mathbb{R}$ is a continuous function, then every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.

Endpoint values are already attained. Suppose $f(a)<y<f(b)$ and define $E=\{x\in [a,b]:f(x)<y\}$. It is nonempty and bounded above, so $c=\sup E$ exists. Continuity at $a$ ensures that $E$ contains points to the right of $a$, and continuity at $b$ ensures that no point sufficiently close to $b$ belongs to $E$. Hence $a<c<b$.

By the definition of the supremum, there are points of $E$ approaching $c$. If $f(c)>y$, continuity would exclude all such nearby points from $E$, a contradiction. If $f(c)<y$, continuity would put some point to the right of $c$ in $E$, again a contradiction. Therefore $f(c)=y$. When $f(a)>f(b)$, apply this argument to $-f$. This proves the intermediate value theorem in all cases.

For a chord of length $1/2$, consider the continuous function $h(x)=f(x+1/2)-f(x)$ on $[0,1/2]$. Since $f(0)=f(1)$,An endpoint zero already supplies the chord. Otherwise the endpoint values have opposite signs, so the intermediate value theorem supplies $h(\alpha )=0$ at some $\alpha \in (0,1/2)$. Thus a horizontal chord of length $1/2$ exists.

For each integer $n>1$, define $h_n(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. The sampled differences telescope:If any sampled value is zero, it gives the required chord. Otherwise there must be both a positive and a negative sampled value. The intermediate value theorem, applied between their sample points, then gives $h_n(\alpha )=0$. Hence a horizontal chord of length $1/n$ exists for every positive integer $n$; for $n=1$, the endpoint chord itself suffices. This proves the horizontal chord of reciprocal-integer length result, without restricting $n$ to powers of two.

For a partition $P$, write $U(f,P)$ and $L(f,P)$ for the upper Darboux sum and lower Darboux sum. The Riemann integrability criterion for bounded $f$ is that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. Thus if the gaps for $D_n$ tend to zero, the Riemann integrability criterion immediately proves that $f$ is Riemann integrable.

Conversely, suppose $f$ is Riemann integrable, and choose a fixed partition $P$ with $U(f,P)-L(f,P)<\epsilon /2$. Let $r$ be the number of its interior division points and choose $M\ge 1$ with $|f|\le M$. Call a cell of $D_n$ bad when its interior contains a division point of $P$. There are at most $r$ bad cells, and their total length is at most $r/n$.

Every other cell lies in a single cell of $P$, so its oscillation is no larger than that of the containing cell. Summing the contributions of these good cells gives at most $U(f,P)-L(f,P)$, while each bad cell has oscillation at most $2M$. Therefore the finite bad-cell estimate for Darboux sums givesFor large enough $n$, the right-hand side is less than $\epsilon$. This proves the uniform-mesh Darboux criterionThis argument does not assume that the partitions $D_n$ are nested; in general they are not.

For the composition, $f$ takes values in $[-M,M]$. Since $g$ is continuously differentiable, the extreme value theorem bounds both $|g|$ and $|g^{\prime }|$ on that interval. In particular, let $K={\sup }_{[-M,M]}|g^{\prime }|<\infty$. The mean value theorem states that a function continuous on the interval between $s,t$ and differentiable in its interior satisfies $g(s)-g(t)=g^{\prime }(\xi )(s-t)$ for some intermediate $\xi$. HenceOn each partition cell, this bounds the oscillation of $g\circ f$ by $K$ times the oscillation of $f$, whether or not the local extrema are attained. It follows thatThe composition is bounded, so the proved uniform-mesh Darboux criterion applies. Thus $g\circ f$ is Riemann integrable. The reusable fact is Lipschitz composition preserves Riemann integrability; continuous differentiability supplies the required bound on the range of $f$.