Past exam of the mathematics course of the University of Cambridge / 2022 / ia

The triangle inequality givesHence the left-hand side of the proposed equation is always at least $|a-b|$. There are no solutions when $c<|a-b|$.

Equality in the triangle inequality holds exactly when the two displacement vectors $a-z$ and $z-b$ point in the same direction. Thus $z$ lies on the closed line segment joining $a$ and $b$. Conversely, every point of that segment satisfiesThe solution set is therefore precisely that segment in the complex plane.

By the focal definition of an ellipse, the solution set is the ellipse with foci $a$ and $b$. Its centre is $(a+b)/2$, its major axis lies along the line through the foci, and its semiaxes areThe sketch is consequently a nondegenerate ellipse symmetric about both the focal line and its perpendicular bisector.

The required root of unity issoTherefore

The logarithms of $1+\omega =e^{i\pi /3}$ areThe definition of complex exponentiation therefore gives all values asSince $1+\omega =\frac{1}{2}+i\frac{\sqrt{3}}{2}$, these can be written

Taking determinants of $M^{T}JM=J$ givesSince $\det J=-1\ne 0$, one has $\det (M{)}^2=1$, and henceBoth values occur, for example for $M=I$ and $M=\mathrm{diag}(1,-1)$.

If $M_1^{T}J{M}_1=J$ and $M_2^{T}J{M}_2=J$, then matrix multiplication and the transpose rule giveThus the product also satisfies the equation.

The equation and part (i) imply that $M_1$ is invertible. Multiplying
$M_1^{T}J{M}_1=J$ on the left by $(M_1^{-1}{)}^T$ and on the right by $M_1^{-1}$ givesHence the set of solutions is closed under products and matrix inverses.

WriteThe first column has unit length for the indefinite quadratic form:Because $p=M_{11}>0$, there is a unique $u\in \mathbb{R}$ such thatThe two columns are orthogonal for the same form and the second has squared length $-1$:The vectors with these properties areUsing the hyperbolic cosine and hyperbolic sine, the two possible families are therefore

The alternating series test says that if $a_n\ge 0$ decreases to $0$, thenconverges. Taking $a_n=n^{-1/2}$ proves convergence of the given series.

It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since $p=1/2\le 1$. Thus the original series has conditional convergence.

For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds $1$, then take the first unused negative term. Next take positive terms until the sum exceeds $2$, then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage $m$ tends to zero, while the preceding partial sum exceeds $m$; consequently these rearranged partial sums tend to $+\infty$. This is a divergent series with exactly the prescribed terms.

For the sum $F=f+g$:

For the product $G=fg$, the answer is no in both cases:

Every displayed $f$ and $g$ is a continuous function and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.

For $A=(A_1,A_2,A_3)$ and $B=(B_1,B_2,B_3)$, the scalar product is the inner productThe cross product isIt is perpendicular to both vectors and has magnitude $|A||B|\sin \theta$.

The vector triple product identity givesApplying the same identity after reversing the outer cross product givesThe formulas differ because the cross product is not associative.

LetLinear independence implies that the scalar triple product $Delta$ is nonzero. Applying the vector triple-product formulas to $(A\times B)\times (C\times D)$ recovers the coefficients of $D$ in the basis $A,B,C$. Equivalently, the reciprocal basis to $A,B,C$ isTaking scalar products with $D$ therefore yields

Because the sphere passes through the origin and has centre $P$, its radius is $|P|$. A point $X$ lies on it exactly whenor equivalentlyWith $P=\alpha A+\beta B+\gamma C$, the three required equations are the Gram matrix systemFor the specified vectors this becomesThusand the centre is

An anticlockwise rotation matrix and reflection in the $x$-axis areThe columns of $R$ are the images of the standard coordinate vectors, while $M(x,y{)}^T=(x,-y{)}^T$.

Direct multiplication givesHenceThis is the defining conjugation relation between a rotation and reflection in a dihedral group.

No. Every power $R^a$ has determinant $1$, whereasMatrices with different determinants cannot be equal.

The relationsimply $RM=M{R}^{-1}$. Moving every occurrence of $M$ to the left and reducing exponents therefore puts every word into one of the normal formsExplicitly, with $\theta =2\pi /n$,andThe $n$ rotations are distinct. The $n$ reflected matrices are also distinct, and no reflected matrix equals a rotation because their determinants are $-1$ and $1$, respectively. Hence there are exactlymatrices, forming the dihedral group.

Let $Av=\lambda v$ for a nonzero eigenvector $v$ of the Hermitian matrix $A$. Hermitian symmetry of the inner product givesSince $lVertv{\Vert }^2>0$, $lambda=\overline{\lambda }$, so every eigenvalue is real.

For an orthogonal projection matrix $P=P^{†}=P^2$,Thus $P$ is a positive semidefinite matrix.

SetThen $v=v_0+v_1$, and idempotence givesMoreover,because both $P$ and $I-P$ are Hermitian. This is the orthogonal decomposition into the kernel and image of the projection.

Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

No. Consider the two orthogonal projectionsTheir product isFor $x=(-1,2{)}^T$,Thus $PQ$ is not a positive semidefinite matrix. It is also not Hermitian; products of orthogonal projections need not remain orthogonal projections unless the factors commute.

For the standard basis vector $e_j$, matrix multiplication selects the $j$th column, soConsequentlyand the columns of $PM$ are $P{c}_1,\dots ,P{c}_n$.

Suppose first that $v,Av,\dots ,A^{n-1}v$ are linearly independent, and defineThen $S$ is invertible. The first $n-1$ columns of $AS$ are the last $n-1$ columns of $S$. By the Cayley-Hamilton theorem,This is exactly the last-column rule for the companion matrix $C$, so

Conversely, suppose $S^{-1}AS=C$, or $AS=SC$, and write the columns of $S$ as $s_1,\dots ,s_n$. Comparing the first $n-1$ columns givesHenceSince $S$ is invertible, its columns are linearly independent. Taking $v=s_1$ therefore makeslinearly independent. Thus $A$ is similar to $C$ exactly when it has the stated cyclic vector.

Suppose $a_n\to L$. Given $\epsilon >0$, choose $N$ such that $|a_k-L|<\epsilon /2$ for $k>N$. ThenThe first term tends to zero because its numerator is fixed, and the second is at most $\epsilon /2$. Thus the Cesaro mean tends to $L$.

The converse is false. For $a_n=(-1{)}^n$, the Cesaro means tend to $0$, while the original sequence alternates between $-1$ and $1$ and does not converge.

First suppose $x_n\to L>0$. Continuity of the natural logarithm gives $\log x_n\to \log L$. Part (a), applied to this sequence, yieldsApplying the continuous exponential function,If $L=0$, then for every $\epsilon >0$ all sufficiently late $x_k$ are below $\epsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\epsilon$; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now supposeThe telescoping product isThereforeand so

A Cauchy sequence is a sequence $(x_n)$ such that for every $\epsilon >0$ there is $N$ for whichThe general principle of convergence, or completeness of the real numbers, states that a real sequence converges if and only if it is Cauchy.

For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,Because the series $\sum a_k$ converges, its tails tend to zero. Also $n-m\ge n/2$, soThe squeeze theorem gives

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

LetThen $h(a)=h(b)=0$. Since $f$ is not linear, $h(c)\ne 0$ for some $c\in (a,b)$. If $h(c)>0$, the mean value theorem on $[a,c]$ gives a point $\xi$ withIf $h(c)<0$, apply it on $[c,b]$ to obtainIn either case $h^{\prime }=f^{\prime }-m$, and therefore