Past exam of the mathematics course of the University of Cambridge / 2023 / ib

Choose bases of $V$ and $W$, of dimensions $v$ and $w$. A linear map is uniquely determined by the arbitrary images of the $v$ basis vectors, each with $w$ coordinates, so

Put $a=\dim A$ and $b=\dim B$, and extend bases of $A,B$ to bases of $V,W$. The condition $\varphi (A)\subseteq B$ forces the $(w-b)a$ matrix entries from $A$ to a complement of $B$ to vanish and imposes no other restriction. Hence

For the last part let $r=\dim (S\cap T)=s+t-v$. ChooseA map in $Y$ sends $S\cap T$ into itself, $S_0$ into $S$, and $T_0$ into $T$, with these choices independent. Therefore

A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

The antipodal action on $S^2$ is free. The surface quotient by a free finite action applies: Around each point choose a small open disc disjoint from its antipodal image; the quotient map restricts to a homeomorphism from that disc onto an open neighbourhood in the quotient. Compactness gives Hausdorffness and second countability descends from the sphere. Thus the quotient is the real projective plane, in particular a topological surface.

For the second quotient write a point away from the poles asThe mapin these cylindrical coordinates extends continuously over the poles and identifies exactly $\varphi$ with $\varphi +\pi$. It therefore induces a continuous bijection $S^2/\approx \to S^2$. The domain is compact and the sphere Hausdorff, so this bijection is a homeomorphism.

The Laurent series isThus zero is an essential singularity.

On $|z|\ge \rho$, put $w=1/z$, so $|w|\le 1/\rho$. The Taylor polynomialsconverge uniformly there to $e^w$. MoreoverFor all sufficiently large $n$, the uniform error is below half this lower bound, so $|f_n(z)|>0$ whenever $|z|\ge \rho$. Every zero therefore lies in $|z|<\rho$.

For a regular constraint $g=0$, an interior constrained extremum satisfiestogether with the constraint; boundary cases must also be checked.

Here maximize $f=xyz$ subject to $x^2+y^2+z^2=L^2$. The Lagrange multiplier equations giveAt a positive maximizer these imply $x=y=z=L/\sqrt{3}$. Boundary points have zero volume, so the maximum is

For any $h$,If the normal equation $A^{T}A{c}^{\ast }=A^{T}y$ holds, the last two terms reduce to $\Vert Ah{\Vert }^2\ge 0$, so $c^{\ast }$ minimizes. Conversely, at a minimizer the directional derivative in every $h$ vanishes, forcing $A^T(A{c}^{\ast }-y)=0$.

If $A$ has full column rank, then for every nonzero $h$,Thus $A^{T}A$ is positive definite and invertible, and the normal equation has the unique solution

For a null parameter space ${\Theta }_0\subseteq \Theta$, the generalized likelihood ratio isThe Wilks theorem states, under standard regularity conditions and under the null, that

The null hypothesis is that mortality is independent of carbolic-acid use, equivalently that the two mortality probabilities are equal. The two-sided alternative is that they differ.

Use the Chi-squared test of independence, equivalently the likelihood-ratio statistic, with Pearson's statisticwhere $E_{ij}$ are the fitted counts under independence using the observed row and column totals. Under the null, $X^2$ is asymptotically ${\chi }_1^2$. A size-$\alpha$ test rejects when

The minimum-cost flow problem chooses, for each directed edge $(i,j)\in E$, a flow $x_{ij}$ and solvessubject toFeasibility requires ${\sum }_i{b}_i=0$.

Set $y_{ij}=x_{ij}-M_{ij}$. Thenand its balance vector isThe objective becomes $\sum C_{ij}{y}_{ij}$ plus the constant $\sum C_{ij}{M}_{ij}$. Translation by $M$ is a bijection between feasible flows and preserves their ordering by cost, so the transformed zero-lower-bound problem is equivalent.

True. By Jordan normal form, the characteristic polynomial records the total size of the blocks for each eigenvalue, while the minimal polynomial records the largest block for each eigenvalue. In dimension three this determines every block partition: for algebraic multiplicity three the possibilities $3$, $2+1$, and $1+1+1$ have largest block sizes three, two, and one; multiplicities one and two are equally immediate. Thus the two matrices have the same Jordan form and are conjugate. This is the characteristic and minimal polynomials determine similarity in dimension three phenomenon.

False. At eigenvalue $2$, algebraic multiplicity two and minimal-polynomial exponent two force one block $J_2(2)$. At eigenvalue $1$, the blocks have total size five and largest size two, so there are only two possible partitions,Hence there are exactly two conjugacy classes satisfying the given polynomial data, represented byandThere cannot be three mutually non-conjugate examples.

True. Since $\alpha$ is invertible, every eigenvalue $\lambda$ is nonzero. If $\alpha$ had a nontrivial Jordan block $\lambda I+N$, then for the stated positive integer $N_0$,would still have a nonzero nilpotent part and could not be diagonalizable. Therefore all Jordan blocks of $\alpha$ have size one. This also proves diagonalizability inherited from an invertible power.

False. The real matrixhas the two distinct complex eigenvalues $i$ and $-i$, so it is diagonalizable over $\mathbb{C}$. It has no real eigenvalue and hence no real eigenbasis, so it is not diagonalizable over $\mathbb{R}$.

True. Suppose $A,B$ are real and $AP=PB$ for an invertible complex matrix $P=X+iY$, with $X,Y$ real. Taking real and imaginary parts givesThe polynomial $q(t)=\det (X+tY)$ is not identically zero, since $q(i)=\det P\ne 0$. Choose a real $t$ outside its finite zero set. Then $Q=X+tY$ is real and invertible, and $AQ=QB$. Thus $A$ and $B$ are conjugate over $\mathbb{R}$, as stated by complex similarity of real matrices implies real similarity.

Assume (i). For $a,b\in R$, write the ideal $(a,b)=(d)$. Then $d=ra+sb$ for some $r,s\in R$. Since $a,b\in (d)$, the element $d$ divides both; and every common divisor of $a,b$ divides the displayed linear combination $d$. Thus $d$ is a greatest common divisor and (ii) holds.

Assume (ii), and let an $R$-submodule $M\subseteq F$ be generated by $x_1,\dots ,x_m$. Choose a common nonzero denominator $d$ and write $x_j=a_j/d$ with $a_j\in R$. Repeated application of (ii) shows thatfor one element $g$: the gcd at each step is a linear combination of the elements processed so far. Thereforeso (iii) holds. This is the finite-ideal property defining a Bézout domain.

Assume (iii). Because $R$ is a Noetherian ring, every ideal $I\subseteq R$ is finitely generated. Regard $I$ as an $R$-submodule of $F$. By (iii), $I=Rx$ for some $x\in F$; since $x\in I\subseteq R$ unless $I=0$, this is a principal ideal of $R$. Thus every ideal is principal and (i) holds.

Together with the previous two implications, this proves (i), (ii), and (iii) equivalent.

If (i) holds, the submodule theorem for free modules over a principal ideal domain says that any submodule $N\subseteq R^n$ is free. Its rank is at most $n$, since tensoring the inclusion with $F$ embeds $N{\otimes }_{R}F$ into $F^n$. Hence $N$ has at most $n$ generators, proving (iv).

Conversely, apply (iv) with $n=1$. Every ideal is an $R$-submodule of $R$ and therefore has one generator, which is (i). Thus all four conditions are equivalent.

Suppose the additive group of the integral domain $R$ is isomorphic to ${\mathbb{Z}}^n$. Every ideal $I$ is then an additive subgroup of a finitely generated free abelian group, so it has finitely many additive generators $x_1,\dots ,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb{Z}$ in $R$. Thus every ideal is finitely generated, proving that $R$ is Noetherian by noetherianity from finite additive rank.

For an example that fails (i)--(iv), takeIt is an integral domain and has additive group ${\mathbb{Z}}^2$. The idealhas index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is ${\mathbb{F}}_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have determinant and absolute indexwhich has no integer solution. Hence $I$ is not principal, as detailed in nonprincipal ideal in the integers adjoined a square root of minus five. Condition (i), and therefore all four equivalent conditions, fails.

A Cauchy sequence $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. A complete metric space is one in which every Cauchy sequence converges to a point of the space.

Every Cauchy sequence is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\ge N$, and putThen every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing sequence of nonempty closed sets with $\mathrm{diam}{F}_n\to 0$. Choose $x_n\in F_n$. Given $\epsilon >0$, choose $N$ with $\mathrm{diam}{F}_N<\epsilon$. For $m,n\ge N$, both points lie in $F_N$, so $d(x_m,x_n)<\epsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. DefineThese sets are nonempty, closed, and decreasing. The Cauchy property implies $\mathrm{diam}{F}_n\to 0$; taking a closure does not change the diameter. Choose $x\in {\bigcap }_n{F}_n$. Since $x_n,x\in F_n$,Thus every Cauchy sequence converges and $M$ is complete. This proves the Cantor intersection theorem characterization.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda \in \Lambda$, the map $T_{\lambda }(x)=T(\lambda ,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^{\ast }(\lambda )$. This defines the required unique function. Fix ${\lambda }_0$. The fixed-point identities and the triangle inequality giveConsequentlyThe numerator tends to zero as $\lambda \to {\lambda }_0$ by the assumed continuity for the fixed point $x^{\ast }({\lambda }_0)$. Hence $x^{\ast }$ is continuous, an instance of continuous dependence of the fixed point of a uniform contraction.

A subset $S\subset {\mathbb{R}}^3$ is a smooth surface if every point has a neighbourhood in $S$ parametrized by a map $X:U\to S$, where $U\subset {\mathbb{R}}^2$ is open, $X$ is a homeomorphism onto that neighbourhood, $X$ is smooth, and $DX$ has rank two everywhere. These are the embedded surface parametrization conditions.

For the given set use local angular intervals in the parametrizationIt is locally one-to-one and has tangent vectorsTheir cross product has magnitudeso the derivative has rank two. The angular charts cover $\Sigma$, proving that it is a smooth surface.

The area between heights $a_0$ and $b_0$ is thereforeBy hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,and squaring gives

If $0<\varphi (t)<r$, then ${\varphi }^{\prime }$ never vanishes, so its sign $\sigma \in \{1,-1\}$ is constant. The last equation givesand henceThus $\sqrt{r^2-\varphi (t{)}^2}=-\sigma (t-t_0)$ for a constant $t_0$, andThe graph lies on a circle of radius $r$, exactly as described by constant strip-area density of a surface of revolution.