Algebra, analysis and differential equations II course of the University of Oxford (AC2)

Source release: Oxford examination archive, member FOI-20210708-1 - Data/4302.pdf. The cover prints formal code 4302, paper label AC2, the title “Algebra, Analysis and Differential equations II”, and Tuesday 19 June in Trinity Term 2012.

Answer availability checked on 28 October 2026: a bounded search of official Oxford sources found no advertisement of complete or selected-part solutions for this exact paper. The general advertisement of even-year solutions dating back to at least 2016 does not establish coverage of 2012.^[ref]

The minimal polynomial $m_T(t)$ is the unique monic polynomial of least positive polynomial degree satisfying $m_T(T)=0$. Such an annihilating polynomial exists without using the Cayley-Hamilton theorem: the vector space $\mathrm{End}(V)$ has dimension of a vector space equal to $n^2$, so $I,T,\dots ,T^{n^2}$ are linearly dependent. Normalize a nonzero annihilating polynomial to be monic and choose one of least degree. Two distinct monic choices of that degree would have an annihilating difference of lower degree, contradicting minimality. A nonzero constant cannot annihilate $T$ because $V\ne 0$.

Choose a basis $B$ and let $A=[T{]}_B$. The characteristic polynomial isIt is independent of the chosen basis. Indeed a change of basis replaces $A$ by a similar matrix $P^{-1}AP$, andby multiplicativity of the determinant. Thus this monic degree-$n$ polynomial is an invariant of the linear transformation itself.

Since $v$ is a cyclic vector and $V\ne 0$, $v\ne 0$. Finite dimension of a vector space forces a first dependence in the sequence $v,Tv,T^{2}v,\dots$. Let $k\ge 1$ be the smallest index with $T^{k}v$ in the linear span of its predecessors. Thenis linearly independent. Its linear span $W$ is invariant under $T$: all but the last basis candidate map to the next, and the last maps to $T^{k}v\in W$. Induction therefore puts every $T^{j}v$ in $W$. Cyclicity gives $W=V$, so $B$ is a basis and $k=n$.

WriteThe matrix of $T$ in this basis is the companion matrixFor $p(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the defining relation says $p(T)v=0$. Since $p(T)$ commutes with $T$, it also annihilates every $T^{j}v$ in $B$, hence $p(T)=0$. No nonzero polynomial of degree less than $n$ can annihilate $v$, because that would contradict linear independence of $B$. Thus $m_T=p$.

Finally the characteristic polynomial of a companion matrix is $p$, by direct determinant expansion. Explicitly, expansion along the first row of $tI-C$ gives the recurrencestarting from $D_1=t+a_0$. It yields $D_n=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$. ConsequentlyThis proves that a cyclic operator has equal minimal and characteristic polynomials by the cyclic basis and a direct determinant calculation, with no appeal to the Cayley-Hamilton theorem.

For $T_1$, choose the third standard basis vector $v=e_3$. Its first three orbit vectors areThey are linearly independent: their matrix of columns has determinant $-1$. They therefore span ${\mathbb{R}}^3$, proving $T_1$ has a cyclic vector, namely $e_3$.

For $T_2$, write $T_2=I+N$ where $N^2=0$. The binomial theorem gives $T_2^{j}v=v+jNv$ for every integer $j\ge 0$. The entire orbit of any $v$ lies in $\mathrm{span}\{v,Nv\}$, which has dimension of a vector space at most two. Therefore $T_2$ has no cyclic vector in the three-dimensional space. Equivalently, its minimal polynomial is $(t-1{)}^2$, while its characteristic polynomial is $(t-1{)}^3$.

Use the inner product convention that the first argument is linear and the second is conjugate-linear. This is the convention corresponding to the conjugated eigenvalue in part (c).

An orthonormal basis $(q_1,\dots ,q_n)$ is a basis with $⟨q_i,q_j⟩={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta. Each vector has norm one and distinct vectors are orthogonal.

The Gram-Schmidt process starts from a basis $(e_1,\dots ,e_n)$ and definesThe coefficients are in this order because the inner product is linear in its first argument. For $i<k$, taking the inner product with $q_i$ gives $⟨u_k,q_i⟩=0$. The residual $u_k$ cannot be zero: otherwise $e_k$ would belong to the linear span of $e_1,\dots ,e_{k-1}$, contradicting the initial basis. Normalization makes $\Vert q_k\Vert =1$, while induction preservesThus the process produces an orthonormal basis of all of $V$.

In an orthonormal basis, every vector $w$ has the coordinate expansion $w={\sum }_i⟨w,e_i⟩e_i$. If $⟨Su,v⟩=⟨u,Rv⟩$ for all $u,v$, apply it to $u=e_i$, $v=e_j$ and take complex conjugates:ConsequentlyConversely, this formula givesLinearity in the first argument and conjugate-linearity in the second extend the equality from basis vectors to all $u,v$. This proves the equivalence.

The displayed formula specifies exactly one image for each basis vector, so it defines a unique linear transformation $S^{\ast }$ and proves the existence and uniqueness of the adjoint on an inner product space.

Conjugate symmetry givesBy uniqueness of the adjoint operator, $\boxed{(S^{\ast }{)}^{\ast }=S}$. Alsoso uniqueness gives $\boxed{(ST{)}^{\ast }=T^{\ast }{S}^{\ast }}$. The reversal of order follows directly from moving the two factors across the inner product one at a time.

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.

A unit in a ring is an element $u$ with a multiplicative inverse in $R$, so $uv=1$ for some $v\in R$.

An irreducible element is a nonzero nonunit $a$ such that every factorization $a=bc$ has at least one of $b,c$ a unit in a ring.

A prime element is a nonzero nonunit $p$ such that $p\mid bc$ implies $p\mid b$ or $p\mid c$. Equivalently its principal ideal $(p)$ is a nonzero prime ideal. These are definitions in the given commutative ring; the equivalence between irreducibility and primality requires additional hypotheses on the ring.

A Euclidean domain is an integral domain $R$ with a function $d:R\setminus \{0\}\to {\mathbb{N}}_0$ such that, for $a\in R$ and nonzero $b\in R$, there exist $q,r\in R$ with $a=qb+r$ and either $r=0$ or $d(r)<d(b)$. This is the Euclidean-ring convention used here.

If an ideal $I$ is zero, it is generated by zero. Otherwise choose nonzero $a\in I$ with $d(a)$ minimal. For any $b\in I$, Euclidean division gives $b=qa+r$, with $r\in I$. A nonzero remainder would have $d(r)<d(a)$, contradicting the choice. Hence $r=0$ and $b\in aR$. The reverse inclusion follows from $a\in I$, provingNow suppose $a$ is an irreducible element. Its principal ideal is proper because $a$ is not a unit. If $aR\subseteq J\subseteq R$, write $J=bR$ by the result just proved. Then $a=bc$ for some $c\in R$. Irreducibility gives either $b$ a unit, so $J=R$, or $c$ a unit, so $bR=aR$. There is no intermediate proper ideal, and therefore $aR$ is a maximal ideal.

For a field $F$, the polynomial ring $F[X]$ is an integral domain and the polynomial division algorithm gives $A=QB+R$ with $R=0$ or $\mathrm{deg}R<\mathrm{deg}B$. Thus the polynomial degree is a Euclidean function, proving $F[X]$ is a Euclidean domain.

The polynomial $X^n-2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with prime two: all coefficients except the leading one are divisible by two, while the constant coefficient is not divisible by four. Together with Gauss's lemma, this establishes irreducibility in $\mathbb{Q}[X]$. Part (b) shows that its principal ideal is a maximal ideal, soLet $\alpha$ denote the residue class of $X$. The polynomial division algorithm gives every residue class a representative of degree less than $n$, so $1,\alpha ,\dots ,{\alpha }^{n-1}$ span $K_n$ over $\mathbb{Q}$. If a linear combination of them vanished, its degree-less-than-$n$ representative would be a multiple of the degree-$n$ polynomial $X^n-2$, and hence would be zero. They are therefore a basis, givingThe quadratic polynomial $X^2+1$ has no rational root, so it is irreducible over $\mathbb{Q}$. The same maximal-ideal argument proves $L=\mathbb{Q}[X]/(X^2+1)$ is a field.

These two quadratic fields are not isomorphic. In $L$, the class of $X$ squares to $-1$. But $K_2$ embeds in $\mathbb{R}$ by sending its class of $X$ to $\sqrt{2}$: evaluation at $\sqrt{2}$ has kernel $(X^2-2)$, by irreducibility and polynomial division. No real number squares to $-1$, so no element of $K_2$ does. A field isomorphism preserves one, minus one and squaring, so it cannot carry $L$ to $K_2$.

The Cauchy integral formula says that if $f$ is holomorphic on the simply connected open set $U$, and $\gamma$ is a positively oriented simple closed curve in $U$, then, for every $z$ in its interior,Here the contour and its enclosed region lie in $U$; the contour can be taken piecewise continuously differentiable. More generally, the same formula needs only holomorphic function on an open neighbourhood of that closed region. The orientation is counterclockwise, so the winding number about an interior point is one.

Write $M=|f(z_0)|$. The local maximum property gives a disk around $z_0$ on which $|f|\le M$, contained in the unit disk. If $M=0$, then $f=0$ on that neighbourhood, and the identity theorem makes $f$ zero on the connected unit disk.

Otherwise multiply $f$ by the unit complex number $\overline{f(z_0)}/M$, obtaining a holomorphic function $g$ with $g(z_0)=M$ and $|g|\le M$ nearby. Applying the Cauchy integral formula on any sufficiently small circle of radius $r$ gives the mean value propertyTake real parts. The continuous function $M-\mathrm{Re}g$ is nonnegative on the circle and has integral zero, so it vanishes everywhere there. Since $|g|\le M$, having real part $M$ forces imaginary part zero. Thus $g=M$ throughout the circle. The zeros of $g-M$ have an accumulation point inside its domain, so the identity theorem gives $f$ constant on the entire unit disk. This proves the local maximum modulus principle directly from the Cauchy integral formula.

The closed unit disk is a compact set, and $|f|$ is continuous on it because $f$ is holomorphic on an open neighbourhood. The extreme value theorem therefore supplies a point where its maximum is attained. If such a point is on the boundary, the conclusion follows. If it is in the interior, part (i) makes $f$ constant on the open disk; continuity then gives the same value on the boundary. ConsequentlyThe assumption that the closed disk lies in $U$ is what makes the boundary values and the compactness argument available.

Because $f(0)=0$, the Taylor series of $f$ at zero shows thatis holomorphic throughout the unit disk. Fix $0<r<1$. On $|z|=r$ we have $|h(z)|<1/r$, and the maximum modulus principle on the closed disk of radius $r$ gives $|h(z)|\le 1/r$ for $|z|\le r$. For any fixed $z$ inside the unit disk we may let $r$ increase to one, obtaining $|h(z)|\le 1$. Hence the Schwarz lemma inequalities areThe derivative bound is the same estimate at $z=0$, using the holomorphic extension of the quotient.