Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 (AC2, XMAT 4302, XMST 4302)

Source: Mathematics release, supplied bundle Part A 2007 11-1.pdf, pages 195–202. The paper is dated 22 June 2010 and serves Mathematics and Mathematics & Statistics Part A.

Bounded exact-paper checks found no matching answer advertisement. All printed parts and their surrounding unnumbered requests are answered. The literal source is retained in the private editable transcription.

Write $n=\dim V$. If $Rv=\lambda v$ with $v\ne 0$, then $0=R^{a}v={\lambda }^{a}v$, hence $\lambda =0$. The characteristic polynomial splits over $\mathbb{C}$ and has degree $n$, soThere is indeed an eigenvector for the eigenvalue zero: a nonzero vector in the last nonzero image of $R$ lies in its kernel of a linear map. For the bound, choose $v$ with $R^{a-1}v\ne 0$, using minimality. The vectors $v,Rv,\dots ,R^{a-1}v$ are linearly independent. Otherwise choose the least $j$ with nonzero coefficient in a relation and apply $R^{a-1-j}$; only its nonzero multiple of $R^{a-1}v$ survives, a contradiction. Therefore $\boxed{a\le n}$.

Suppose $S^b=0$. Use the increasing chain of vector subspacesChoose a basis successively, extending a basis of each kernel of a linear map to one of the next kernel of a linear map. If a basis vector first enters at level $j$, its image under $S$ belongs to $\ker S^{j-1}$, spanned entirely by earlier basis vectors. Thus every matrix column has nonzero entries only above its diagonal: the matrix is strictly upper triangular.

Conversely, for such a matrix and its ordered basis $e_1,\dots ,e_n$, one has $S⟨e_1,\dots ,e_j⟩\subseteq ⟨e_1,\dots ,e_{j-1}⟩$. Iterating $n$ times kills every vector. Hence $\boxed{S^n=0}$, establishing both directions without assuming a Jordan form.

The polynomials $t^c$ and $(t-1{)}^d$ are coprime. Bézout's identity gives polynomials $u,v$ with $u(t)t^c+v(t)(t-1{)}^d=1$. SetFor every $x$, the decompositionhas first summand in $V_0$ and second in $V_1$, since $T^c(T-I{)}^d=0$ and polynomial expressions in $T$ commute. If $x$ belongs to both kernels, the same Bézout identity gives $x=0$. Thus $V=V_0\oplus V_1$. Both vector subspaces are invariant under $T$.

On $V_0$, the restriction of $T$ is nilpotent; on $V_1$, the restriction of $T-I$ is nilpotent. Apply (b) separately, then concatenate the two bases. This yieldswhere $A,N$ are strictly upper triangular. Empty blocks are allowed if one subspace is zero. In particular $C=I+N$ has precisely the required diagonal.

Let $D=T^{\ast }T-T{T}^{\ast }$. This operator is self-adjoint, andIf condition (iii) holds, this quadratic form vanishes. For a real inner product, polarization givesTherefore $D=0$, which is condition (i). This supplies the nontrivial reverse implication needed for the equivalence.

By the definition of the adjoint,If (i) holds, the expression is zero for all $x,y$, giving (ii). Conversely, if this expression vanishes for every $x$, nondegeneracy of the inner product gives $(T^{\ast }T-T{T}^{\ast })y=0$ for every $y$, so (ii) also implies (i).

Taking $y=x$ in (ii) gives (iii). Together with (iii)$\Rightarrow$(i) from part (i) and (i)$\Rightarrow$(ii) from part (ii), this completes

For a normal operator, (a) gives $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^{\ast }}$.

For real $\lambda$, the adjoint of $T-\lambda I$ is $T^{\ast }-\lambda I$. Expanding both products shows that their difference is $T^{\ast }T-T{T}^{\ast }=0$. Thus $T-\lambda I$ is normal, and applying the kernel of a linear map conclusion to it givesThis proves equality of each real eigenspace, hence the same eigenvalues and eigenvectors on the stated real vector space. If neither has real eigenvalues, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline{\lambda }$ for the adjoint; no claim of the same complex eigenvalues is needed here.

In fact the adjoint invariance follows from the orthogonal decomposition and the invariance under $T$, without using the normal operator property. For $x\in V_i$ and $y\in V_j$ with $j\ne i$,since $Ty\in V_j$. Hence $T^{\ast }x$ is orthogonal to every other summand and therefore lies in $V_i$. Moreover, for $x,y\in V_i$, the adjoint identity restricted to that subspace shows

Restricting $T^{\ast }T=T{T}^{\ast }$ to $V_i$, using (i), gives $T_i^{\ast }{T}_i=T_i{T}_i^{\ast }$, so the global normal operator property implies the normal operator property of every restriction. Conversely, if the restrictions are normal and the adjoint preserves each summand, then for $x={\sum }_i{x}_i$,Thus $\boxed{T\text{ is normal}⟺\text{every }{T}_i\text{ is normal}}$, with condition (i) automatic under the stated decomposition.

Interpret membership as having a representation with denominator coprime to $30$, equivalently having this property in lowest terms. Clearly $0,1\in R$. If $a/b,c/d\in R$ with admissible denominators, thenThe product $bd$ is still coprime to $30$. These are elements of $R$, proving that $\boxed{R\text{ is a subring of }\mathbb{Q}}$.

Take a reduced fraction $a/b\in R$. If $a\ne 0$, its only possible inverse in $\mathbb{Q}$ is $b/a$. Since $\gcd (a,b)=1$, this is in lowest terms up to sign and belongs to $R$ exactly when $\gcd (a,30)=1$. Zero is not invertible. ThereforeEquivalently, these are the rational numbers whose valuations at $2,3,5$ are all zero.

The intersection $J=I\cap \mathbb{Z}$ is an additive subgroup of $\mathbb{Z}$, and $nj\in I\cap \mathbb{Z}$ for $n\in \mathbb{Z}$, $j\in J$. Thus it is an ideal of $\mathbb{Z}$, necessarily $d\mathbb{Z}$ for a unique $d\ge 0$.

Because $d\in I$, one has $dR\subseteq I$. Conversely, if $a/b\in I$ with $b$ coprime to $30$, multiplication by $b\in R$ gives $a\in I\cap \mathbb{Z}=d\mathbb{Z}$. Write $a=dk$; then $a/b=d(k/b)\in dR$. Thus $I=dR$.

Every positive integer factors as $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }u$ with $\gcd (u,30)=1$. The factor $u$ is a unit, so it does not change the ideal. The complete distinct list isThey are distinct because the intersection of the displayed nonzero ideal with $\mathbb{Z}$ is $2^{\alpha }{3}^{\beta }{5}^{\gamma }\mathbb{Z}$: admissible denominators cannot cancel any of these prime powers. The triple $(0,0,0)$ gives $R$ itself.

The zero ideal is a prime ideal because $R$ is an integral domain. Each of $(2),(3),(5)$ is a prime ideal: for $p$ among these three, reduction $a/b\mapsto \overline{a}{\overline{b}}^{-1}$ is a surjective ring homomorphism $R\to {\mathbb{F}}_p$ with kernel of a ring homomorphism $(p)$, so its quotient ring is a field.

Any other nonzero proper ideal in (c) has generator $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }$ with total exponent at least two. Choose one prime factor $p$ and put $e=d/p$. Then $pe=d\in (d)$ but neither $p$ nor $e$ belongs to $(d)$, as seen from the required prime exponents. Such an ideal is not prime. Thus with the standard proper-ideal convention the answer isThe definition printed in this paper omits properness. Literally under that definition $R$ itself also satisfies the implication, so its literal complete list is the preceding four ideals together with $\boxed{R}$. Both interpretations are thereby accounted for.

Write $f=u+iv$ and $z=x+iy$. The complex derivative computed along a real increment is $f^{\prime }=u_x+i{v}_x$, while along an imaginary increment it is $f^{\prime }=v_y-i{u}_y$. Equality givesThese are the Cauchy–Riemann equations at every point where $f$ is holomorphic. The directional calculations also establish the existence of those partial derivatives. More explicitly, $f(z+h)-f(z)=f^{\prime }(z)h+o(|h|)$ gives both real-direction formulas directly. If a converse is desired, continuous first partial derivatives together with these equations make the real differential multiplication by $u_x+i{v}_x$, hence give complex differentiability; the necessity above is what is requested.

The allowed theorem makes $f^{\prime }$ holomorphic and hence continuous; applying it repeatedly gives continuous derivatives of every order. Since $u_x=\Re f^{\prime }$, $v_x=\Im f^{\prime }$, $u_y=-\Im f^{\prime }$ and $v_y=\Re f^{\prime }$, the real and imaginary parts have continuous second partial derivatives. Differentiate the Cauchy–Riemann equations and commute mixed derivatives:Thus $\boxed{\Delta u=\Delta v=0}$, with the required $C^2$ regularity, proving harmonicity.

Write $f=u+iv$ and evaluate all derivatives of $\psi$ at $(u,v)$. The real chain rule givesThe Cauchy–Riemann equations give equal squared lengths $u_x^2+u_y^2=v_x^2+v_y^2=|f^{\prime }{|}^2$ and zero cross term $u_x{v}_x+u_y{v}_y=0$. Part (ii) removes the last two terms. ThereforeThe composition is $C^2$, so it is harmonic. No assumption that $f^{\prime }$ is nonzero is required.

The complex exponential function $E(z)=e^z$ maps the strip bijectively and holomorphically to the upper half-plane: its modulus is $e^{\Re z}>0$ and its argument is $\Im z\in (0,\pi )$. Its inverse is the holomorphic logarithm $L(w)=\log |w|+i\mathrm{Arg}w$ with $\mathrm{Arg}w\in (0,\pi )$.

Part (iii) applies to both maps. The required inverse correspondences areThey compose to the identity because $L\circ E$ and $E\circ L$ do. This proves injectivity and surjectivity, not merely a map between the two sets.

For any $0<r<1$, a Laurent series converges uniformly on the circle $|z|=r$, so termwise contour integration is valid. Multiplication by $z^{-k-1}$ and integration gives, for every integer $k$,Indeed, the integral of $z^{j-k-1}$ is $2\pi i$ for $j=k$ and zero otherwise. These coefficients depend only on $f$, so any two Laurent expansions must agree coefficient by coefficient. This proves uniqueness, including all terms of the principal part.

A removable singularity means that $f$ has a holomorphic extension to the full disk, assigning a finite value at zero. Equivalently its Laurent series has no negative-power terms. One direction follows by the Taylor expansion of the extension; the other by the convergent nonnegative-power series. Thus $\boxed{a_k=0\text{ for all }k<0}$ is the coefficient criterion.

A pole of order $n$ means $f(z)=z^{-n}g(z)$ with $g$ holomorphic near zero and $g(0)\ne 0$. Equivalently, the principal part is finite and starts with $\boxed{a_{-n}{z}^{-n},a_{-n}\ne 0}$, while $a_k=0$ for $k<-n$. In particular $n$ is the exact order, not merely an upper bound.

An essential singularity is an isolated singularity that is neither removable nor a pole. In terms of the Laurent series this means $\boxed{\text{infinitely many negative-power coefficients are nonzero}}$: no finite multiplication by a power of $z$ removes the singularity.

For $|z|<1$, differentiate the geometric series to obtain $(1-t{)}^{-2}={\sum }_{k\ge 0}(k+1)t^k$. Since $(z-i{)}^2=-(1-z/i{)}^2$,The convergence region is precisely $0<|z|<1$, the specified punctured disk; the other pole is on its boundary at $i$.

If $f=z^{-n}g$ with $g(0)\ne 0$, continuity gives a small disk on which $|g(0)|/2\le |g(z)|\le 3|g(0)|/2$. Multiplying by $|z{|}^{-n}$ gives the two required bounds.

Conversely, set $g(z)=z^{n}f(z)$ on the punctured disk. The upper bound makes $g$ bounded, so the allowed removable-singularity theorem gives a holomorphic extension. The lower bound gives $|g(0)|\ge C_1>0$ by continuity. Therefore $f=z^{-n}g$ has a pole of exact order $n$. The lower bound is essential: the upper bound alone only proves order at most $n$.

On $D=\mathbb{C}\setminus [0,\infty )$ choose the continuous argument $0<\arg z<2\pi$, and putThis is holomorphic on the slit plane for every fixed $w$. The upper boundary value at $x>0$ is $x^w$, whereas the lower boundary value is $e^{2\pi iw}{x}^w$. This difference is exactly what the keyhole contour measures.

Use the branch in (i) and the positively oriented slit-annulus contour. Its upper ray runs from $r$ to $R$, its outer circle counterclockwise, its lower ray from $R$ back to $r$, and its inner circle clockwise. Initially keep their arguments at $\alpha$ and $2\pi -\alpha$ as in the figure, with $0<\alpha <\pi /2$; let $\alpha \downarrow 0$ for fixed $0<r<1<R$. Both poles $i,-i$ lie inside.

Write $\sigma =\Re w\in (-1,1)$. On the outer circle the absolute integral is $O(R^{\sigma -1})$, since $|z^w|\le e^{2\pi |\Im w|}{R}^{\sigma }$ and $|z^2+1|\ge R^2-1$. On the inner circle it is $O(r^{\sigma +1})$. Both tend to zero. The real integral converges absolutely because its endpoint behaviors are $x^{\sigma }$ and $x^{\sigma -2}$.

The two residues, with the chosen arguments, areThe residue theorem and the two ray orientations therefore giveFor $w\ne 0$ in this strip we may cancel $1-e^{i\pi w}$; its only possible zero here is $w=0$. ConsequentlyAt $w=0$ the same result follows directly from $I(0)={\int }_0^{\infty }(1+x^2{)}^{-1}dx=\pi /2$, or by dominated convergence. The denominator has no zero in the open strip.

Set $w=i$, which satisfies $\Re w=0$, and take real parts of the absolutely convergent integral. Since $x^i=\cos (\log x)+i\sin (\log x)$ and $\cos (i\pi /2)=\cosh (\pi /2)$,Absolute integrability justifies moving the real-part operation through the integral.

If $Au+Bv=0$, evaluation at zero gives $A=0$ and differentiation at zero gives $B=0$. Thus the solutions are linearly independent. For $w=u{v}^{\prime }-u^{\prime }v$,so $\boxed{w(x)=\exp (-{\int }_0^{x}p(s)ds)>0}$.

Variation of parameters sets $y=A(x)u+B(x)v$ with $A^{\prime }u+B^{\prime }v=0$, $A^{\prime }{u}^{\prime }+B^{\prime }{v}^{\prime }=f$. Solving gives $A^{\prime }=-vf/w$ and $B^{\prime }=uf/w$. Zero integration constants enforce both initial conditions. ThereforeFor fixed $t$, this kernel is a linear combination of homogeneous solutions, so $K_{xx}+p{K}_x+qK=0$. On the diagonal $K(t,t)=0$ and $K_x(t,t)=(v^{\prime }(t)u(t)-u^{\prime }(t)v(t))/w(t)=1$. Differentiating the integral twice now verifies the forced equation as well as $y(0)=y^{\prime }(0)=0$. Standard uniqueness of the linear initial-value problem makes this the solution, for continuous $f$.

If $u,v$ have period $T>0$, their derivatives do also, hence $w$ is periodic. The kernel is periodic in its first variable. With $f=v$, split the integral at $x$ and use this periodicity to getAt $x=0$, the right side is $-{\int }_0^{T}v(t{)}^2/w(t)dt<0$, because $w>0$ and $v^{\prime }(0)=1$ ensures $v$ is not identically zero. Thus $\boxed{y(T)\ne y(0)}$, so $y$ cannot have period $T$. This direct sign argument also rules out any apparent cancellation between the two terms.

Use a parameter $t$. The characteristic equations areFor a graph of a function $z=f(x,y)$ satisfying the equation, solve the first two equations after substituting $z=f(x,y)$. The chain rule and the partial differential equation give $df(x(t),y(t))/dt=f+bx-ay$, exactly the third equation. Uniqueness for the resulting linear ordinary differential equation system shows that its characteristic through the initial point agrees with this curve on the graph of a function. This proves invariance as long as the graph of a function remains defined along the curve. A claim about the entire curve additionally requires a surface continued along that flow; a restricted graph of a function patch alone cannot imply global continuation.

Let $\mathbf{k}=(a,b,c)$, $\mathbf{\ell }=(l,m,n)$ and $\mathbf{r}=(x,y,z)$. Then $\dot{\mathbf{r}}=\mathbf{r}-\mathbf{k}\times \mathbf{r}$. Since $|\mathbf{k}|=1$,Along the characteristic starting from $s\mathbf{\ell }$ these give $\mathbf{k}\cdot \mathbf{r}=e^{t}s(\mathbf{k}\cdot \mathbf{\ell })$ and $|\mathbf{r}|=e^t|s|$. Consequently the characteristic surface generated by the given line satisfiesTo see that the whole displayed cone is generated, not just contained in it, decompose $\mathbf{\ell }=\kappa \mathbf{k}+\mathbf{v}$ with $\mathbf{v}\perp \mathbf{k}$, $\kappa =\mathbf{k}\cdot \mathbf{\ell }$. The full solution isWhen $0<|\kappa |<1$, variation of $t$ gives all azimuths about the axis, while the freely chosen $s$ supplies every positive radius on both halves. This is precisely the two-sided circular cone, withThe edge cases matter: $\kappa =0$ gives the plane $ax+by+cz=0$ (semi-angle $\pi /2$), and $|\kappa |=1$ generates only the axis (semi-angle zero), not a two-dimensional surface. In the latter case the initial line is itself characteristic, so it does not determine a unique integral surface. The nondegenerate cone is smooth away from its vertex and can be used as a graph of a function only on patches whose tangent plane is not vertical. Thus the printed geometric conclusion has its ordinary characteristic-surface meaning in the nondegenerate case, with these domain and degeneracy qualifications; it cannot be asserted for every global smooth graph of a function without them.

For an azimuth-independent product $u=R(r)\Theta (\theta )$, multiply Laplace's equation by $r^2/(R\Theta )$ where the factors are nonzero. The radial and angular expressions must equal opposite constants:The resulting differential identities extend across isolated zeros by continuity. Put $\rho =\cos \theta$. Then ${\Theta }^{\prime }=-\sin \theta P^{\prime }$ andThis gives $((1-{\rho }^2)P^{\prime }{)}^{\prime }+\lambda P=0$ on $(-1,1)$.

A nonzero angular factor must be regular at both poles: it must be bounded as $\rho \to \pm 1$ and extend smoothly as an axisymmetric function on the sphere. Equivalently one may use the regular endpoint solution at each end; the logarithmic endpoint solutions are excluded. These two endpoint conditions select $\boxed{\lambda =n(n+1),n=0,1,2,\dots }$ and the multiples of $P_n$. It is not sufficient to impose no condition at the endpoints, since local solutions exist for arbitrary $\lambda$. One way to see the quantization is the regular Legendre series about one endpoint: unless it terminates at $\lambda =n(n+1)$, continuation to the other endpoint has a logarithmic singularity.

Direct differentiation verifiesIndeed $R=r^m$ gives $m(m+1)=\lambda$, so the two powers for degree $n$ are $r^n$ and $r^{-n-1}$.

For the shell assume $a>1$. The outer datum decomposes as ${\rho }^2=\frac{1}{3}{P}_0(\rho )+\frac{2}{3}{P}_2(\rho )$. Normalize each radial solution to vanish at $1$ and equal $1$ at $a$; superposition givesEvery term is harmonic in the shell and regular at the angular poles; substitution at each boundary gives precisely the two prescribed data. For uniqueness among functions harmonic in the shell and continuous on its closure, subtract any two solutions. Their difference is harmonic, zero on both boundary spheres, and by the maximum principle is both at most zero and at least zero on the bounded shell. It vanishes identically. Thus the classical Dirichlet solution is unique. The usual continuity at the boundary is the interpretation of the stated boundary conditions; allowing singular boundary behavior would remove that hypothesis.