Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 (AC1, XMAT 4301, XMST 4301)

Source: released examination bundle, Part A 2007 11-1.pdf, pages 190–194. All nine questions are covered. No mark allocations are printed.

Every vector decomposes as $v=Ev+(v-Ev)$, where $Ev\in \mathrm{im}E$ and $E(v-Ev)=0$. If $w$ lies in both the image and kernel, write $w=Eu$ to obtain $w=E^{2}u=Ew=0$. Thusand $E$ is the projection onto its image along its kernel. It is the identity on its image and zero on its kernel. Let $N=\dim V$ and $r=\mathrm{rank}E$. A basis adapted to this direct sum gives matrix $\mathrm{diag}(I_r,0_{N-r})$. Since $E$ is not the identity, $N-r>0$, and therefore

In the adapted basis from part (a), the matrix has $r$ diagonal entries equal to one and $N-r$ equal to zero. With the characteristic polynomial convention ${\chi }_E(X)=\det (XI-E)$,The characteristic polynomial is independent of the choice of basis.

If $Ev=\lambda v$ with $v\ne 0$, applying $E^2=E$ gives ${\lambda }^{2}v=\lambda v$, hence $\lambda \in \{0,1\}$. The kernel is nonzero since $E\ne I$, so zero always occurs. One occurs exactly when the image is nonzero. Thus the eigenvalue set is $\{0\}$ if $E=0$, and $\{0,1\}$ otherwise. Their algebraic multiplicities are $N-r$ and $r$ respectively.

The polynomial $X(X-1)$ annihilates $E$ by idempotence, so the minimal polynomial divides it. If $E=0$, its monic polynomial of smallest positive degree is $X$. If $E\ne 0$, both eigenvalues zero and one occur, so both factors must divide every annihilating polynomial. Therefore

For real scalars $\alpha ,\beta$ and matrices $A,C,B$, distributivity of matrix multiplication and linearity of the trace giveThus $\phi$ is linear in its first argument.

Expand the trace by matrix entries:Hence $\phi (A,B)=\phi (B,A)$. Combined with part (a), this makes $\phi$ a symmetric bilinear form.

For $n\ge 2$, take the matrix whose leading $2\times 2$ block is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and whose other entries are zero. Its square has leading block $-I_2$, soThis contradicts the positive definiteness required of an inner product. Even the nonzero matrix unit $E_{12}$ has $\phi (E_{12},E_{12})=0$.

The printed assertion has an exception: when $n=1$, $\phi (a,b)=ab$ is an inner product on $\mathbb{R}$. Thus the requested failure requires $n\ge 2$, a restriction absent from the printed question.

Interpret diagonalizability over the specified underlying field $\mathbb{R}$. Write $A=P\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)P^{-1}$ with real eigenvalues. Invariance of the trace under similarity givesIf $A\ne 0$ is diagonalizable, at least one eigenvalue is nonzero, proving $\phi (A,A)>0$. Real diagonalizability is essential: the rotation block from part (c) is diagonalizable over $\mathbb{C}$ but has eigenvalues $i,-i$ and negative trace of its square.

Let $E_{ji}$ be the matrix unit having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\mathrm{tr}(A{E}_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. ThusThe matrix trace pairing is therefore a nondegenerate bilinear form, despite its failure to be positive definite when $n\ge 2$.

First extract the polynomial content from $f$:The primitive quartic is Eisenstein at the prime $2$: every nonleading coefficient is divisible by $2$, while the constant term is not divisible by $4$. Eisenstein's criterion makes it irreducible over $\mathbb{Q}$, and Gauss's lemma makes it irreducible in $\mathbb{Z}[X]$. The constant factor $2$ is itself irreducible in $\mathbb{Z}[X]$.

The polynomial $g$ is already irreducible over $\mathbb{Q}$: it is Eisenstein at $3$. Its nonleading coefficients, including the missing zero coefficients, are divisible by $3$, and $-21$ is not divisible by $9$. Thus $g$ is its own sole nonconstant irreducible factor over $\mathbb{Q}$.

Over the field ${\mathbb{Z}}_3$, grouping terms givesThe first two factors are linear and therefore irreducible. The quadratic takes the values $1,2,2$ at $0,1,2$ respectively, so it has no root in ${\mathbb{Z}}_3$. A degree-two polynomial over a field is reducible exactly when it has a root, proving its irreducibility.

All three factorizations are unique up to order and multiplication by units. The ring $\mathbb{Z}$ is a unique factorization domain, hence so is $\mathbb{Z}[X]$ by Gauss's lemma. A polynomial ring over a field is a Euclidean domain, so both $\mathbb{Q}[X]$ and ${\mathbb{Z}}_3[X]$ have unique factorization. The units are $\pm 1$ in the first ring, nonzero rational constants in the second, and $1,2$ in the third.

In the subspace topology, $B$ is relatively open in $A$ if $B=A\cap U$ for an open subset $U$ of the ambient Euclidean space. Equivalently, every $b\in B$ has a radius $\epsilon >0$ with $A\cap B_{\epsilon }(b)\subseteq B$. It is relatively closed if its complement $A\setminus B$ is relatively open, equivalently $B=A\cap F$ for an ambient closed set $F$.

For the upper half of the open disk, $B$ is closed in $A$, but not open in $A$. It is the intersection with the closed half-plane $\mathrm{Im}z\ge 0$. A relative neighborhood of zero contains points of negative imaginary part, so relative openness fails.

The rational cut is both open and closed in $\mathbb{Q}$. It is relatively open as $\mathbb{Q}\cap (-\infty ,\sqrt{2})$. Since $\sqrt{2}$ is irrational, its relative complement is $\mathbb{Q}\cap (\sqrt{2},\infty )$, also relatively open. Thus this cut is a clopen set in the subspace topology of the rationals.

A set is compact if every open cover of it has a finite subcover. Let $K_1,\dots ,K_m$ be compact and let $\mathcal{U}$ cover their union. For each $j$, the same family covers $K_j$, so compactness supplies a finite subfamily ${\mathcal{U}}_j$ covering $K_j$. The union of these finitely many finite subfamilies is finite and covers ${\bigcup }_j{K}_j$. Hence a finite union of compact sets is compact.

An arbitrary union need not be compact. Each singleton $\{n\}$, $n\ge 1$, is compact, but their union $\mathbb{N}$ has the open cover $\{(n-1/3,n+1/3):n\ge 1\}$ with no finite subcover. Every member covers only one positive integer, so finitely many leave infinitely many integers uncovered.

The complex line integral along the smooth parametrized curve isFor the reversed curve, the chain rule gives ${\tilde{\gamma }}^{\prime }(t)=-{\gamma }^{\prime }(a+b-t)$. Substitution $s=a+b-t$ yieldsConsequently reversing orientation negates the complex line integral.

An entire antiderivative is $F(z)=z\sin z+\cos z$, since differentiation cancels the sine terms and leaves $F^{\prime }(z)=z\cos z$. The curve runs from $-i$ to $i$. Both terms of $F$ are even, so $F(i)=F(-i)$, and the complex line integral gives

Along the circle, $z=z_0+r{e}^{it}$ and $dz=ir{e}^{it}dt$. Expanding the nonholomorphic integrand givesAfter multiplication by $ir{e}^{it}$, all nonconstant integer-frequency exponentials have zero integral over $[0,2\pi ]$. The remaining constant term is $i{r}^2{z}_0$. ThereforeThis direct parametrization avoids applying a holomorphic-antiderivative argument to $|z{|}^2$, which is not holomorphic.

For the positively oriented circle $C=\{z:|z-z_0|=r\}$, the Cauchy derivative formula statesThe hypothesis that the closed disk lies in the open holomorphy domain ensures the formula applies on and inside this circle.

On the circle, the integrand in the Cauchy derivative formula has modulus at most $M/r^{k+1}$, and the circle has length $2\pi r$. The ML inequality therefore givesThe maximum $M$ exists because the modulus of the continuous function $f$ attains a maximum on the compact circle. These are the Cauchy estimate.

Suppose, for a contradiction, that $f$ has no zero in the open disk. The strict boundary inequality implies no boundary zero either. Thus $1/f$ is holomorphic on a neighborhood of the closed disk: compactness and continuity keep $f$ nonzero in a sufficiently small neighborhood. Apply the $k=0$ Cauchy estimate to $1/f$:This contradicts the assumed strict inequality. If $f(z_0)=0$, the conclusion is immediate without taking a reciprocal. Therefore $f$ has a zero in the open disk. This is a local form of the minimum modulus principle.

The product $(\tanh s)y(s)$ is continuous, so the fundamental theorem of calculus makes the integral continuously differentiable. Differentiating the integral equation and then evaluating at zero gives $y^{\prime }=(\tanh x)y$, $y(0)=1$. Since $(\log \cosh x{)}^{\prime }=\tanh x$, an integrating factor is $1/\cosh x$. Thus $(y/\cosh x{)}^{\prime }=0$, andPut $L(x)=\log \cosh x$, so $L(0)=0$ and $L^{\prime }=\tanh x$. The claimed Picard iterate formula holds for $n=0$. Assuming it at $n$, termwise integration of the finite sum givesThis proves by inductionThe same calculation works for negative $x$ because it uses an antiderivative rather than a monotonicity substitution. On every bounded interval $|x|\le R$, $0\le L(x)\le \log \cosh R$, so comparison with the exponential-series tail also proves uniform convergence there.

Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the autonomous differential equation makes its trajectory a simple closed curve $C$, bounding a region $D$ by the Jordan curve theorem. The Green theorem givesAlong the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the Bendixson-Dulac criterion in this case.

For the second-order equation set $y=x^{\prime }$, giving $x^{\prime }=y$, $y^{\prime }=-y-\epsilon y^3-\sin x$. Choose the Dulac function $\rho =-1$. ThenThus there is no nonconstant periodic solution for $\epsilon >0$. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H^{\prime }=-y^2-\epsilon y^4$, whose integral over a period could vanish only if $y\equiv 0$.

For $p$ in a half-plane of absolute convergence, interchange the integrals using Fubini's theorem:For positive real $p$, integration by parts gives $I_n={\int }_0^{\infty }{e}^{-pt}{t}^{n}dt=n{I}_{n-1}/p$, starting from $I_0=1/p$. Hence $I_n=n!/p^{n+1}$, orThe formula extends throughout $\mathrm{Re}p>0$ by the same integration argument. Write $X=\mathcal{L}x$. The integral term is the convolution of $t$ with $x(t)$, whose Laplace transform is $X/p^2$. The derivative transforms to $pX-x(0)=pX-1$. Thereforeand elementary factorization givesInverting the two polynomial transforms yieldsDirect substitution verifies $x^{\prime }=t$ and ${\int }_0^t(t-s)(1+s^2/2)ds=t^2/2+t^4/24$, as required. Uniqueness also follows without transform assumptions: differentiating the equation twice gives $x^{(3)}+x=1+t^2/2$ with initial data $x(0)=1$, $x^{\prime }(0)=0$, $x^{\prime \prime }(0)=1$, a linear initial-value problem with a unique solution.