Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 (AC1, XMAT 4301, XMST 4301)

Source: released examination bundle, Part A 2007 11-1.pdf, pages 259–264. The printed title is Algebra, Analysis and Differential Equations I. The paper is cross-listed for mathematics and mathematics and statistics and contains no printed mark allocations.

The annihilator of a vector subspace is the vector subspace of the dual space consisting of the linear functionals that vanish on $U$:It is a vector subspace because every linear combination of functionals vanishing on $U$ still vanishes there.

For the basis $e_1,\dots ,e_n$, define the dual basis byUniqueness of the coordinates makes each $e_i^{\prime }$ well-defined and linear. If ${\sum }_i{a}_i{e}_i^{\prime }=0$, evaluation at $e_j$ gives $a_j=0$, so these linear functionals are linearly independent. For any $\lambda \in V^{\prime }$, evaluation on each basis vector givesBoth sides agree on a basis, hence on every linear combination. Thus they also span the dual space and are a basis of it.

Let ${\epsilon }_i$ denote the coordinate linear functional dual to the $i$th vector of the standard basis. Writing a vector in the supplied basis givesSuccessive differences recover the coefficients. Hence the dual basis isThese formulas directly give $v_i^{\prime }(v_j)={\delta }_{ij}$ and exemplify the dual basis of cumulative coordinate sums.

The span $U$ is the coordinate plane $\{(x_1,x_2,0,0{)}^T\}$. Consequently its annihilator of a vector subspace is $U^{\circ }=\mathrm{span}\{{\epsilon }_3,{\epsilon }_4\}$. The dual map is composition with $T$, soIt follows thatEquivalently, $T^{\prime }(a{\epsilon }_3+b{\epsilon }_4)$ sends $x$ to $a{x}_3$.

The Gram-Schmidt process subtracts projections onto directions already constructed and then normalizes. Define recursivelywhere the empty sum for $k=1$ is zero. Assuming the preceding $e_j$ are orthonormal, taking the inner product of $w_k$ with $e_i$, $i<k$, gives zero. Moreover $w_k\ne 0$: otherwise $v_k$ would belong to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Normalization is therefore legitimate and produces an orthonormal basis. At each stage the first $k$ constructed vectors have the same span as the first $k$ original vectors.

A real inner product must be symmetric. Here ${\beta }_a(e_1,e_2)=-2$ and ${\beta }_a(e_2,e_1)=-a$, forcing $a=2$. For this value,This is strictly positive for every nonzero vector. Bilinearity is already given, so symmetry and positive definiteness prove

Use the inner product with $a=2$. The required first vector $v_1=(1/2,0{)}^T$ already has squared norm $4(1/2{)}^2=1$. Apply the Gram-Schmidt process to $v_1$ followed by $e_2=(0,1{)}^T$. Since $\beta (e_2,v_1)=-1$, its orthogonal remainder is $w=e_2+v_1=(1/2,1{)}^T$. Its squared norm is $\beta (w,w)=1-2+3=2$. Therefore an orthonormal basis containing the specified vector isThe second vector may also be replaced by its negative.

The Chinese remainder theorem for two comaximal two-sided ideals $I,J$ of a ring states that the natural ring homomorphism inducesIndeed, the map $r\mapsto (r+I,r+J)$ has kernel of a ring homomorphism $I\cap J$. Choose $i\in I$, $j\in J$ with $i+j=1$. Given target representatives $a,b\in R$, the element $aj+bi$ is congruent to $a$ modulo $I$ and to $b$ modulo $J$, proving surjectivity. The first isomorphism theorem for rings gives the result. For a commutative ring, $I\cap J=IJ$: if $x\in I\cap J$, then $x=xi+xj\in IJ$, while the reverse inclusion follows from the ideal properties.

Take $R=\mathbb{Z}$, $I=a\mathbb{Z}$ and $J=b\mathbb{Z}$. Since $a,b$ are coprime integers, Bézout's identity gives $ra+sb=1$, so $I+J=\mathbb{Z}$. Their intersection is $ab\mathbb{Z}$: it consists of integers divisible by both $a$ and $b$, and coprimality makes this equivalent to divisibility by $ab$. Thus the Chinese remainder theorem givesas a ring isomorphism. This includes $a=1$ or $b=1$, where the corresponding factor is the zero ring.

Let $N$ count everyone present, including the host. The grouping, seating and departure conditions give the modular congruencesThe first two imply $N=39+45k$, and oddness forces $k$ even. By the Chinese remainder theorem, the complete solution isBecause not every invitee attended, at most $y-1$ guests came, so $N\le y$. For $y=120$, the only positive possibility is 39 people. It is feasible: seven groups of five and one of four; three tables of ten and one of nine; nineteen departing pairs and the host.

For $y=150$, both 39 and 129 people are possible. The latter uses twenty-five groups of five and one of four, three tables of ten and eleven of nine, and sixty-four departing pairs. Thus the larger invitation count destroys uniqueness; it does not force a different attendance.

In the subspace topology on $S$, a subset $A$ is open in $S$ precisely when $A=S\cap O$ for some open set $O\subseteq \mathbb{C}$. A subset $B$ is closed in $S$ precisely when $S\setminus B$ is open in $S$, equivalently $B=S\cap F$ for some closed set $F\subseteq \mathbb{C}$.

A compact space is one for which every open cover has a finite subcover. Here the cover may use relatively open subsets of $S$, or equivalently ambient open subsets covering it. The Heine-Borel theorem states thatClosedness and boundedness here refer to the ambient complex plane, not merely closedness in $S$ itself.

A connected space $S$ cannot be expressed as $S=A\cup B$ with $A,B$ disjoint nonempty relatively open sets. Equivalently, the only subsets simultaneously relatively open and closed are $\varnothing$ and $S$. A decomposition into such $A,B$ is a separation of the topological space.

A path-connected space $S$ has, for every pair $z_0,z_1\in S$, a continuous path $\gamma :[0,1]\to S$ with $\gamma (0)=z_0$ and $\gamma (1)=z_1$. Every path-connected space is a connected space, because a separation of the target would pull a joining path apart and contradict connectedness of an interval.

The integers are unbounded in the complex plane, so the Heine-Borel theorem shows they are not compact. They are also not connected: $\{0\}$ and $\mathbb{Z}\setminus \{0\}$ are disjoint nonempty relatively open sets covering $\mathbb{Z}$. Each integer is isolated, so the induced subspace topology is discrete.

The disc is a closed set and is bounded in the complex plane, so it is compact by the Heine-Borel theorem. It is convex: the triangle inequality gives $|(1-t)z_0+t{z}_1|\le 2$ for all $t\in [0,1]$ when both endpoints lie in it. Straight segments therefore supply paths, so it is path-connected and hence connected.

The union is bounded but is not a closed set: the points $-1+1/n$, $n\ge 2$, belong to the open disc and tend to $-1$, which lies in neither disc. Thus it is not compact by the Heine-Borel theorem.

Nevertheless it is connected, indeed path-connected. Every point of the open disc joins $0$ by a straight segment. The segment from $0$ to $1$ stays in the union, since its points before the endpoint belong to the open disc and the endpoint $1$ belongs to the closed disc. Every point of that closed disc joins $1$ inside it by convexity. Concatenating these paths joins any two points of the union. The fact that the two discs are disjoint does not imply disconnection: their contact point belongs to one of them. This illustrates connectedness at a one-sided contact point.

An isolated singularity at $a$ means that the function is holomorphic on some punctured disc $0<|z-a|<r$, while it is undefined or fails to be holomorphic at $a$. The punctured neighbourhood contains no other singularities; this isolation is needed before applying the usual removable/pole/essential classification.

A removable singularity at $a$ is an isolated singularity for which one can assign a value at $a$ to obtain a holomorphic function on the whole disc. Equivalently its Laurent series has no negative powers. A finite limit $L$ supplies the extension value $f(a)=L$.

A pole of order $m\ge 1$ at $a$ meansEquivalently the lowest nonzero power in its Laurent series is $(z-a{)}^{-m}$. In particular, $(z-a{)}^{m}f(z)$ tends to a finite nonzero limit.

The possible singular points are $i,-i,1$. Near $i$, put $h(z)=1/((z+i)(z-1{)}^2)$; thenThe numerator vanishes at $i$, so the apparent pole cancels and $i$ is removable. Its extension value is $h^{\prime }(i)=1/4+3i/8$, obtained from $h^{\prime }/h=-1/(z+i)-2/(z-1)$.

At $-i$, the first term has a simple pole with nonzero residueThe subtracted term is holomorphic there, so $-i$ is a simple pole. At $1$,so $1$ is a double pole. There are no other finite singularities.

Away from zero, singularities occur exactly when $e^{1/z}=-1$, namelyFor $D(z)=z(e^{1/z}+1)$, differentiation at $z_k$ gives $D^{\prime }(z_k)=1/z_k\ne 0$. Thus every $z_k$ is a simple pole, with residue $z_k$.

As $|k|\to \infty$, these poles accumulate at zero. Therefore zero is a non-isolated singularity. Every punctured disc around it contains poles, so the function is not holomorphic on any punctured disc there. In particular zero cannot be classified as an isolated essential singularity; the isolated-singularity classification does not apply. The displayed poles and their accumulation point zero exhaust the finite singularities.

The contour integral along a piecewise smooth path is defined by parametrization:with a sum over the smooth pieces. The residue theorem states that for a positively oriented simple closed contour and a function holomorphic on and inside it apart from finitely many isolated singularities $a_j$ inside,More generally a closed contour uses the winding number of each singularity as its weight.

Set $z=e^{it}$, so $dt=dz/(iz)$ and $\cos t=(z+z^{-1})/2$. The required real integral becomesThe denominator factors as $(2z^2+1)(z^2+2)$. Only the simple poles $z=\pm i/\sqrt{2}$ lie inside the unit circle. At either one,The residue theorem now gives

With $y=xv$, direct differentiation givesThe homogeneous ordinary differential equation therefore has general solution $y=x(A\cos x+B\sin x)$. Chooseso $y_1(\pi /2)=0$ and $y_2(\pi )=0$. Their Wronskian in the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ is $W(x)=x^2$. There is no nonzero homogeneous solution satisfying both boundary conditions, since the first forces $B=0$ and the second forces $A=0$.

For the operator as printed, whose leading coefficient is $x^2$, the Green's function isIt satisfies the endpoint boundary conditions, is continuous across $x=s$, and its first derivative has jump $1/s^2$. Multiplying that jump by the leading coefficient $s^2$ produces the unit source required by a Green's function. This factor matters because the operator was not normalized to leading coefficient one; it is an instance of a Green function under dependent-variable rescaling.

Thus $y(x)={\int }_{\pi /2}^{\pi }G(x,s)f(s)ds$. For $f(s)=s^3$, split the integral at $s=x$ to obtainThereforeThe substitution above gives $v^{\prime \prime }+v=1$, so the forcing is indeed $x^3$, and both endpoint values vanish.

Both components vanish atThe Jacobian matrix of the vector field isIts eigenvalues classify these equilibria by linearization of a dynamical system:All real parts are nonzero, so the linear classifications determine the local stability of the nonlinear autonomous system.

At the origin, the stable direction is the $y$-axis and the unstable direction is the $x$-axis. At $(1,0)$ the eigendirections are $(1,0)$ for $-1$ and $(1,-2)$ for $-3$. At $(0,1)$,has only one eigendirection, the $y$-axis; its linearization of a dynamical system has solutions $\delta x=A{e}^{2t}$, $\delta y=(B-At)e^{2t}$, explaining the improper node. At $(4,3)$, stable and unstable tangent vectors are respectively $(4,5-\sqrt{13})$ and $(4,5+\sqrt{13})$.

For the phase portrait, the axes are invariant, and the remaining nullclines are $y=x-1$ and $y=x/2+1$. Off these lines, the complete directional information isIn the positive quadrant, arrows point right above $y=x-1$ and left below it, and point up above $y=x/2+1$ and down below it. On the positive $x$-axis, trajectories tend to $(1,0)$. On the $y$-axis, $0<y<1$ flows toward the origin and $y>1$ flows upward. These signs and the saddle directions give the following sketch; the plotted curves illustrate the phase portrait, with the marked saddle curves tangent to the calculated eigenvectors.

Introduce a characteristic parameter $s$ and label each point of the initial segment by $r\in [0,1]$. The method of characteristics givesIntegrating gives the parametric solutionThe projected characteristic curves satisfy $y=r(2-x)$, so each lies on a straight line through $(2,0)$. Its branch from the data always has $x<2$; the common vertex is approached only as $s\to -\infty$.

Eliminating $s,r$ using $e^s=2-x$ and $r=y/(2-x)$ yieldsIts derivatives give $(x-2)z_x+y{z}_y=y$, and substitution of $x=1$ gives the prescribed value one.

The Jacobian determinant of $(s,r)\mapsto (x,y)$ is $-e^{2s}\ne 0$, and the inverse just displayed is unique. Hence the data determine the solution exactly onThis is the domain determined by a finite segment under a radial characteristic flow. Its interior is $x<2$, $0<y<2-x$; the two boundary rays come from the endpoints of the initial segment. Every point there lies on exactly one characteristic from the data. Outside this sector, a characteristic either misses the initial segment or lies on the branch $x>2$, which cannot reach it. Thus the formula may be extended beyond this sector, but such an extension is not uniquely specified by the given data.