Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 (AC1, XMAT 4301, XMST 4301)

The minimal polynomial $m_T$ is the unique monic polynomial of least degree with $m_T(T)=0$. Existence follows from a linear dependence among powers of $T$ in the finite-dimensional vector space $\mathrm{End}(V)$; polynomial division proves uniqueness and that it divides every annihilating polynomial. The characteristic polynomial is ${\chi }_T(t)=\det (tI-[T{]}_{\mathcal{B}})$, independent of basis because a change of basis conjugates the matrix.

WriteIn the cyclic basis, the companion matrix isFor $q(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the displayed vector relation gives $q(T)v=0$. Polynomial evaluation commutes with $T$, so $q(T)T^{j}v=T^{j}q(T)v=0$ for every basis vector. Thus $q(T)=0$.

No nonzero polynomial of degree below $n$ can annihilate $T$, since applying it to the cyclic vector $v$ would contradict independence of $v,Tv,\dots ,T^{n-1}v$. Hence $m_T=q$. Finally, direct expansion of $\det (tI-C)$, or induction along the last column, gives ${\chi }_T(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0=q(t)$. ThereforeThe argument computes the determinant and annihilator directly and does not invoke the Cayley-Hamilton theorem.

A projection is a linear operator $E$ satisfying $E^2=E$; it need not be an orthogonal projection without an inner product. Pairwise annihilation givesThus their sum is a projection.

The identities $E_{i}P=E_i$ imply $T(I-P)=0$. Since $T$ is nonsingular, $P=I$. Alsoso $\mathrm{im}{E}_i\subseteq V_i$. Conversely, if $Tx={\lambda }_{i}x$, applying $E_j$ gives $({\lambda }_j-{\lambda }_i)E_{j}x=0$. For $j\ne i$ the eigenvalues are distinct, hence $E_{j}x=0$. Since $P=I$, $x=E_{i}x$, proving $\mathrm{im}{E}_i=V_i$.

Every vector is ${\sum }_i{E}_{i}x$. If ${\sum }_i{v}_i=0$ with $v_i\in V_i$, applying $E_j$ gives $v_j=0$. ThusThese operators are the spectral projections of a diagonalizable matrix, expressed without choosing an eigenbasis.

For a self-adjoint operator on a finite-dimensional complex inner product space, every eigenvalue is real and eigenspaces for distinct eigenvalues are orthogonal. The spectral theorem gives an orthonormal eigenbasis. Accordinglyand the $E_i$ are orthogonal projections satisfying $E_i^{\ast }=E_i$. Since $T$ is nonsingular, none of its eigenvalues is zero. Its representing matrix in the orthonormal eigenbasis is real diagonal.

Reduction modulo the prime $p$ preserves integer addition, multiplication, and one:Every congruence class has an integer representative, so the ring homomorphism is onto. Its kernel is exactly $p\mathbb{Z}=(p)$. The quotient is a field because every class not divisible by $p$ has a multiplicative inverse by Bézout's identity.

For the coefficientwise extension to the polynomial ring, addition is preserved term by term. If $f={\sum }_i{a}_i{x}^i$ and $g={\sum }_j{b}_j{x}^j$, the coefficient of $x^k$ in their product is ${\sum }_{i+j=k}{a}_i{b}_j$. Reducing it modulo $p$ gives ${\sum }_{i+j=k}{\bar{a}}_i{\bar{b}}_j$, the coefficient in $\hat{\varphi }(f)\hat{\varphi }(g)$. Thus multiplication and one are preserved as well. Lift each coefficient of a polynomial over ${\mathbb{Z}}_p$ to an integer to prove surjectivity. The kernel isIndeed, its elements are precisely the polynomials whose coefficients are all divisible by $p$.

In the binomial theorem expansion of $(x-1{)}^p$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$: the numerator contains the prime factor $p$, while $j!(p-j)!$ does not for $1\le j<p$. HenceFor $p=2$, the last equality uses $1=-1$ in characteristic two. This is an instance of the Frobenius endomorphism.

Write $z=x+iy$, so the real and imaginary components are $u=x^2$ and $v=y^2$. The Cauchy-Riemann equations require $u_x=v_y$, namely $2x=2y$; the other equation is $u_y=-v_x=0$. The first relation fails off the line $x=y$, so the function is not holomorphic on $\mathbb{C}$. It is complex differentiable at points of that line, but a line contains no open neighborhood on which the equations hold.

The reciprocal map is holomorphic for $z\ne 0$, and the complex exponential is entire. Its denominator vanishes exactly whenAll these points are excluded from $U$. The set $U$ is open: it is the inverse image of $\mathbb{C}\setminus \mathbb{Z}$ under the reciprocal map on $\mathbb{C}\setminus \{0\}$. A reciprocal of a nonvanishing holomorphic function is holomorphic. Therefore $f$ is holomorphic throughout $U$.

The set $U$ has two open connected components, the left and right half-planes. On each the function is constant in a neighborhood of every point, so its complex derivative is zero everywhere. Thus $f$ is holomorphic on $U$. A holomorphic function can take different constant values on different components of a disconnected domain.

At any point on the imaginary axis, approaching from the left gives limit $-1$, while the assigned value and the right-hand limit are $1$. Thus the function is discontinuous on that axis. Complex differentiability implies continuity, so $f$ is not holomorphic on $\mathbb{C}$. It is holomorphic separately in the two open half-planes.

The hyperbolic cosine is entire and $\cosh 0=1$. Continuity makes it nonzero in a neighborhood of zero, so its reciprocal is holomorphic there and has a convergent Taylor series. Since $\cosh$ and its reciprocal are even, all odd coefficients vanish.

Multiply the proposed expansion by $\cosh z$:Coefficient comparison gives $c_2+1/2=0$ and $c_4+c_2/2+1/24=0$, henceThe zeros of $\cosh z$ solve $e^{2z}=-1$, so they are $z=i(\pi /2+k\pi )$, $k\in \mathbb{Z}$. The nearest ones are $\pm i\pi /2$, and they are simple because $\sinh$ is nonzero there. The hyperbolic secant has genuine poles at these points and is holomorphic in the disk of radius $\pi /2$. The radius of convergence of its Taylor series is therefore

Let $F(z)=e^{i\lambda z}/(z-i)$. For $\lambda >0$, close the interval $[-R,R]$ by the upper semicircle, counterclockwise. Its only enclosed pole is $z=i$, with residue $e^{-\lambda }$. The residue theorem givesOn $z=R{e}^{i\theta }$, $|e^{i\lambda z}|=e^{-\lambda R\sin \theta }$ and $|z-i|\ge R-1$. Hence the arc is bounded byby the supplied estimate, with parameter $\lambda R$.

For $\lambda <0$, close below, with clockwise orientation. The pole at $i$ lies outside, and the exponential decays there. Parametrizing the lower arc yields the same bound with $|\lambda |R$ in place of $\lambda R$. Its contribution again tends to zero, and no residue remains. Therefore

This is the contour-decay mechanism behind Jordan's lemma; the pole contributes only when the decaying closure encloses it.

Rewrite the linear equation as $(x^{2}y{)}^{\prime }=0$. The initial value gives the unique solutionThe Picard iteration is its integral equationThe proposed formula holds for $n=0$. If it holds at $n$, termwise integration of the finite sum and $d(\log t)/dt=1/t$ giveThis proves the expression by induction. The exponential series then gives $y_n(x)\to e^{-2\log x}=x^{-2}$. Convergence is also uniform on every compact subinterval of $(0,\infty )$, since $|\log x|$ is bounded there and the exponential majorant series converges. This is Picard iteration for a simple-pole linear differential equation.

Write the principal part as $e^{2y}{u}_{xx}-u_{yy}$. The characteristic curves satisfyor $dx/dy=\pm e^y$. Integration gives $x+e^y=\text{constant}$ and $x-e^y=\text{constant}$. Thus take $\xi =x+e^y$, $\eta =x-e^y$. Their Jacobian determinant is $-2e^y\ne 0$, and their image is $\xi >\eta$.

For $u(x,y)=U(\xi ,\eta )$, the chain rule givesThe first-derivative terms cancel exactly. The canonical form of a second-order partial differential equation isOn $y=0$, $F(x+1)+G(x-1)=x$. Differentiating this condition in $x$ and using $u_y=1$ givesHence $F^{\prime }=1$ and $G^{\prime }=0$. Substitution into the value condition fixes their combined constant and yieldsDirect substitution verifies the PDE and both initial data. The cancellation illustrates a hyperbolic equation reduced by an exponential characteristic coordinate.

Under the usual piecewise smoothness and exponential-order conditions, integration by parts in the Laplace transform givesThe boundary term at infinity vanishes in the transform's half-plane of convergence. The value at zero supplies the initial-data term.

The convolution theorem for Laplace transforms stateswhen $x(t)={\int }_0^{t}y(s)z(t-s)ds$ and the transforms converge appropriately. One proves it by changing variables $t=s+r$ in the double integral over $0\le s\le t<\infty$, which becomes the product domain $s,r\ge 0$. Absolute convergence justifies the interchange.

Using $\sin t=(e^{it}-e^{-it})/(2i)$ and integrating the two exponentials for $\mathrm{Re}p>0$,This also follows directly by two integrations by parts in the Laplace transform.

Applying the derivative rule twice and using both zero initial data givesso $\bar{x}(p)=1/(p^2+1{)}^2$. By the convolution theorem for Laplace transforms,Use the product-to-sum identity $\sin s\sin (t-s)=\frac{1}{2}[\cos (2s-t)-\cos t]$ to integrate:It satisfies $x(0)=x^{\prime }(0)=0$, and differentiation gives $x^{\prime \prime }+x=\sin t$. The term proportional to $t\cos t$ is the resonant response of the forced harmonic oscillator.