Past exam of the mathematics course of the University of Cambridge / 2017 / ia

By Euler's formula, $z=e^{i\theta }$. Factoring out the half-angle givesSince $0\le \theta <\pi$, the cosine is positive. ThereforeOn the Argand diagram, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta /2$; resolving either side along the diagonal gives its length $2\cos (\theta /2)$.

Similarly,Thus, for $0<\theta <\pi$,At $\theta =0$, $1-z=0$, whose complex argument is undefined. Geometrically, $|1-z|$ is the chord joining the two unit-circle points $z$ and $1$. The isosceles triangle with vertex angle $\theta$ has half-chord $\sin (\theta /2)$, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

Componentwise,soRepeated application gives $(A^k{)}^T=(A^T{)}^k$. Transposing the matrix exponential term by term therefore yieldsMultiplication of the two power series givesHenceIf $e^{tA}$ is an orthogonal matrix for every real $t$, thenIts coefficient of $t$ must vanish, sothat is, $A$ is a skew-symmetric matrix.

Because $(a_n)$ is a monotone sequence, every one of its first $n$ terms is at most $a_n$, soOn the other hand, every term from $a_n$ through $a_{2n}$ is at least $a_n$, and henceBoth the lower bound and upper bound tend to $x$ because the Cesaro means converge to $x$. The squeeze theorem gives

Convergence at $z_0$ implies that the terms $a_n{z}_0^n$ form a bounded sequence: $|a_n{z}_0^n|\le M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, thenComparison with the convergent geometric series $\sum M{r}^n$ proves absolute convergence. The case $z_0=0$ is vacuous.

The radius of convergence is the number $R\in [0,\infty ]$ such that a power series converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.

TakeThen $\sum v^n/n$ is the convergent alternating harmonic series, whereas $\sum w^n/n$ is the divergent harmonic series. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that

The cross product $x\times y$ is the vector perpendicular to $x$ and $y$, of magnitude $|x||y|\sin \alpha$ and orientation fixed by the right-hand rule. In suffix notation,where ${\epsilon }_{ijk}$ is the Levi-Civita symbol. ThereforeUsinggivesHence the vector triple product identity is

The identity in part (a), together with $|a|=1$, givesFor every $n\ge 1$, $x_n$ is perpendicular to $a$, because it is a scalar multiple of $a\times x_{n-1}$. ThereforeAlso $|x_1|=\lambda |a\times x_0|$ and $|x_2|={\lambda }^2|a\times x_0|$. Induction using the two-step recurrence yields

The recurrence $x_{n+2}=-{\lambda }^2{x}_n$ shows that every odd-indexed position vector is a scalar multiple of $x_1$, while every positive even-indexed position vector is a scalar multiple of $x_2$. HenceMoreover $x_2=\lambda a\times x_1$, so these two straight lines through the origin are perpendicular.

Let $d_n=x_{n+1}-x_n$ be the direction vector of the segment $X_n{X}_{n+1}$. Consecutive vectors $x_n,x_{n+1}$ are perpendicular and $|x_{n+1}|=\lambda |x_n|$. Using $x_{n+2}=-{\lambda }^2{x}_n$,Thus adjacent segments are perpendicular. AlsosoIf $0<\lambda <1$, the norm formula in part (i) gives $x_n\to 0$. If $\lambda =1$, then $x_{n+2}=-x_n$: the points $X_1,X_2,-X_1,-X_2$ repeat cyclically and form a square centered at the origin.

Put $L=|x_n|$ and choose perpendicular unit vectors $e,f$ such that $x_n=Le$ and $x_{n+1}=\lambda Lf$. Direction vectors for the two lines areandTheir scalar product and norms giveTherefore the acute angle between the straight lines satisfies

A matrix and its matrix transpose have the same characteristic polynomial, becauseThus they have the same eigenvalues, including algebraic multiplicities.

The proposed sum identity is false. Forboth eigenvalues ${\lambda }_i$ are zero, whereashas eigenvalues $0,1$. HenceEquivalently, the two sides are generally $\mathrm{tr}(A^{T}A)$ and $\mathrm{tr}(A^2)$, which need not agree.

The product identity always holds, sinceTherefore

If $p=m\times n$, then $n^{T}p=0$, and hence $Bp=p$. Also $Bm=m$ because $m\perp n$. Thus $p$ and $m$ are linearly independent eigenvectors with eigenvalue one.

For every vector $v$,so $B-I$ is nonzero but has square zero:Consequently all three eigenvalues are one. The eigenspace iswhich has dimension two. It cannot supply three linearly independent eigenvectors, so

Multiplication givesThe vector $p=m\times n$ is an eigenvector with eigenvalue one. On the plane spanned by the orthonormal pair $m,n$, the matrix isIts eigenvalues are the roots of ${\mu }^2-3\mu +1$, namelyFor either root, an eigenvector is $m+({\mu }_{\pm }-1)n$. ThereforeThese positive eigenvalues are the squared singular values of $B$.

If $A^T=-A$, then the scalar $x^{T}Ax$ equals its own transpose and henceso it vanishes.

Conversely, suppose $x^{T}Ax=0$ for every real $x$. Taking $x=e_i$ gives $A_{ii}=0$. Taking $x=e_i+e_j$ then gives $A_{ij}+A_{ji}=0$. Thus every entry satisfies $A^T=-A$, and

A real skew-symmetric matrix satisfies $A^{†}=A^T=-A$, so it is a skew-Hermitian matrix. If $Ax=\lambda x$, thenTaking the complex conjugate and using $A^{†}=-A$ shows that this scalar is the negative of its conjugate. Henceso every eigenvalue is purely imaginary or zero.

If $Ax=\lambda x$ and $Ay=\mu y$, thenSince ${\lambda }^{\ast }=-\lambda$, this becomes $(\mu -\lambda )x^{†}y=0$. Distinct eigenvalues therefore have orthogonal eigenvectors:

In odd dimension,so $\det A=0$. Since $A$ is real, its kernel contains a nonzero real vector $a$.

The plane $a^{\perp }$ is invariant under $A$, because $a\cdot Ab=-(Aa)\cdot b=0$. On this two-dimensional plane, $Ab$ is perpendicular to $b$. Put $|Ab|=\theta |b|$ with $\theta >0$. Then $A^{2}b$ is parallel to $b$, andsoThe matrix exponential fixes the kernel vector:Separating even and odd powers in its series and using $A^{2j}b=(-1{)}^j{\theta }^{2j}b$ givesThus $e^A$ acts as a rotation through angle $\theta$ on $a^{\perp }$ and fixes its axis $\mathbb{R}a$.

Write $y=(y_1,y_2,y_3{)}^T$. From the coordinate formula for the cross product,If $y=c_1{e}_1$, this becomesThus $c_1\ne 0$ gives rank three and a zero-dimensional kernel, while $c_1=0$ gives rank one and kernel dimension two.

If $y\cdot e_1=0$, then $y_1=0$. When $y\ne 0$, the equations $Tx=0$ force $x_1=0$ and $y_3{x}_2-y_2{x}_3=0$, soWhen $y=0$, again $\mathrm{rank}T=1$ and $\dim \ker T=2$. These dimensions also agree with the rank-nullity theorem.

First computeThe vector $\ell =(1,-2,4{)}^T$ lies in the kernel of $(AB{)}^T$, since ${\ell }^{T}AB=0$. If $ABx=d$ is solvable, multiplying by ${\ell }^T$ gives the necessary compatibility conditionTherefore

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then there exists $c\in [a,b]$ such that $f(c)=y$.

A function is differentiable at $a$ when the finite limitexists. A derivative need not be continuous. For example,is differentiable everywhere and $f^{\prime }(0)=0$, but for $x\ne 0$,which has no limit at zero.

The mean value theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfies

Set $h(x)=f(x)-yx$. ThenThe extreme value theorem gives a minimum of $h$ on $[a,b]$. If it occurs in $(a,b)$, the two-sided difference quotient gives $h^{\prime }(c)=0$. If it occurs at $a$, either $h^{\prime }(a)=0$ or a negative right derivative contradicts minimality; the endpoint $b$ is analogous. Hence some $c\in [a,b]$ satisfiesThis proves the Darboux theorem for derivatives in the stated case without assuming that $f^{\prime }$ is continuous.

Because a differentiable function is continuous, defineThe fundamental theorem of calculus gives $F^{\prime }(x)=f^{\prime }(x)+f(x)$. At the endpoints,whileApplying the result just proved to $F$ yields $d\in [a,b]$ with

The function $f$ is continuous at $y_0$ when for every $\epsilon >0$ there is a $\delta >0$ such thatSuppose $f$ has a local minimum at $c$ and is differentiable there. For sufficiently small $h>0$,Both one-sided limits equal $f^{\prime }(c)$, so it is both nonnegative and nonpositive. Therefore

The upper bound follows immediately from convexity, sinceFor the lower bound, writeApplying convexity and rearranging givesChoose a fixed $r_0>0$ with $[y_0-r_0,y_0+r_0]\subset (a,b)$. Applying these bounds with $r=r_0$ and $r=-r_0$ shows that the difference $f(y_0+h)-f(y_0)$ is trapped between quantities of order $|h|/r_0$. Both tend to zero, so every convex function on an open interval is continuous.

Now suppose $c$ is a local minimum but some $x$ satisfies $f(x)<f(c)$. For sufficiently small $\lambda >0$, the point $z=(1-\lambda )c+\lambda x$ lies in the neighbourhood on which $c$ is minimal, whereas convexity givesa contradiction. ThusDifferentiability at the minimizer is unnecessary: $f(x)=|x-c|$ is convex, has its global minimum at $c$, and is not differentiable there.