Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1

Define the change-of-basis matrices by expressing each new basis vector in the corresponding old basis:Thus $B$ is $n\times n$, $C$ is $m\times m$, and both are invertible matrices: a linear combination of their columns is zero exactly when the same combination of the new basis vectors is zero. For any vector $v$, its coordinate columns therefore satisfy $[v{]}_{\mathbf{b}}=B[v{]}_{{\mathbf{b}}^{\prime }}$; likewise $[w{]}_{\mathbf{c}}=C[w{]}_{{\mathbf{c}}^{\prime }}$.

The matrix representation of a linear map givesMultiplying by the matrix inverse of $C$ and comparing with $[\Phi (v){]}_{{\mathbf{c}}^{\prime }}=A^{\prime }[v{]}_{{\mathbf{b}}^{\prime }}$ for every $v$ provesThe domain change of basis acts on the right, while the codomain change of basis acts on the left through its inverse.

The polar form of a complex number and the roots of unity givesince $512=2^9$. Using the complex conjugate to compute the squared complex modulus,The residue $3j-k$ runs through every class modulo $9$, already with $j=0$. The cosine is even, so the nine angles give five distinct values, represented by $r=0,1,2,3,4$. They are distinct because cosine is strictly decreasing on $[0,\pi ]$ and these representative angles lie in that interval. HenceEvery listed value is attained by choosing $z_1=1$ and $z_2=2e^{2\pi ir/9}$.

For any complex number $w$, $\mathrm{Re}w\le |w|$, while the complex modulus is multiplicative and unchanged by complex conjugation. ConsequentlyBoth sides are nonnegative, so taking square roots proves the triangle inequality:The proposed further inequality fails for perpendicular vectors in the complex plane. Taking $z_1=1$, $z_2=i$ gives

A power series has radius of convergence $R$ when it converges absolutely for every $|z|<R$ and diverges for every $|z|>R$. Its behaviour on $|z|=R$ is not specified by this definition. When $R=0$, it converges at the centre and diverges at every nonzero point; an infinite radius of convergence means absolute convergence at every finite $z$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for all $n$ at some $|z|>R$. Choose a complex number $w$ with $R<|w|<|z|$. ThenThe comparison test for series with the convergent geometric series $\sum M{q}^n$ yields absolute convergence at $w$, contradicting the definition of $R$. Thus terms of a power series are unbounded outside its convergence disc:For the given power series, the ratio test gives, when $z\ne 0$,Hence it converges absolutely for $|z|<1$ and its terms fail to tend to zero for $|z|>1$ (the ratios are eventually larger than some number greater than $1$). Therefore $\boxed{R=1}$. In fact, on $|z|=1$ it also converges absolutely by comparison with the convergent p-series $\sum n^{-3}$.

Pairing the first and last terms of the arithmetic progression gives ${\sum }_{k=1}^{n}k=n(n+1)/2$. Therefore the limit of a sequence is

The sequence is at least $1$. Fix $\epsilon >0$. By the binomial theorem, for $n\ge 2$,once $(n-1){\epsilon }^2>2$. Taking positive $n$th roots gives $1\le n^{1/n}<1+\epsilon$ for all sufficiently large $n$. This proves directly from the definition of a limit of a sequence that

Since $0<a\le b$, we have $b^n\le a^n+b^n\le 2b^n$. Taking positive $n$th roots yieldsFor every $\epsilon >0$, the binomial theorem gives $(1+\epsilon {)}^n\ge 1+n\epsilon >2$ eventually, so $2^{1/n}\to 1$. The squeeze theorem therefore gives

The matrix trace is the sum of the diagonal entries:Using the definitions of matrix multiplication and matrix transpose,This is the symmetry of the real Frobenius inner product.

The matrix trace calculation in (i) givesA sum of nonnegative real numbers is zero exactly when each is zero. Thus $\boxed{\mathrm{tr}(A^{T}A)=0⟺A=0}$, so the real Frobenius inner product is positive-definite.

Write $\alpha =\mathrm{tr}(A^{T}A)$, $\beta =\mathrm{tr}(B^{T}B)$ and $\gamma =\mathrm{tr}(A^{T}B)$. If $B\ne 0$, then $\beta >0$, and applying the same nonnegativity to $A-tB$ givesTaking $t=\gamma /\beta$ yields $\alpha -{\gamma }^2/\beta \ge 0$. If $B=0$, the claimed inequality is immediate. Hence the Cauchy-Schwarz inequality here isWhen $B\ne 0$, equality holds exactly when $\mathrm{tr}((A-(\gamma /\beta )B{)}^T(A-(\gamma /\beta )B))=0$, hence exactly when $A=(\gamma /\beta )B$. Together with the case $B=0$, the complete equality condition is $\boxed{A,B\text{ are linearly dependent}}$, including the case that either is zero.

Writing $X=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ and using the Frobenius inner product, the three linear equations areThe first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the vector subspace has the unique expressionThe two coefficient matrices satisfy all three constraints. Their linear independence follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a basis for this orthogonal complement is

In an oriented orthonormal coordinate system, the dot product and cross product areThe scalar triple product is $a\cdot (b\times c)$; equivalently it is the determinant of the matrix whose columns are $a,b,c$. The three vectors have linear independence when

Let $U=(a_1{a}_2{a}_3)$ and $V=(b_1{b}_2{b}_3)$ be column matrices. Since $(U^{T}V{)}_{ij}=a_i\cdot b_j$, we have $S=U^{T}V$. Multiplicativity of the determinant and its invariance under matrix transpose yieldA square matrix has maximal matrix rank exactly when its determinant is nonzero, equivalently when its columns have linear independence. A product of two real numbers is nonzero exactly when both factors are nonzero. Hence $\boxed{\mathrm{rank}S=3⟺(a_i{)}_{i=1}^3\text{ and }(b_j{)}_{j=1}^3\text{ are both independent}}$.

The same argument works in every dimension. Set $U_n=(c_1\cdots c_n)$ and $V_n=(d_1\cdots d_n)$. Thenso $\boxed{\mathrm{rank}T=n⟺(c_i{)}_{i=1}^n\text{ and }(d_j{)}_{j=1}^n\text{ are both independent}}$.

For the last construction, choose two standard basis vectors $e_1,e_2$ in ${\mathbb{R}}^n$ and take $c_j=e_1+j{e}_2$ for $j=1,\dots ,n$. If $\alpha c_i+\beta c_j=0$ with $i\ne j$, its first two coordinates give $\alpha +\beta =0$ and $i\alpha +j\beta =0$, hence $\alpha =\beta =0$. Every pair therefore has linear independence. Every triple lies in the two-dimensional linear span of $e_1,e_2$ and has linear dependence. More explicitly, for distinct $i,j,k$,a nontrivial dependence. Thus $\boxed{c_j=e_1+j{e}_2(1\le j\le n)\text{ works for every }n>2}$.

By the definition of an eigenvalue, there is a nonzero eigenvector $v$ with $Av=\lambda v$. Induction using matrix multiplication givesstarting from $r=1$. Thus the same nonzero eigenvector establishes

If $\lambda =0$, then $Av=0$ for a nonzero eigenvector $v$, contradicting the existence of a matrix inverse. Hence $\lambda \ne 0$. Multiplying $Av=\lambda v$ by $A^{-1}$ gives $v=\lambda A^{-1}v$, so

The scalar matrix $tI$ commutes with $A$, soTaking determinants and using their multiplicativity gives, with the specified characteristic polynomial convention,

For $t\ne 0$, factor the matrix asThe determinant of a scalar multiple of an $n\times n$ matrix gains the scalar's $n$th power. ThusThe right side extends to a polynomial at $t=0$: multiplying ${\chi }_A(t^{-1})$ by $t^n$ removes all negative powers. Its value there is $(\det A{)}^{-1}$, since the leading coefficient of ${\chi }_A(s)=\det (A-sI)$ is $(-1{)}^n$. This equals ${\chi }_{A^{-1}}(0)$, so the identity holds at zero with that interpretation.

For the relation between eigenvalues and the characteristic polynomial,The first equivalence is the definition of an eigenvector; the others use the kernel of a linear map and the criterion for an invertible matrix to have nonzero determinant. Since ${\chi }_A$ has degree $n$, the fundamental theorem of algebra gives $n$ roots over $\mathbb{C}$ counted with multiplicity. The multiplicity of a root is its algebraic multiplicity, and the eigenspace is exactly $\ker (A-\lambda I)$; its dimension is the geometric multiplicity, which need not equal the algebraic multiplicity.

Finally, factor $p(s)=s^4-\mu$ over $\mathbb{C}$ as $p(s)={\prod }_{j=1}^4(s-{\lambda }_j)$, including repeated roots when $\mu =0$. Substituting $A$ into this polynomial is valid because the resulting factors commute. If $\mu$ is an eigenvalue of $A^4$, thenAt least one factor must be zero, so its ${\lambda }_j$ is an eigenvalue of $A$ and satisfies ${\lambda }_j^4=\mu$. Thus the reverse direction of this instance of the spectral mapping theorem isThis argument also covers $\mu =0$, when the conclusion is that $A$ is singular and has eigenvalue zero.

An orthogonal matrix satisfies $R^{T}R=I$, so it preserves the Euclidean norm. For a nonzero real eigenvector $v$,Therefore $\boxed{\lambda =\pm 1}$. Geometrically, eigenvalue $1$ fixes every vector on the linear span of $v$; eigenvalue $-1$ reverses every vector on that line. In either case the line is preserved as a set.

For $M$, solving $(M-I)v=0$ gives $y=x$, $z=y$, $x=z$. For $N$, the equations are $-2y-2z=0$ and $-2z=0$, hence $y=z=0$. For $P$, let $a=(1,1,1{)}^T$ and $J=a{a}^T$, so that $P=I-\frac{2}{3}J$. Then $(P-I)v=0$ is equivalent to $x+y+z=0$. The resulting eigenspaces and their bases areSolving all the equations shows that these lists contain the maximum possible numbers of independent eigenvectors with eigenvalue $1$.

The columns of $M$ form an orthonormal basis, so $M$ is an orthogonal matrix. Also $\det M=1$. Its fixed axis is the line through $a$, and on the perpendicular plane it acts as a two-dimensional rotation. If $\theta$ denotes the rotation angle, its matrix trace is $1+2\cos \theta$. Since $\mathrm{tr}M=0$, we obtain $\cos \theta =-1/2$. To fix the orientation relative to $u=a/\sqrt{3}$, note that $M{e}_1=e_3$ andThe component of $e_1$ perpendicular to $u$ has squared length $2/3$, and this scalar triple product equals $(2/3)\sin \theta$. Hence $\sin \theta =-\sqrt{3}/2$. ThusEquivalently, its unsigned angle is $2\pi /3$, or its positive angle is $2\pi /3$ about the opposite oriented axis.

For $P$, the reflection in a hyperplane formula givesIt fixes the plane $a\cdot v=0$ pointwise and maps $a$ to $-a$, so $\boxed{P\text{ reflects in }x+y+z=0}$. These actions also show that $P$ is an orthogonal matrix. In contrast, the second column of $N$ has squared length $5$, so $N$ is not orthogonal and cannot represent either of these Euclidean motions.

A diagonalizable matrix must have a basis of eigenvectors over the field in question. For $M$, a direct determinant computation gives ${\chi }_M(t)=1-t^3$. Its eigenvalues are $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. Only one is real, with one-dimensional eigenspace, so $M$ is not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$ the three eigenvalues are distinct, and the vectors $(1,\lambda ,{\lambda }^2{)}^T$ for $\lambda =1,\omega ,{\omega }^2$ give an eigenbasis. The upper triangular matrix $N$ has only eigenvalue $1$, of algebraic multiplicity $3$, but its eigenspace has dimension $1$ over either field. Hence $N$ is not diagonalizable over either field. For $P$, the two real plane vectors displayed above, together with $a$, form an eigenbasis, with eigenvalues $1,1,-1$. They remain independent over $\mathbb{C}$. Therefore

The ratio test says that a series $\sum u_n$ whose terms are eventually nonzero converges absolutely if $\mathrm{limsup}|u_{n+1}/u_n|<1$; in particular, it suffices to have $|u_{n+1}/u_n|\le q<1$ eventually, since then its terms are bounded by a convergent geometric series. The comparison test for series states that, for nonnegative terms, comparison with a larger convergent series gives convergence, and comparison with a smaller divergent series gives divergence. For $u_n=n!/n^n$,By the binomial theorem, $(1+1/n{)}^n\ge 1+n(1/n)=2$. Thus $u_{n+1}/u_n\le 1/2$ for every $n\ge 1$, and induction gives $0<u_n\le 2^{1-n}$. The comparison test for series with $\sum 2^{1-n}=2$ proves

The limit comparison test says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their series either both converge or both diverge. This follows because, eventually, $(L/2)v_n\le u_n\le (3L/2)v_n$, and the comparison test for series applies in both directions.

Here comparison with the harmonic series givesFor completeness, the natural logarithm grows slowly enough because for $x\ge 1$,so $(\log n{)}^2/n\le 16/\sqrt{n}\to 0$. The harmonic series diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence

The alternating series test states that if $b_n\ge 0$ decreases to zero, then $\sum (-1{)}^n{b}_n$ converges. Indeed, its even and odd partial sums are monotone in opposite directions and bound each other, while their difference tends to zero. Here $b_n=1/(1+\sqrt{n})$ decreases to zero, so the series converges.

It fails absolute convergence, since for $n\ge 1$,The p-series $\sum n^{-1/2}$ diverges. This also follows by comparison with the divergent harmonic series, because $n^{-1/2}\ge n^{-1}$. The comparison test for series therefore shows that $\sum b_n$ diverges. Thus

For $s=\pm 1$, the identity $s/(2+s)=-1/3+(2/3)s$ givesThe alternating series test applies to $\sum (-1{)}^n/n$ because $1/n$ decreases to zero; its partial sums consequently have a finite limit. The harmonic series $\sum 1/n$ diverges to $+\infty$. Applying the displayed identity to each finite partial sum yieldsTherefore $\boxed{\text{the series diverges to }-\infty }$. Alternating signs alone do not satisfy the alternating series test: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.

One precise version of the Taylor theorem with Lagrange remainder is as follows. Let $n\ge 0$, let $f$ and its derivatives through order $n$ be continuous on the closed interval between distinct points $a$ and $x$, and suppose $f^{(n+1)}$ exists throughout its interior. Then some $\xi$ strictly between $a$ and $x$ satisfiesHere the endpoint derivatives are understood as restrictions of derivatives defined near the interval, or as their continuous one-sided extensions. No continuity of $f^{(n+1)}$ is required. For $x=a$ the equality is immediate with zero remainder.

To prove the formula, write the Taylor polynomial as $P_n(t)={\sum }_{k=0}^n{f}^{(k)}(a)(t-a{)}^k/k!$ and defineThen $h(a)=h(x)=0$, and $h^{(j)}(a)=0$ for $j=1,\dots ,n$. The Rolle theorem gives a zero ${\xi }_1$ of $h^{\prime }$ strictly between $a$ and $x$. If $n\ge 1$, apply the Rolle theorem to $h^{\prime }$ between $a$ and ${\xi }_1$, using $h^{\prime }(a)=0$, to get a zero ${\xi }_2$ of $h^{\prime \prime }$. Repeating produces a point ${\xi }_{n+1}$ strictly between $a$ and $x$ with $h^{(n+1)}({\xi }_{n+1})=0$. Each application is justified by continuity on the relevant closed interval and differentiability in its interior; the same construction works when $x<a$.

Since $P_n$ has degree at most $n$,Substituting this value of $K$ into its definition proves the displayed Taylor theorem with Lagrange remainder.

The assumed differentiability makes $e$ continuous. The identity $e^{\prime }=e$ then implies inductively that $e$ has continuous derivatives of every order andFix $x\ne 0$. By the extreme value theorem, the continuous function $|e|$ is bounded on the closed interval between $0$ and $x$; let $M_x$ be its maximum. Applying the Taylor theorem with Lagrange remainder about $0$ gives some ${\xi }_n$ between $0$ and $x$ such thatFor the last factor, successive ratios are $|x|/(n+2)$, which are at most $1/2$ once $n$ is sufficiently large. A geometric sequence bound therefore proves that $|x{|}^{n+1}/(n+1)!\to 0$. The partial sums converge to $e(x)$, soAt $x=0$ the equality follows directly from $e(0)=1$. The proof obtains this Taylor series solely from the differential identity, continuity on a fixed finite interval, and the Taylor remainder estimate.

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.

For $n=1$, differentiability at $x_0$ means existence of the finite two-sided limit of a function ${\lim }_{h\to 0}(f(x_0+h)-f(x_0))/h$. More generally, higher-order differentiability at a point is defined recursively: for $n\ge 2$, the derivatives $f^{\prime },\dots ,f^{(n-1)}$ must exist throughout some open interval containing $x_0$, and $f^{(n-1)}$ must be differentiable at $x_0$. With $f^{(0)}=f$, the final condition isThe neighbourhood requirement makes the numerator meaningful for every sufficiently small nonzero $h$. It does not require $f^{(n)}$ to exist at other points of that neighbourhood, or to be continuous at $x_0$.

Since $f(0)=0$, its difference quotient isThe absolute value of cosine is at most $1$, so this quotient lies between $0$ and $|h|$. The squeeze theorem proves

The zeros of $\cos (\pi /x)$ occur atLet $g(x)=\cos (\pi /x)$. At these nonzero points,The definition of the derivative gives $g(x_k+h)=g^{\prime }(x_k)h+o(|h|)$, henceThe sign function is constant sufficiently near the nonzero point $x_k$, and $(x_k+h{)}^2=x_k^2+O(h)$. As $f(x_k)=0$, multiplication yieldsIts right and left one-sided derivatives are therefore $\pi \mathrm{sign}(x_k)$ and $-\pi \mathrm{sign}(x_k)$ respectively. They differ, so $f$ is not differentiable at any $x_k$. In particular, $x_k\to 0$ as $k\to +\infty$, provingThe absolute value at a simple zero mechanism creates unequal one-sided derivatives. It also shows why the existence of $f^{\prime }(0)$ alone does not imply higher-order differentiability at a point of order two.