Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1

The fixed-point equation isA unique solution at $z=i$ means that this polynomial isComparing coefficients givesThis is the repeated-root case of prescribed fixed points of a Möbius transformation.

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For roots $3i$ and $1+i$, the fixed-point polynomial must beComparing this with $z^2+(1-a)z-b$ gives

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Write $z=x+iy$. Rationalizing givesAn argument of $\pi /4$ requires equal positive real and imaginary parts. HenceCompleting squares yieldsThus the constant-argument locus of a Möbius transformation is the part above the real axis of the circle with centre $(1,1)$ and radius $\sqrt{2}$. Its endpoints $z=0$ and $z=2$ are excluded because the quotient is respectively undefined and zero.

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The $j$th column isAll three columns therefore lie in the two-dimensional span of $v$ and $1$, so

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For $x=(x_1,x_2,x_3{)}^T$,The vectors $v$ and $1$ are linearly independent, and $a\ne 0$. Thus $Ax=0$ exactly whenSolving gives

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Every column lies in $\mathrm{span}\{v,1\}$. Conversely, differences of two columns are nonzero multiples of $1$, while any one column together with $1$ recovers $v$. HenceandThis is an instance of the rank of a matrix with affine columns.

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The rank-nullity theorem states that for a linear map with finite-dimensional domain,Here the domain is ${\mathbb{R}}^3$, and the preceding parts givewhich verifies the theorem.

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The $j$th column of $B$ isThus the image is contained in $\mathrm{span}\{v,1\}$. Since $n>1$, $v$ is not constant; since $m>1$ and $a\ne 0$, two columns have a nonzero constant difference. The two spanning vectors therefore both lie in the image, so $\mathrm{rank}B=2$. By rank-nullity theorem,

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Convergence of $\sum a_n$ implies $a_n\to 0$. Hence $0<a_n\le 1$ for all sufficiently large $n$, and then $a_n^2\le a_n$. The comparison test, with the finitely many initial terms treated separately, proves thatThis is the squares of a summable positive sequence lemma.

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For every $N$, the Cauchy-Schwarz inequality givesThe right-hand side is bounded independently of $N$. Since the summands are positive, the partial sums increase to a finite limit, proving the geometric means of two summable positive sequences result.

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Another application of the Cauchy-Schwarz inequality givesWhen $p>1/2$, the second factor is bounded because $2p>1$, so the series converges.

At the endpoint, take $a_1=1$ and, for $n\ge 2$,Then $\sum a_n$ converges, whilewhose series diverges. This proves both parts of weighted square roots of a summable sequence.

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Suppose the series converges at $w\ne 0$. Then its terms $a_n{w}^n$ are bounded, say by $M$. Whenever $|z|<|w|$,so comparison with a geometric series proves absolute convergence at $z$.

Letwith the natural values $0$ or $\infty$ allowed. If $|z|<R$, choose a convergent point $w$ with $|z|<|w|\le R$; the preceding argument gives convergence at $z$. If $|z|>R$, convergence at $z$ would contradict the definition of $R$. Hence the series converges for $|z|<R$ and diverges for $|z|>R$. No universal assertion is possible on $|z|=R$.

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By the Cauchy-Hadamard theorem, putThe squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_n{z}^n$ and $\sum a_n^2{z}^n$ have radius one.

For the coefficients $a_n^{a_n}$, the relevant root terms areEquality can again occur. For example, takeThen $a_n\to \infty$, $a_n^{1/n}\to 1$, andThus both $\sum a_n{z}^n$ and $\sum a_n^{a_n}{z}^n$ have radius one. These examples illustrate the radius of convergence after powering coefficients.

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Both sides are antisymmetric in $j,k$ and in $p,q$. If either pair has equal indices, both sides vanish. It therefore suffices to take $j\ne k$ and $p\ne q$. The sum over $i$ is nonzero exactly when $(p,q)=(j,k)$ or $(k,j)$; it is respectively $1$ or $-1$. This is exactlythe contraction of two Levi-Civita symbols.

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Expanding a sum of squares gives the $n$-dimensional Lagrange identity for the cross product:The left-hand side is nonnegative, so

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In components, part (i) givesTherefore the vector triple product is

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PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.

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Since $P=P^{\prime }=0$, it suffices to map the two edge vectors. Let the columns of the required matrix be $u_1,u_2,u_3$. The condition $MQ=Q^{\prime }$ givesThe condition $MR=R^{\prime }$ then givesand henceThese are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0{)}^T$ produces the orthogonal matrixIts columns are orthonormal, and direct substitution verifies $MP=P^{\prime }$, $MQ=Q^{\prime }$, and $MR=R^{\prime }$.

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Write $a=(a_1,a_2,a_3,a_4{)}^T$. The reflection in a hyperplane orthogonal to $a$ sendssoIt acts as $-1$ on $\mathrm{span}\{a\}$ and as $1$ on the three-dimensional hyperplane $a^{\perp }$. Therefore

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For $a=(a_1,a_2,a_3{)}^T$, the cross-product matrix isIts diagonal is zero, so $\mathrm{tr}A=0$. Also $Aa=a\times a=0$, and $a\ne 0$, so $A$ has a nontrivial kernel. Hence

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Expanding $\det A^{(n)}$ along its last row gives the last diagonal contribution $a_n{d}_{n-1}$. The only other nonzero entry is $c_{n-1}$; expanding its minor along the last column contributesThus the tridiagonal determinant recurrence isso

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The recurrence becomesIts characteristic polynomial isTherefore $d_n=(\alpha +\beta n)(-1{)}^n$. The initial conditions give $\alpha =1$ and $\beta =1$, so

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The matrix trace isUsing matrix multiplication and interchanging dummy indices,

The proposed noncyclic interchange is false. With matrix units,one hasTrace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.

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The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.

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Consider the real polynomialThe hypothesis says $p(i)\ne 0$, so $p$ is not identically zero. A nonzero polynomial has only finitely many roots, whereas $\mathbb{R}$ is infinite. Hence there is a real $\lambda$ withThus $A+\lambda B$ is invertible, as summarized by the real parameter avoiding a singular matrix pencil.

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Write the complex similarity matrix aswith $X,Y$ real. From $C=SD{S}^{-1}$ we obtain $CS=SD$, and equating real and imaginary parts givesThereforefor every real $t$.

Now $p(t)=\det (X+tY)$ is a real polynomial andThus some real $t$ has $p(t)\ne 0$. Taking $T=X+tY$, we have a real invertible matrix satisfying $CT=TD$, orThis proves real similarity from complex similarity.

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Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, supposeis a shortest nontrivial relation, where $A{v}_j={\lambda }_j{v}_j$ and the ${\lambda }_j$ are distinct. Applying $A-{\lambda }_{m}I$ givesa shorter nontrivial relation, a contradiction.

Thus $n$ distinct eigenvalues provide a basis of $n$ eigenvectors, proving that $A$ is diagonalizable over $\mathbb{C}$. If $A$ is real and all its eigenvalues are real, each real singular matrix $A-\lambda I$ has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over $\mathbb{R}$. This is the distinct eigenvalues imply diagonalizability criterion.

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The characteristic polynomial ${\chi }_A(t)=\det (tI-A)$ has degree $n$. By the fundamental theorem of algebra, it has a complex root $\lambda$. Henceso $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfiesThus $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector.

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The adjoint matrix of $A$ is its conjugate transpose:

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A complex matrix $U$ is unitary matrix whenIf $Uv=\lambda v$ with $v\ne 0$, unitarity preserves norms, soTherefore every eigenvalue satisfies

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Suppose $A$ is diagonalized by a unitary matrix:where $D$ is diagonal. ThenSince the diagonal matrices $D$ and $D^{†}$ commute,Thus $A$ is a normal matrix.

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First note that if $A$ is normal, thenfor every $v$. Hence an eigenvector $v$ with $Av=\lambda v$ also satisfiesNormalize $v$ and extend it to an orthonormal basis. Let $U$ have these basis vectors as columns. The first column of $U^{†}AU$ is $(\lambda ,0,\dots ,0{)}^T$. For every $w\perp v$,so the first row also has no off-diagonal entries. ThereforeComparing the two block products in the normality identity shows that $B{B}^{†}=B^{†}B$, so $B$ is normal.

The result is trivial in dimension one. Applying the induction hypothesis to $B$ and adjoining the eigenvector $v$ gives an orthonormal eigenbasis for $A$. This proves the unitary diagonalization of a normal matrix.

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The proof is by induction on $n$. Choose a unit eigenvector $v$ of $A$, which exists by part (a), and extend it to an orthonormal basis. In that basis,By induction, a unitary change of basis on the last $n-1$ coordinates makes $C$ upper triangular. Extending that change by the identity on the first coordinate leaves the displayed zero block intact. Their product is unitary and makes all of $A$ upper triangular. This proves Schur triangularization.

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LetThis permutation matrix is unitary. The actionsgivewhich is upper triangular.

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The ratio test says that ifthen $\sum a_n$ converges absolutely. If the ratio has a limit $L>1$, the series diverges.

For the first assertion, choose $q$ with $L<q<1$. For all sufficiently large $n$,Iteration bounds the tail by a constant multiple of the convergent geometric series $\sum q^n$. If $L>1$, choose $q$ with $1<q<L$. Eventually $|a_{n+1}|\ge q|a_n|$, so $a_n$ cannot tend to zero; the series therefore diverges.

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First suppose $L>0$. Taking logarithms,Divide by $n$. Since the summands tend to $\log L$, their Cesàro means also tend to $\log L$. Henceand exponentiation gives $a_n^{1/n}\to L$.

If $L=0$, then for every $\epsilon >0$ the ratios are eventually at most $\epsilon$, so $a_n\le C{\epsilon }^n$ for a suitable constant $C$. ThusLetting $\epsilon \downarrow 0$ proves the ratio limit implies root limit in all cases.

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SetThenThe ratio test therefore proves

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For the sequence in part (iii), part (ii) givesTaking reciprocals yields the factorial-over-power series limit

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The function $f$ is differentiable at $a$ when there is a real number $L$ such thatEquivalently, the difference quotient has a finite limit, and

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WriteDivide by $h$ and let $h\to 0$. Differentiability implies continuity, so $g(a+h)\to g(a)$. Therefore the product rule gives

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No. Take $a=0$ andThen $f$ is differentiable andSince $x|x|/x=|x|\to 0$, the product is differentiable at zero withHowever, $g$ has one-sided derivatives $-1$ and $1$ and is not differentiable at zero. This is a nondifferentiable factor with a differentiable nondegenerate product.

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Differentiability lets us writeand, with $k\to 0$,Substitute $k=f(a+h)-f(a)=f^{\prime }(a)h+r(h)$. Since $k=O(h)$, the remainder $s(k)$ is $o(h)$. Thusand the chain rule is