MATH40003 · Practice Paper

MATH40003 2026 Exam Standard Full Paper E

Revision questions and worked solutions, presented read-only. This review interface was prepared after the recorded study period.

Status: Completed
Questions completed: 4 / 4
Suggested time: 120 minutes
Source date: 2026-08-05

Question 120 marks

Parameter-dependent rank and reconstruction of the inverse

For \(\alpha\in F\), define \(T_\alpha:F^4\to F^4\) by \[ T_\alpha(a,b,c,d)^T=(a-b,\ b-c,\ c-d,\ \alpha(a+b+c+d))^T. \] (a) Write the matrix of \(T_\alpha\). [3 marks] (b) Assume \(2\) is invertible in \(F\). Determine \(\ker T_\alpha\) and \(\operatorname{rk}T_\alpha\) separately for \(\alpha=0\) and \(\alpha\ne0\). [6 marks] (c) Under the assumptions \(\alpha\ne0\) and \(2\) invertible, solve \(T_\alpha x=y\) explicitly and hence write a formula for \(T_\alpha^{-1}\). [6 marks] (d) Now suppose \(2=0\) in \(F\). Determine the rank and exhibit a non-zero kernel vector for every \(\alpha\). Explain why the answer differs from part (b). [5 marks]

Worked solution and marking guidance

**(a)** \[ [T_\alpha]=\begin{pmatrix} 1&-1&0&0\\0&1&-1&0\\0&0&1&-1\\ \alpha&\alpha&\alpha&\alpha \end{pmatrix}. \] **(b)** The first three homogeneous equations imply \(a=b=c=d=t\). The final equation is \(4\alpha t=0\). If \(2\) is invertible, then so is \(4\). For \(\alpha\ne0\), \(t=0\), so the kernel is zero and the rank is \(4\). For \(\alpha=0\), the kernel is \(\operatorname{span}\{(1,1,1,1)^T\}\) and the rank is \(3\). **(c)** Write \(y=(y_1,y_2,y_3,y_4)^T\). From the first three equations, \[ c=d+y_3,\quad b=d+y_3+y_2,\quad a=d+y_3+y_2+y_1. \] The last equation gives \[ 4d+3y_3+2y_2+y_1=y_4/\alpha, \quad d=\frac{y_4/\alpha-y_1-2y_2-3y_3}{4}. \] Substitution in the preceding three formulae is \(T_\alpha^{-1}(y)\). **(d)** In characteristic \(2\), \(4=0\), and \((1,1,1,1)^T\) is always in the kernel. The first three rows remain independent. The last row is dependent because \((1,1,1,1)=r_1+r_3\) when \(-1=1\); hence the rank is \(3\) for every \(\alpha\). The step “divide by 4” in (b) is exactly what fails.

Question 220 marks

Polynomial-space involution and operator ranks (E)

Let \(V=F[x]_{\le 5}\), with ordered basis \(\mathcal B=(1,x,\ldots,x^5)\). Define \(Jp(x)=p(-x)\) and let \(D^2=D\circ D\). **Variant-specific requirement.** This variant deliberately uses the second derivative. Prove the commuting identity directly rather than quoting the first-derivative anticommutation. Consequently parity is preserved, not exchanged. Compare the coefficients k(k-1) on even and odd monomials and list the separate rank losses in characteristics 3 and 5. (a) Prove that the displayed involution is linear, show that its square is the identity, and write its matrix in \(\mathcal B\). [4 marks] (b) Assume \(2\ne0\) in \(F\). Find bases of its \(+1\) and \(-1\) eigenspaces \(V_+\) and \(V_-\), and prove \(V=V_+\oplus V_-\). [6 marks] (c) Write the matrix of the differential operator used in this question and prove \(D^2J=JD^2\). Deduce the corresponding mapping relation between \(V_+\) and \(V_-\). [5 marks] (d) Determine the ranks of the restrictions to \(V_+\) and \(V_-\), including every exceptional field characteristic caused by a coefficient appearing in the derivative. [5 marks]

Worked solution and marking guidance

**(a)** \[ [J]_{\mathcal B}=\operatorname{diag}(1,-1,1,-1,1,-1) \] in \((1,x,\ldots,x^5)\). The map is linear and \(J^2=I\). **(b)** For \(2\ne0\), \[ V_+=\operatorname{span}\{1,x^2,x^4\},\qquad V_-=\operatorname{span}\{x,x^3,x^5\},\qquad V=V_+\oplus V_-. \] **(c)** Since \(D^2x^k=k(k-1)x^{k-2}\), \[ [D^2]_{\mathcal B}=\begin{pmatrix} 0&0&2&0&0&0\\0&0&0&6&0&0\\0&0&0&0&12&0\\ 0&0&0&0&0&20\\0&0&0&0&0&0\\0&0&0&0&0&0 \end{pmatrix}. \] Directly, \[ (D^2Jp)(x)=p''(-x)=(JD^2p)(x), \] so \(D^2\) preserves each eigenspace. **(d)** On \(V_+\), the coefficients \(k(k-1)\) are \(0,2,12\): the image is \(\operatorname{span}\{1,x^2\}\), rank 2, except in characteristic 3, where it is \(\operatorname{span}\{1\}\), rank 1. On \(V_-\), they are \(0,6,20\): the image is \(\operatorname{span}\{x,x^3\}\), rank 2, except in characteristic 3, where it is \(\operatorname{span}\{x^3\}\), rank 1, and characteristic 5, where it is \(\operatorname{span}\{x\}\), rank 1.

Question 320 marks

Spectral theorem and realification (E)

(a) Let \[ A=\begin{pmatrix}2&1&0\\1&2&0\\0&0&1\end{pmatrix}\in M_3(\mathbb R). \] (i) Find all eigenvalues, with algebraic multiplicities. [3 marks] (ii) Construct an orthogonal matrix \(Q\) and diagonal \(\Lambda\) such that \(A=Q\Lambda Q^T\). The columns of Q must be stated explicitly. [4 marks] (iii) Construct a real matrix B satisfying \(A=BB^T\), and justify why the construction is valid. [3 marks] (b) For n=1 identify unit complex numbers with planar rotations. Then prove the same realification statement for general n and explain why commutation with J expresses complex linearity. Give all block identities explicitly and prove both the stated direction and any requested converse. [10 marks]

Worked solution and marking guidance

**(a)(i)** The eigenvalues are \((3,1,1)\). They may be obtained from the characteristic polynomial or from the visible invariant directions. **(a)(ii)** One valid ordered orthogonal eigenbasis gives \[ Q=\begin{pmatrix}1/\sqrt2&1/\sqrt2&0\\1/\sqrt2&-1/\sqrt2&0\\0&0&1\end{pmatrix},\qquad \Lambda=\operatorname{diag}(3,1,1). \] Direct multiplication verifies \(Q^TQ=I\) and \(AQ=Q\Lambda\), so \(A=Q\Lambda Q^T\). **(a)(iii)** All diagonal entries of \(\Lambda\) are non-negative. Hence \[ B=Q\Lambda^{1/2} \] satisfies \(BB^T=Q\Lambda^{1/2}\Lambda^{1/2}Q^T=A\). **(b)** Write \(Z=X+iY\). The unitary identity is equivalent to \[ X^TX+Y^TY=I,\qquad X^TY=Y^TX. \] For \(M=\begin{pmatrix}X&-Y\\Y&X\end{pmatrix}\), block multiplication therefore gives \(M^TM=I\). With \(J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}\), direct block multiplication gives \(MJ=JM\) (with the equivalent sign convention both sides change consistently). Conversely, writing \(Q=\begin{pmatrix}A&B\\C&D\end{pmatrix}\) and comparing \(QJ\) with \(JQ\) gives \(D=A\) and \(C=-B\), up to the same fixed sign convention. Thus Q is the realification of \(A-iB\). Its orthogonality equations are exactly the real and imaginary parts of the unitary equation. Realification also respects multiplication because the block product reproduces complex multiplication, and consequently respects inverses. For \(n=1\), \(e^{i\theta}\) becomes the usual rotation matrix.

Question 420 marks

Permutations, sign and conjugacy (E)

Let \(\sigma\in S_11\) be given in two-line notation by \[ \sigma=\begin{pmatrix} 1&2&3&4&5&6&7&8&9&10&11\\ 3&4&5&6&7&2&9&10&1&8&11 \end{pmatrix}. \] **Variant-specific requirement.** Construct one concrete permutation τ different from σ but with the same cycle type by relabelling every entry in the non-trivial cycles. Then write an explicit conjugating permutation g and verify gσg^{-1}=τ on the generators of each cycle. (a) (i) Write \(\sigma\) as a product of disjoint cycles, including the fixed points when recording its cycle type. [4 marks] (ii) Write it as a product of transpositions and compute \(\operatorname{sgn}(\sigma)\). [4 marks] (b) Let G be a group. Prove that conjugacy is an equivalence relation, naming the conjugating element in the reflexive, symmetric and transitive steps. Also prove \(gx^mg^{-1}=(gxg^{-1})^m\) for every \(m\ge1\). [6 marks] (c) Prove that two permutations of \(S_n\) are conjugate if and only if they have the same cycle type. Your proof must construct a conjugating permutation in the reverse direction. Apply the result to state the cycle type of the conjugacy class containing \(\sigma\). [6 marks]

Worked solution and marking guidance

**(a)(i)** Following each orbit gives \[ \sigma=(1 3 5 7 9)(2 4 6)(8 10). \] There are 1 fixed point(s), so the complete cycle type is \([5, 3, 2, 1]\). **(a)(ii)** Use \((a_1\ldots a_k)=(a_1a_k)\cdots(a_1a_2)\) on every non-trivial cycle. This uses \(7\) transpositions in total; hence \(\operatorname{sgn}(\sigma)=-1\). Any equivalent transposition decomposition earns full credit. **(b)** Reflexivity uses \(e x e^{-1}=x\). If \(gxg^{-1}=y\), then \(g^{-1}yg=x\), proving symmetry. If \(gxg^{-1}=y\) and \(hyh^{-1}=z\), then \((hg)x(hg)^{-1}=z\), proving transitivity. For powers, expand the product: adjacent factors \(g^{-1}g\) cancel, or use induction, giving \(gx^mg^{-1}=(gxg^{-1})^m\). **(c)** Conjugating a cycle replaces every entry a by g(a), so cycle lengths, including fixed points, are preserved. Conversely, if σ and τ have the same cycle type, pair cycles of equal length and define g to send the j-th entry of each σ-cycle to the j-th entry of the paired τ-cycle. This defines a bijection and direct evaluation gives \(g\sigma g^{-1}=\tau\). Thus the class containing the printed σ is exactly the set of permutations with cycle type \([5, 3, 2, 1]\). **Required variant check.** Take \(g=(1\ 2)(3\ 4)(5\ 6)(7\ 8)(9\ 10)\), fixing 11. Then \(\tau=g\sigma g^{-1}=(2\ 4\ 6\ 8\ 10)(1\ 3\ 5)(7\ 9)(11)\ne\sigma\). For example \(g\sigma g^{-1}(2)=g\sigma(1)=g(3)=4=\tau(2)\), and the same position-by-position calculation verifies every cycle.