MATH40003 Revised Historical Coverage Practice Paper 1

Question 1 Historical basis: 2026 Q1; supplementary basis: 2022 Q1 (field characteristic) and 2020 Q4 (direct sums). Gap: earlier papers repeated this combination of methods too often, so it is retained only once in this set. Difficulty calibration: 8 marks for routine matrix/kernel calculations, 6 for standard structural applications, 4 for proving the intersection, 2 for characteristic counterexamples; approximately 30 minutes. Recognition signal: two linear maps in opposite directions, requests for kernel/image/intersection/direct sum, and field conditions on invertibility of 2 or 3. First key step: solve the homogeneous equations for ker D, then express im G as coordinate constraints; the intersection is obtained by solving both sets of constraints simultaneously. The standard matrices are [D] =   −1 1 0 0 0 0 0 −1 1 0 0 0 −1 0 1   , [G] =   1 0 0 −1 0 0 0 1 0 0 0 1 0 −1 −1   . Solving D(a, b, c, d, e)T = 0 gives b = a, d = c, e = c, so ker D = span{(1, 1, 0, 0, 0)T , (0, 0, 1, 1, 1)T }, dim ker D = 2. The map G is injective because its first, third and fourth coordinates are s, t, u. Hence ker G = 0 and dim im G = 3. Every vector in the image satisfies the two displayed equations. Conversely , if x1 + x2 = 0 and x3 + x4 + x5 = 0, then x = G(x1, x3, x4)T . Now let v ∈ ker D ∩ im G. Write v = (a, a, c, c, c) = ( s, −s, t, u, −t − u). Then 2s = 0, u = t, and −t − u = t, so 3t = 0. If 6 is invertible, s = t = u = 0, hence v = 0. The dimensions are 2 and 3, so the zero intersection gives F 5 = ker D ⊕ im G. Rank-nullity gives rk D = 5 − 2 = 3 . If 2 = 0 , (1, 1, 0, 0, 0) = G(1, 0, 0) is a non-zero intersection vector . If 3 = 0 , (0, 0, 1, 1, 1) = G(0, 1, 1) is one. Marking points: 4 marks matrices; 4 kernel; 4 image/constraints; 4 zero intersection; 2 dimen- sion/rank; 2 characteristic examples. Common errors: Only comparing dimensions without first proving the intersection is zero; dividing by 2 or 3 without checking the field; proving only one inclusion for the image description. Final self-check: Multiply the two kernel basis vectors by D; multiply the two characteristic ex- amples by both D and the stated G inputs; verify 2 + 3 = 5 . 2/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 2 Historical basis: 2022 Q2 (transpose operator and change of basis) and 2021 Q3 (idempotent/nilpotent); the difficulty structure is compared with 2026 Q2, but the objects and course method sequence come from earlier formal papers. Approximately 30 minutes. Recognition signal: matrix spaces, transpose, symmetric/skew-symmetric subspaces, change of basis, projections and square-zero operators. First key step: find the images of the four standard basis matrices; read the later subspaces and projections directly from T (A) = ±A. For A, B ∈ V and λ ∈ F , transpose preserves addition and scalar multiplication. In the basis E, [T ]E =   1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1   . If 2 ̸= 0, then V + = {( a b b d )} = span{E11, E22, E12 + E21}, V − = {( 0 b −b 0 )} = span{E12 − E21}, and their intersection is zero. If 2 = 0 , T + I = T − I, so V − = V + and the intersection has dimension 3. For 2 ̸= 0, the matrix in the basis B is [T ]B = diag(1, 1, 1, −1), and the change-of-basis matrix from B-coordinates to E-coordinates is CE←B =   1 0 0 0 0 0 1 1 0 0 1 −1 0 1 0 0   . Using T 2 = I, P 2 = P, K 2 = K, P K = KP = 0, P + K = I. Moreover im P = V +, ker P = V −, im K = V − and ker K = V +. Thus V = V + ⊕ V −. Finally E2 12 = 0, so N 2(A) = E2 12AE2 12 = 0. Therefore (I − N )(I + N ) = I − N 2 = I, and (I − N )−1 = I + N . Marking points: 3 linearity/matrix; 5 symmetric/skew cases; 4 basis-change calculation; 5 projec- tions and direct sum; 3 nilpotent inverse. Common errors: Forgetting that V − changes in characteristic 2; writing the change-of-basis matrix with rows instead of columns; asserting a projection is idempotent without expanding T 2 = I. Final self-check: Check CE←Bej = [bj]E; check P (A) + K(A) = A; multiply (I − N )(I + N ). 3/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 3 Historical basis: 2026 Q3 ；supporting history: 2020 Q4(b), 2022 Q5 and 2025 Q5. This is the only retained near-current spectral/block-structure question. 30 minutes. Recognition signal: a real symmetric positive-definite matrix followed by a complex unitary matrix written as real blocks. First key step: For part (a), exploit the invariant x-z plane and the vector e2. For part (b), expand Z∗Z = I into real and imaginary parts before multiplying blocks. The vectors (1, 0, 1)T , (1, 0, −1)T and (0, 1, 0)T are eigenvectors with eigenvalues 7, 3, 3. Take Q =   1/ √ 2 1/ √ 2 0 0 0 1 1/ √ 2 −1/ √ 2 0   , Λ = diag(7, 3, 3). Then QT Q = I and A = QΛQT . One valid square-root factor is B = Q diag( √ 7, √ 3, √ 3), because BB T = QΛQT . If Z = C + iD is unitary , then CT C + DT D = I, C T D − DT C = 0. Direct block multiplication gives R(Z)T R(Z) = I2n. A second direct multiplication gives JR(Z) = R(Z)J. Conversely , write R = ( A B C D ) . The equation J R = RJ gives C = −B and D = A, so R = R(A + iB). Orthogonality yields AT A + BT B = I, A T B − BT A = 0, which is exactly (A + iB)∗(A + iB) = I. Hence A + iB is unitary . Marking points: 10 marks spectral computation/factor; 3 realification orthogonality; 3 commuta- tion; 4 converse. Common errors: Using Q−1 without noting Q−1 = QT ; taking square roots of matrix entries instead of eigenvalues; reversing a sign in the block transpose. Final self-check: Check all three eigenpairs; check QT Q = I; expand the four blocks of RT R. 4/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 4 Historical basis: 2026 Q4 ；supporting history: 2023 Q5, 2024 Q5 and 2025 Q6. This retains one current-style group-theory ladder . 30 minutes. Recognition signal: first compute a concrete two-line permutation, then prove conjugacy properties from definitions and classify cycle types. First key step: trace each unvisited element to obtain disjoint cycles; in the general proof use the fact that conjugation simply relabels the points in a cycle. Tracing the images gives σ = (1 4 7 2)(3 6 9)(5 8)(10) . For example, (1 4 7 2) = (1 2)(1 7)(1 4) , (3 6 9) = (3 9)(3 6) , and (5 8) is already a transposition. There are six transpositions in total, so sgn (σ) = +1 . Conjugacy is reflexive using e, symmetric using g−1, and transitive because h(gxg −1)h−1 = (hg)x(hg)−1. For a cycle, direct evaluation at τ (aj) proves τ (a1 · · · ak)τ −1 = (τ (a1) · · · τ (ak)). Thus conjugation preserves all cycle lengths. Conversely , if two permutations have the same cycle type, define a bijection τ sending the entries of each cycle of the first permu- tation, in order , to the entries of the corresponding cycle of the second. Then τ στ−1 = ρ. For cycle type [4, 3, 2, 1] in S10, the number is 10! 4 · 3 · 2 = 151200, since the cycle lengths are distinct. Marking points: 8 concrete permutation marks; 3 equivalence relation; 2 cycle conjugation formula; 5 classification; 2 count. Common errors: Applying cycles left-to-right; omitting fixed points when discussing cycle type; proving only the “conjugate implies same type ”direction. Final self-check: Reconstruct the two-line map from the cycles; parity also equals (−1)10−4 where 4 is the number of cycles including the fixed point; check the count denominator . 5/5