MATH40003 Revised Historical Coverage Practice Paper 6

Question 1 Historical basis: 2025 Q3, 2026 Q2 and 2022 field-characteristic work. It changes the wording and cycle lengths to test entry recognition, not memory of the current paper . 30 minutes. Recognition signal: a function space plus a permutation with cycles of lengths 2,3,1; ask fixed space, differ- ence image and characteristic exceptions. First key step: Use delta coordinates. Analyse I − T separately on each cycle of τ . Evaluation gives V ∼= F 6 and the delta functions form a basis. Since T δj = δτ −1(j), the matrix is the permutation matrix of τ −1 = (1 2)(3 5 4) . A fixed function is constant on each cycle, so ker(I − T ) = span{δ1 + δ2, δ 3 + δ4 + δ5, δ 6}. On each non-trivial cycle, differences have coordinate sum zero; on the fixed point the image coordinate is zero. Thus the stated image containment holds. Rank-nullity gives rank 6 − 3 = 3 , equal to the dimension of the stated subspace, hence equality . A vector in both spaces has the form (a, a, b, b, b, c) and must satisfy 2a = 0, 3b = 0, c = 0. Therefore the intersection is zero if the characteristic is neither 2 nor 3; it is spanned by (1, 1, 0, 0, 0, 0) in characteristic 2; and by (0, 0, 1, 1, 1, 0) in characteristic 3. Marking points: 4 isomorphism/basis; 4 operator matrix; 4 fixed space; 4 image/rank; 4 character- istic intersection. Common errors: Using τ instead of τ −1 in the matrix; assuming the fixed space has dimension one; dividing by the cycle length without checking characteristic. Final self-check: Check one basis delta function from each cycle; verify the sum constraints; substitute the characteristic intersection vectors. 2/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 2 Historical basis: 2020 Q4(a), 2021 idempotents and 2026 characteristic exceptions. The polynomial has three distinct roots only when 2 is invertible. 30 minutes. Recognition signal: an operator satisfying A3 = A and the field condition that 2 is invertible. First key step: Factor x3 − x = x(x − 1)(x + 1) and construct the three Lagrange interpolation projectors. From A3 = A, (A2)2 = A4 = A2, so A2 is idempotent. If A2v = 0 , then Av = A3v = A(A2v) = 0 , proving the kernel equality . Also im A2 ⊆ im A, while Av = A2(Av) gives the reverse inclusion. Direct polynomial calculation modulo A3 = A gives P0 + P+ + P− = I, P 2 i = Pi, P iPj = 0 ( i ̸= j). Moreover im P0 = ker A, im P+ = ker(A − I), im P− = ker(A + I). The sum of pairwise orthogonal projections is the identity , so their images form the re- quired direct sum. Choosing bases of the three eigenspaces gives a basis of eigenvectors, hence A is diagonalisable. In characteristic 2, take A = ( 1 1 0 1 ) . Writing A = I + N with N 2 = 0, one has A2 = I and A3 = A. It is not diagonalisable because its only eigenvalue is 1 but A ̸= I (equivalently , its minimal polynomial is (t − 1)2). Marking points: 4 A2 facts; 10 projectors; 3 direct sum/diagonalisation; 3 characteristic-two exam- ple. Common errors: Using the projectors when 2 is not invertible; proving the images are con- tained in eigenspaces but not the reverse; claiming repeated eigenvalue alone means non-diagonalisable. Final self-check: Expand every projector polynomial using A3 = A; apply P+ to a vector with Av = v; square the characteristic-two matrix. 3/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 3 Historical basis: 2026 Q3(a), 2021 Q4(b), 2022 Q4(c) and 2025 Q5(a). It mixes three historical entry signals while preserving the 10/10 computation-proof balance. 30 minutes. Recognition signal: symmetric matrix followed by unit-sphere norm and a general QR existence proof. First key step: Use the evident eigenvector (1, 1, 1) and its orthogonal complement; for QR, apply Gram-Schmidt to the columns. The eigenvalue 4 has unit eigenvector u1 = (1, 1, 1)T / √ 3. The orthogonal complement has eigenvalue 1; take u2 = (1, −1, 0)T / √ 2, u 3 = (1, 1, −2)T / √ 6. Thus Q = (u1|u2|u3) and A = Q diag(4, 1, 1)QT . A valid factor is B = Q diag(2, 1, 1). For the norm formula, write a unit vector as v = ∑ ciui in an orthonormal eigenbasis. Then ∥Sv∥2 = ∑ λ2 i c2 i , giving both bounds and equality at eigenvectors. For an invertible matrix C, its columns are independent. Apply Gram–Schmidt to obtain orthonormal columns qi. Each original column cj lies in span (q1, . . . , qj), so the coefficient matrix R is upper triangular . Choosing the sign of each qi gives positive diagonal entries. Then C = QR. Marking points: 6 diagonalisation; 3 square-root factor; 5 norm theorem; 6 QR existence. Common errors: Using non-orthogonal eigenvectors in Q; confusing the matrix square root with entrywise square roots; failing to explain why R is upper triangular . Final self-check: Check ui·uj = δij; check eigenpairs; verify each column of QR is the corresponding original column. 4/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 4 Historical basis: 2025 Q6 direct A/B coverage, with the added Z2 entry from 2023 Q6. It is the final transfer test for reading group descriptions. 30 minutes. Recognition signal: concrete permutation products, maximum order, cyclicity of four groups, and a simplified subgroup test under a finite-order assumption. First key step: For the product, apply the rightmost cycle first. For cyclicity , use a structural invariant: commutativity , rank, or an explicit generator . The product is (1 5 2 6)(3 4) , so its order is lcm (4, 2) = 4 . Its sign is (−1)3+1 = +1. The maximum order in S9 is 20, attained by disjoint cycles of lengths 5 and 4. Checking partitions of 9 shows no larger least common multiple. The group S5 is not cyclic because it is non-abelian. The group Z × Z is not cyclic: no single pair generates both (1, 0) and (0, 1). The shear-matrix group is cyclic, generated by the matrix with upper-right entry 1, and is isomorphic to (Z, +). The rotations of a regular m-gon form a cyclic group generated by rotation through 2π/m. Finally choose h ∈ H. Since h has finite order n, closure under multiplication puts hn = e and hn−1 = h−1 in H. Thus H contains the identity , is closed under inverses, and is already closed under multiplication; it is a subgroup. Marking points: 4 product/order/sign; 4 maximum order; 5 cyclicity; 7 subgroup theorem. Common errors: Applying permutation factors left-to-right; saying S5 is not cyclic only because it lacks an element of order 120 without justifying possible orders; forgetting integer negative powers in the shear group. Final self-check: Reconstruct the product on all six points; check the least common multiple for the relevant partitions; multiply two shear matrices. 5/5