MATH40003 2026 Practice Paper 8

MA TH40003 Linear Algebra and Group Theory Tony 2026 Summer Resit MATH40003 – Complete Solutions and Marking Guide Mock Paper 8: Final comprehensive unseen practice paper Each question carries 20 marks. The mark splits are indicative and reward mathematically equivalent arguments. Question 1 – Solution and marking guide Put u1 = (1, 0, 1, 0)t, u 2 = (0, 1, 0, 1)t, k 1 = (1, 0, −1, 0)t, k 2 = (0, 1, 0, −1)t. (a) [T ] =   1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1   . [2] (b) The equations for the kernel are a + c = 0 and b + d = 0, hence ker T = span{k1, k2}. Every output is (r, s, r, s)t, and both u1, u2 occur as outputs, so im T = span{u1, u2}. Thus rk T = 2 and nullity T = 2 . These statements remain valid in characteristic 2; only the relation between the two subspaces changes. [6] (c) Applying T twice gives T 2(a, b, c, d)t = 2(a + c, b + d, a + c, b + d)t = 2T (a, b, c, d)t. Thus P 2 = P . If 2 is invertible, then u1, u2, k1, k2 form a basis: the first two span im T , the last two span ker T , and their intersection is zero because (r, s, r, s) ∈ ker T implies 2r = 2s = 0. Therefore F 4 = im T ⊕ ker T. In the ordered adapted basis (u1, u2, k1, k2), [T ] = diag(2, 2, 0, 0), [P ] = diag(1, 1, 0, 0). [5] (d) If 2 = 0 , the identity T 2 = 2 T gives T 2 = 0 , so im T ⊆ ker T . Both spaces have dimension 2, hence they are equal; equivalently ki = ui in characteristic 2. Moreover (I − T )(I + T ) = I − T 2 = I = (I + T )(I − T ), so (I − T )−1 = I + T (which equals I − T in characteristic 2). [4] (e) If 2 ̸= 0 , T acts as multiplication by 2 on im T and as 0 on ker T , so its eigenvalues are 2 and 0, and the direct-sum eigenbasis shows that it is diagonalisable. If 2 = 0 , T 2 = 0, so the only possible eigenvalue is 0. Since T ̸= 0, it cannot be diagonalisable: a diagonalisable operator with only eigenvalue 0 is the zero operator. [3] Tony template: an operator polynomial gives a projection or a nilpotent operator depending on the characteristic. Recognition signal: T 2 = cT, with kernel/image/direct sum requested and characteristic treated separately. Standard start: calculate explicit bases for the kernel and image, then check whether c is invertible; divide by c to get an idempotent when it is invertible, and obtain a nilpotent when c = 0. Reusable structure: an idempotent gives V = im P ⊕ ker P; a nonzero nilpotent operator cannot be diagonalisable; if N 2 = 0 then (I − N )−1 = I + N. Common lost marks: treating −1 as different from 1 in characteristic 2; asserting equality from im T ⊆ ker T without comparing dimensions. 1

MA TH40003 Linear Algebra and Group Theory Tony 2026 Summer Resit Question 2 – Solution and marking guide Let n = dim V . (a) Each Ur is the kernel of the linear map T r, hence is a subspace. If v ∈ Ur, then T rv = 0 , so T r+1v = T (0) = 0 and v ∈ Ur+1. Also U0 = ker I = {0}. [4] (b) The dimensions 0 = dim U0 ≤ dim U1 ≤ · · · ≤ n form a nondecreasing integer sequence. Among U0, . . . , Un+1, two consecutive dimensions must be equal; since one space contains the other, Uk = Uk+1 for some k ≤ n. If Um = Um+1 and x ∈ Um+2, then T x ∈ Um+1 = Um, so T m+1x = 0 and x ∈ Um+1. Thus Um+2 = Um+1. Induction gives Uk = Uk+j for all j ≥ 1. [5] (c) Stability gives ker T 2k = U2k = Uk = ker T k. If x ∈ ker T k ∩ im T k, write x = T ky. Then 0 = T kx = T 2ky, so y ∈ ker T 2k = ker T k, and therefore x = T ky = 0. [4] (d) By rank–nullity for T k, dim ker T k + dim im T k = n. Together with the zero intersection from part (c), this implies V = ker T k ⊕ im T k. [3] (e) For the given matrix, ker T = span{e1}, ker T 2 = span{e1, e2}. On span{e3, e4}, T is a 90◦ rotation and is invertible, while the upper-left block is nilpotent of degree 2. Hence ker T 3 = ker T 2, so the smallest stabilising value is k = 2. Also T 2 =   0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 −1   , so ker T 2 = span{e1, e2}, im T 2 = span{e3, e4}. [4] Tony template: stabilisation of the kernel chain → a Fitting-type direct-sum decomposition. Recognition signal: ker T, ker T 2, . . ., finite dimension, eventual stabilisation, and a direct sum of kernel and image. Standard start: use the inclusions to express dimensions as a bounded nondecreasing integer sequence; once adjacent spaces agree, first apply T to x to propagate the equality. Reusable structure: x = T ky and T kx = 0 ⇒ T 2ky = 0; stabilisation reduces ker T 2k to ker T k. Common lost marks: stating dimension stabilisation without explaining that inclusion and equal dimensions imply equality of spaces; forgetting to use U2k = Uk when proving the intersection is zero. 2

MA TH40003 Linear Algebra and Group Theory Tony 2026 Summer Resit Question 3 – Solution and marking guide Set λ+ = 2 + √ 2, λ 0 = 2, λ − = 2 − √ 2. (a) det(λI − A) = ( λ − 2) ( (λ − 2)2 − 2 ) , so the eigenvalues are λ+, λ0, λ−. Corresponding eigenspaces are Eλ+ = span{(1, √ 2, 1)t}, E λ0 = span{(1, 0, −1)t}, E λ− = span{(1, − √ 2, 1)t}. [6] (b) Normalising the displayed eigenvectors gives u+ =   1 2 1√ 2 1 2   , u 0 =   1√ 2 0 − 1√ 2   , u − =   1 2 − 1√ 2 1 2   . They are orthonormal. Hence, with Q = (u+ | u0 | u−), D = diag(λ+, 2, λ−), we have QtQ = I and A = QDQt. [4] (c) All three eigenvalues are positive. Define B = Q diag( √ λ+, √ 2, √ λ−)Qt. Then B is symmetric positive definite and B2 = QDQt = A. Since Bt = B, also BB t = B2 = A. [3] (d) Write a unit vector as v = c+u+ + c0u0 + c−u−, where c2 + + c2 0 + c2 − = 1. Then ∥Av∥2 = λ2 +c2 + + 4c2 0 + λ2 −c2 −. Therefore sup ∥v∥=1 ∥Av∥ = λ+ = 2 + √ 2, inf ∥v∥=1 ∥Av∥ = λ− = 2 − √ 2. Equality occurs at v = ±u+ and v = ±u− respectively. [4] (e) By the spectral theorem, S = Q diag(µ1, . . . , µn)Qt with every µi > 0. Thus S−1 = Q diag(µ−1 1 , . . . , µ−1 n )Qt has positive eigenvalues and is positive definite. Also det S = ∏ i µi > 0. [3] Tony template: real symmetric matrix → orthonormal eigenbasis → apply functions to the eigenvalues. Recognition signal: symmetric, orthogonal diagonalisation, square root, and maximum/minimum norms on the unit sphere. Standard start: find all eigenvalues and eigenspaces and normalise; then f (A) replaces each λi by f (λi). Reusable structure: A = QDQt implies A1/2 = QD1/2Qt and A−1 = QD−1Qt; extremal norms on unit vectors come from the largest and smallest |λi|. Common lost marks: putting eigenvectors in columns without normalising; taking a matrix square root entry by entry; the infimum may be 0 for a singular matrix and is strictly positive here because the matrix is positive definite. 3

MA TH40003 Linear Algebra and Group Theory Tony 2026 Summer Resit Question 4 – Solution and marking guide (a) Following the images gives σ = (1 4 7 10)(2 5 8)(3 6)(9). One transposition decomposition is σ = (1 10)(1 7)(1 4)(2 8)(2 5)(3 6). There are six transpositions, so sgn (σ) = 1 . Its order is lcm(4, 3, 2, 1) = 12 . [8] (b) Reflexivity follows from exe−1 = x. If gxg −1 = y, then g−1yg = x, giving symmetry. If gxg −1 = y and hyh−1 = z, then (hg)x(hg)−1 = h(gxg −1)h−1 = z, which gives transitivity. Thus conjugacy is an equivalence relation. [3] (c) If σ = (a1 . . . ar) is a cycle and τ ∈ Sn, then τ στ−1 = (τ (a1) . . . τ (ar)). Applying this to every disjoint cycle shows that conjugation merely relabels the entries and preserves every cycle length, including the number of fixed points. Conversely, suppose σ and ρ have the same cycle lengths. Order their disjoint cycles so corresponding cycles have equal length, and define a permutation τ by sending the jth entry of each cycle of σ to the jth entry of the corresponding cycle of ρ. This defines a bijection of {1, . . . , n}. The displayed formula then gives τ στ−1 = ρ. [6] (d) The partitions of 4 and the corresponding class sizes are cycle type description size [1, 1, 1, 1] identity 1 [2, 1, 1] one transposition ( 4 2 ) = 6 [2, 2] two disjoint transpositions 3 [3, 1] one 3-cycle ( 4 3 ) · 2 = 8 [4] one 4-cycle (4 − 1)! = 6 The sizes sum to 24 = |S4|. [3] Tony template: compute concrete permutations → conjugation relabels points → classify cycle types. Recognition signal: two-line notation, transpositions, sign/order, followed by conjugacy or partitions. Standard start: begin at the smallest unvisited point and trace its entire cycle; write conjugation as τ (a1 · · · ar)τ −1 = (τ a1 · · · τ ar). Reusable structure: preservation of cycle lengths proves necessity; construct τ by matching positions in corresponding cycles to prove sufficiency; count by choosing supports and removing repeated arrangements. Common lost marks: omitting fixed points and therefore giving an incomplete cycle type; reversing the order of the transposition factorisation; proving only that conjugacy implies equal type and omitting the converse construction. 4