22 MATH40003 Practice Paper 2 ANSWERS

English review edition prepared on 4 October 2026 from a preserved source copy. It is a later presentation, not the historical study interface. Source checks and difficulty judgements describe the original material author's own process; they do not indicate Imperial College London endorsement.

Source SHA-256: b6a9fecf44921aa093db98616c4d6d36e8bc6eefee311ef4307e5a369fac2463
Source date: 2026-08-06

Source page 1

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Mathematical typesetting and diagrams are retained below. Chinese prose is replaced by the English passages that follow, in source reading order.

English passage 1 (11% down the source page) Historical basis: 2025 Q1 (direct A coverage). The matrix size and parameter pattern are changed, but the entry signal, case split and row/column-space methods are the official 2025 methods. 30 minutes.

English passage 2 (16% down the source page) Recognition signal: a matrix with parameters; first REF/RREF, then rank and row/column spaces by parameter case, and finally proofs of space inclusions induced by matrix products.

English passage 3 (20% down the source page) First key step: Do not row-reduce immediately . First inspect pivot positions and separate the cases b = 0 and d = 0.

English passage 4 (61% down the source page) Marking points: 3 REF; 3 RREF; 6 rank/bases; 6 three inclusions; 2 counterexample.

English passage 5 (64% down the source page) Common errors: Missing the non-REF case b = 0, d ̸= 0 when computing rank; using pivot columns of a row-reduced matrix instead of the corresponding original columns; giving a coun- terexample with incompatible sizes.

English passage 6 (69% down the source page) Final self-check: Check every proposed row-space basis is independent; compute AB and BA explicitly in the counterexample.

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Mathematical typesetting and diagrams are retained below. Chinese prose is replaced by the English passages that follow, in source reading order.

English passage 1 (11% down the source page) Historical basis: 2025 Q2 (all high-mark parts directly covered). Course method: subspace test, finite basis argument, and recursive construction of an independent sequence. 30 minutes.

English passage 2 (15% down the source page) Recognition signal: an increasing chain of subspaces, an infinite union and eventual stabilisation.

English passage 3 (17% down the source page) First key step: For addition in the union, put both vectors into the same UN using N = max(n1, n2). For stabilisation, use a finite basis of the union or monotone dimensions.

English passage 4 (38% down the source page) Marking points: 2 counterexample; 5 union proof; 1 definition; 6 stabilisation; 3 polynomial chain; 3 converse.

English passage 5 (42% down the source page) Common errors: Trying induction on the finite unions; asserting the union equals V ; using limits or convergence of subspaces.

English passage 6 (46% down the source page) Final self-check: In the stabilisation proof, verify every basis vector of the union lies in one common UN; in the converse, verify each inclusion is strict.

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Mathematical typesetting and diagrams are retained below. Chinese prose is replaced by the English passages that follow, in source reading order.

English passage 1 (11% down the source page) Historical basis: 2025 Q3 direct A coverage. Course method: construct the inverse coordinate map, transfer the vector-space axioms through a linear bijection, and compute the induced permu- tation matrix by acting on delta functions. 30 minutes.

English passage 2 (16% down the source page) Recognition signal: an abstract set of functions followed by an evaluation map and a permutation of the domain.

English passage 3 (18% down the source page) First key step: Construct Φ−1 explicitly . For the operator matrix, compute T (δj) = δr−1(j), not δr(j).

English passage 4 (58% down the source page) Marking points: 6 vector space/isomorphism; 3 basis; 4 matrix; 3 invertibility data; 4 fixed/image.

English passage 5 (60% down the source page) Common errors: Assuming a bijection alone creates the stated vector-space structure; using r instead of r−1; proving only containment for the image without the dimension argument.

English passage 6 (64% down the source page) Final self-check: Multiply the displayed matrix by each standard basis vector; check every column sum of I −[T] is zero and its nullity is one.

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Mathematical typesetting and diagrams are retained below. Chinese prose is replaced by the English passages that follow, in source reading order.

English passage 1 (11% down the source page) Historical basis: 2025 Q4, including the 9-mark rank-by-minors proof. Course method: characteristic polynomial/eigenspaces, then independent rows and columns to construct a non-zero minor. 30 minutes.

English passage 2 (16% down the source page) Recognition signal: one part is a concrete eigenspace calculation, the other a proof in both directions relating all k-order minors to rank.

English passage 3 (18% down the source page) First key step: For the theorem, prove the equivalent positive statement: rank at least k iff some k-minor is non-zero.

English passage 4 (55% down the source page) Marking points: 2 matrix; 8 eigenanalysis; 6 rank theorem; 4 parameter application.

English passage 5 (57% down the source page) Common errors: Showing selected rows are independent only inside a submatrix without connect- ing them to the original matrix; saying distinct eigenvalues imply diagonalisation but not giving eigenspaces; at t = 2, failing to prove rank is at least two.

English passage 6 (63% down the source page) Final self-check: Check AP = P D; verify both directions of the minor theorem; compute one non-zero 2 × 2 minor at t = 2.