MATH40003 Revised Historical Coverage Practice Paper 2

Question 1 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. 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. First key step: Do not row-reduce immediately . First inspect pivot positions and separate the cases b = 0 and d = 0. The first row has pivot in column 1. If b ̸= 0 , row 2 has pivot in column 2, and row 3 may be non-zero or zero, so the matrix is in REF for any d. If b = 0 , row 2 has pivot in column 3, so REF requires d = 0. Thus REF iff b ̸= 0 or (b, d) = (0 , 0). For RREF , if b ̸= 0 and d ̸= 0 , row 2 has a non-zero entry in the pivot column of row 3, so RREF is impossible. If b ̸= 0, d = 0, RREF requires b = 1 and a = 0. If b = d = 0, the matrix is already RREF for arbitrary a, c. Hence (b, d) = (0 , 0) or (a, b, d) = (0 , 1, 0). The rank is 3 iff bd ̸= 0, and otherwise it is 2. In the rank-three case the three rows form a row-space basis and columns 1,2,3 form a column-space basis. In all other cases rows 1 and 2 form a row-space basis. For the column space: if b ̸= 0 , d = 0 , use columns 1 and 2; if b = 0 (any d), use columns 1 and 3. If Bv = 0 , then ABv = 0 , proving ker B ⊆ ker AB. Each row of AB is a linear combination of rows of B, and each column of AB is A applied to a column of B, proving the two space inclusions. A counterexample is A = ( 0 1 0 0 ) , B = ( 1 0 0 0 ) , for which AB = 0 but BA = A ̸= 0. Marking points: 3 REF; 3 RREF; 6 rank/bases; 6 three inclusions; 2 counterexample. 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. Final self-check: Check every proposed row-space basis is independent; compute AB and BA explicitly in the counterexample. 2/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 2 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. Recognition signal: an increasing chain of subspaces, an infinite union and eventual stabilisation. 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. Take the two coordinate axes in F 2 for part (a). Their union is not closed under addition. Let U = ∪ Un. It contains 0. If u ∈ Ui and v ∈ Uj, then both lie in UN for N = max(i, j), so u + v ∈ UN ⊆ U . Scalar closure is immediate. A vector space is finite-dimensional if it has a finite basis (equivalently , a finite spanning set). If V is finite-dimensional, so is U = ∪ Un. Choose a basis u1, . . . , ur of U . Each ui lies in some Uni; with N = max ni, the whole basis lies in UN . Therefore UN = UN +1 = · · · = U . In F [x], take Un = F [x]≤n. Then xn+1 ∈ Un+1 \ Un. Conversely , if V were infinite-dimensional, recursively choose vn+1 /∈ span(v1, . . . , vn) and set Un = span(v1, . . . , vn). This chain never stabilises, a contradiction. Marking points: 2 counterexample; 5 union proof; 1 definition; 6 stabilisation; 3 polynomial chain; 3 converse. Common errors: Trying induction on the finite unions; asserting the union equals V ; using limits or convergence of subspaces. 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. 3/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 3 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. Recognition signal: an abstract set of functions followed by an evaluation map and a permutation of the domain. First key step: Construct Φ−1 explicitly . For the operator matrix, compute T (δj) = δr−1(j), not δr(j). The inverse of Φ sends (a1, a2, a3, a4)T to the unique function with values ai. Pointwise oper- ations give Φ(f + g) = Φ( f ) + Φ(g), Φ(λf ) = λΦ(f ), so V is a vector space and Φ is an isomorphism. The zero element is the zero function. The delta functions map to the standard basis of F 4, so they form a basis. Since r−1 = (1 4 3 2) , T δ1 = δ4, T δ 2 = δ1, T δ 3 = δ2, T δ 4 = δ3, so [T ] =   0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0   . The operator is invertible, hence ker T = 0, im T = V , and rk T = 4. The fixed functions are the constants, so ker(I − T ) = span(δ1 + δ2 + δ3 + δ4). Every vector in im (I − T ) has coordinate sum zero because the sum is unchanged by T . Rank-nullity gives rk (I − T ) = 3 , the same as the dimension of the sum-zero subspace, so equality holds. Marking points: 6 vector space/isomorphism; 3 basis; 4 matrix; 3 invertibility data; 4 fixed/image. 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. 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. 4/5 MATH40003 - Linear Algebra and Group Theory Revised Historical Coverage

Question 4 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. Recognition signal: one part is a concrete eigenspace calculation, the other a proof in both directions relating all k-order minors to rank. First key step: For the theorem, prove the equivalent positive statement: rank at least k iff some k-minor is non-zero. The matrix is A =   2 1 0 1 2 0 0 0 4   . The eigenspaces are E3 = span(1, 1, 0)T , E 1 = span(1, −1, 0)T , E 4 = span(0, 0, 1)T . Taking these eigenvectors as columns gives a valid diagonalising matrix P and P −1AP = diag(3, 1, 4). If a k × k submatrix has non-zero determinant, its selected columns are independent, so the whole matrix has rank at least k. Conversely , if rank is at least k, choose k independent columns. The resulting matrix has rank k, so it has k independent rows. The intersection of those selected rows and columns is a non-singular k × k submatrix. This proves the equivalence. Finally det Ct = t − 2. Thus rk Ct = 3 for t ̸= 2 . At t = 2 , the third row is the sum of the first two and the leading 2 × 2 minor is non-zero, so the rank is 2. Marking points: 2 matrix; 8 eigenanalysis; 6 rank theorem; 4 parameter application. 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. Final self-check: Check AP = P D; verify both directions of the minor theorem; compute one non-zero 2 × 2 minor at t = 2. 5/5