MATH40003 2025 Past Paper · 2026 Four Question Practice Scope

Solutions 1. (a) We must have b =c = 0, andd = 1 ord = 0, to be row-echelon. There is no condition ona. 2, A (b) In addition to the conditions in part (a), whend = 1, we need alsoa = 0. (Note: if the student missed the possibilityd = 0 in part (a), they should not be penalised again here.) 2, A (c) Note that the matrix has rank at least one because the first row is nonzero. The rank is exactly one if and only if the second row is a multiple of the first row. For this to be the case, we need the multiple to be zero since the last entry,0, is only the zero multiple of−1. So the rank is exactly one if and only ifb =c =d = 0. 2, B In the rank one case, then, a basis for the row space is just the first, nonzero row. 1, A In this case, the first row is nonzero (but not the second), so a basis for the column space is the vector ( 1 0 ) . 1, A Finally, in the rank two case, so that some ofb,c,d are nonzero, then the two rows form a basis for the row space, as they are linearly independent. 1, C The column space is all ofR2 so a basis is given by ( 1 0 ) , ( 0 1 ) . 1, C (d) (i) Each row of A +B is the sum of a row ofA and a row ofB, so in the sum of the row spaces of A and B. As the rows ofA +B span the row space (by definition), the inclusion follows. 2, B (ii) The null space consists of vectorsv such thatBv = 0. 1, A Now ifB = 0 then AB = 0 too. 1, A (iii) Write A = (aij), and let the rows ofB be denotedB1,...,B m. Then thei-th row ofAB, by the definition of matrix multiplication, isai1B1 +···+aimBm. This is in the row space. 3, B (Alternative solution: The row space is the perpendicular space to the null space, i.e., the vectors with zero dot product with the null space. So since nullspace(AB)⊇nullspace(B), we get the opposite inclusion on row spaces.) (iv) This is false. A counterexample is given byA = ( 0 0 0 1 ) , B = ( 0 1 0 0 ) . 3, C (Total: 20 marks)

2. (a) You could take V = R2 and U1 = span{e1}and U2 = span{e2}(the x- and y-axes). Then e1,e 2∈U1∪U2 but e1 +e2 /∈U1∪U2, so the union is not a subspace. 3, A un(b) •First note that0V∈⋃ nUn since 0V∈U1. •Next, ifu∈⋃ nUn and λ∈R then, by definition of the union,u∈Un for somen. SinceUn is a subspaceλu∈Un, and thereforeλu∈Un. •Finally, ifu1,u 2∈U, then u1∈Un1 and u2∈Un2 for some n1,n 2. If n = max{n1,n 2}, then since the sequence is increasing,Un1,U n2 ⊆Un, so u1,u 2∈Un. Since Un is a subspace u1 +u2∈Un, and sou1 +u2∈U. 5, B (c) V is finite dimensional if it is the span of a finite set of vectors, or equivalently, if it admits a finite basis (either answer is fine). 1, A un (d) By part (b) the union U = ⋃ nUn is a subspace ofV. It is a fact from lectures that a subspace of a finite dimensional vector space is itself finite dimensional. ThereforeU is finite dimensional, and there are vectorsu1,...,u k∈U such thatU = span{u1,...,u k}. By definition of the union each ui is in Uni for some ni. Let N = max{n1,...,n k}. Then for any M ≥N, we have UN⊆UM⊆U = span{u1,...,u k}⊆UN. SoUN =UN +1 =UN +2 =··· An alternative proof is to more heavily use some dimension theory from the lectures: Say that dim(V ) = n. Since each Ui⊆V is a subspace, it is a fact from lectures thatUi must also be finite dimensional withdim(Ui)≤n. By the same fact from lectures,dim(Ui)≤dim(Ui+1). This means we have inequalities dim(U1)≤dim(U2)≤···≤dim(Ui)≤···≤n. From this we can see that there must be anN such that dim(UN ) = dim(UN +1) =··· Now, it is also a fact from lectures that ifU is a subspace ofU′, and dim(U ) = dim( U′) then U =U′. From this it follows thatUN =UN +1 =··· 6, D un(e) One natural example is Un = R[x]≤n = { polynomials of degree≤n } . These satisfy R[x]≤1⊆R[x]≤2⊆···and R[x]≤n̸= R[x]≤n+1 for alln. 5, B (Total: 20 marks)

3. un (a) This is just saying that a function is uniquely determined by its values on the domain. In more detail, the inverse to this map is the map sending(a,b,c,d )∈R4 to the unique functionf such that f (0, 1) = a,f (1, 0) = b,f (0,−1) = c, andf (−1, 0) = d. It is clear that the composition in either order of these two maps is the identity. 3, A un (b) Since φ:V→R4 is a bijection, provided we haveφ(f +g) = φ(f ) +φ(g) and φ(λf) = λφ(f ), then the axioms forV to be a vector space will each of them follow from the axioms ofR4, for 0 the elementφ−1(0). Moreover, by definitionφis then an isomorphism. So we check this, writing for convenience the elements ofR4 as transposed tuples(a,b,c,d )t: φ(f +g) = (f (0, 1) +g(0, 1),f (1, 0) +g(1, 0),f (0,−1) +g(0,−1),f (−1, 0) +g(−1, 0))t. This clearly equals φ(f ) +φ(g) = (f (0, 1),f (1, 0),f (0,−1),f (−1, 0)) + (g(0, 1),g (1, 0),g (0,−1),g (−1, 0))t. Similarly, φ(λf) = (λf(0, 1),λf(1, 0),λf(0,−1),λf(−1, 0))t =λφ(f ). The zero element is the zero function, sinceφsends it to(0, 0, 0, 0)t. 6, D (c) Since φis an isomorphism, a basis forV is obtained asφ−1 of a basis ofR4. Using the standard basis of R4, we get the basis ofV given by the functionsf0,1,f 1,0,f 0,−1, andf−1,0. 3, D (d) The basis is permuted by this rotation and we get the following permutation matrix, using the given ordering of the basis:   0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0  . In more detail,fa,b◦R =fa′,b′ where (a′,b′) = R−1(a,b ) (note the inverse!). ThenT (f0,1) = f1,0, T (f1,0) = f0,−1, T (f0,−1) = f−1,0, andT (f−1,0) = f0,1. 4, B seen ⇓(e) The transformation T is invertible, with inverse given by the clockwise rotation byπ/2. Accordingly the kernel is zero, the image is all ofV, and the rank is the dimension ofV, which is four. 4, A(Total: 20 marks)

4. (a) (i) The images of the vectors inE are T (e1) =   1 1 −1  ,T (e2) =   0 5 −6  ,T (e3) =   2 1 −2  . Thus A =   1 0 2 1 5 1 −1 −6 −2  . 3, A (ii) The characteristic polynomial ofA is det(XI 3−A) = (X−3)(X−2)(X + 1). Thus there are three eigenvalues:−1, 2, 3. 3, A The eigenspace for−1 is E−1 = ker(−I3−A) = ker   −2 0 −2 −1 −6 −1 1 6 1   = Span      1 0 −1      . The eigenspace for2 is E2 = ker   1 0 −2 −1 −3 −1 1 6 4   = Span   2 −1 1  . The eigenspace for3 is E3 = ker   2 0 −2 −1 −2 −1 1 6 5   = Span      1 −1 1      . 3, B There is a basis ofR3 consisting of eigenvectors ofA. HenceA is diagonalisable. 2, A (b) Note to the markers: give (at least) partial credit for solutions that are written in less detail than the ones below. The main idea behind both proofs below is that a matrix has non-zero determinant if and only if its rows (and columns) are linearly independent un (i) We prove that there is a square submatrix of orderk with non-zero determinant if and only if the rank ofA is at leastk. Suppose there is a square submatrixM of orderk with non-zero determinant. LetM′be the (n×k)-submatrix consisting of the columns chosen forM. Then the rank ofM′is k because thek rows chosen forM are linearly independent asM has rankk and these form a maximal set of linearly independent rows forM′. Therefore, there are at leastk linearly independent columns inA and so its rank is at leastk. 3, B Conversely, suppose the rank is at leastk, thenA has (at least)k linearly independent columns. LetM′bean( n×k)-submatrixof Aconsistingofasetof k linearlyindependent columns. This matrix has rank k so it hask linearly independent rows. If these are chosen and the rest is deleted, we get a square submatrix of orderk with non-vanishing determinant. 2, C un (ii) The submatrices in this part of the question are a subset of those in the previous part, so we need only prove that if their determinant vanishes,rkA≤k. We prove the contrapositive, that is, if the rank ofA is at leastk + 1, then one of the submatrices in the question is non-singular. Following the hint, the rows ofA containing M as a submatrix are linearly independent but cannot span the entire row space ofA, so there is another row that is not in the span of these. Considering thesek + 1 rows alone, gives a submatrix, ofA of rankk + 1. 2, D LetM′be the submatrix of rankk + 1 formed by the rows ofM and the additional row (as explained above). Similarly to what done before (but this time using columns), the column space ofM′has dimensionk + 1. The columns containingM (in M′) span a space of dimensionk, so there must be another column inM′that is not in the span of those containingM. Adjoining this column to those containingM gives the sought non-singular submatrix of orderk + 1 containing M. 2, D (Total: 20 marks)

5. (a) (i) By the Gram-Schmidt process we have w1 =v1, w2 =v2−v2·w1 ∥w1∥2w1 =   0 0 3  , w3 =v3−v3·w1 ∥w1∥2w1−v3·w2 ∥w2∥2w2 =   4 0 0  , Normalising the vectors above, we find the basis B =      0 1 0  ,   0 0 1  ,   1 0 0      4, A (ii) Calling the vectors ofB u1, u2, andu3 respectively, we get that v1 =u1, v 2 = 4u1 + 3u2, v 3 =−u1 + 4u2 + 4u3 3, A (iii) By the above, it is immediate to see that Q =   0 0 1 1 0 0 0 1 0  , R =   1 4 −1 0 3 4 0 0 4  . 3, B (b) (i) We use the subgroup test. LetH = On(Z). Clearly H̸=∅as it contains the identity matrix. If two matrices with integer entries are multiplied, the result is another matrix with integer entries, soH is closed under multiplication. It is also closed under taking inverses because, for the matrices inOn(R), the inverse is just the transpose and if A∈Mn(Z), thenAT∈Mn(Z). Thus, ifA∈H =On(R)∩Mn(R), A−1 =AT∈H. 3, A un (ii) A permutation matrix has integer entries and its columns are pair-wise orthonormal because they are just a rearrangement of the standard basis. HenceOn(Z) contains all the permutation matrices. 3, C un (iii) Let a1,...,a n be the columns ofA∈On(Z). Then for alli,j ∈{1,...,n} ai·aj =    1 i =j 0 i̸=j. Moreover for all i∈ {1,...,n}we have ai·ai = ∑n k=1a2 ik = 1 . This means that for each column there only is one non-zero entry and its value is either1 or−1. For i,j ∈{1,...,n}with i̸=j, 0 = ∑n k=1aikajk, this implies that the non-zero entry ofai is not in the same row as the non-zero entry ofaj. We conclude thatOn(Z) consists of all those matrices that have exactly one entry equal to1 or−1 in each row and each column with all other entries equal to0. Suppose we are choosing a matrix inOn(Z). We have 2n choices for the first column. After this, we have2(n−1) choices for the second column because we are allowed to put±1 in any of then−1 rows where the first column is0. Similarly there are2(n−2) choices for the third column. This way there are exactly2(n−i + 1) choices for theith column and so the order of the group is2n·n!. 4, D Alternatively one can argue that ifH is the subgroup of permutation matrices (which has ordern!), the left cosets ofH are exactly diag(a1,...,a n)H where ai∈{−1, 1}. So there are exactly as many left cosets ofH as there are choices for diag(a1,...,a n), that is2n. (Total: 20 marks)

seen ⇓6. (a) (i) This permutation is (15)(24), hence its order is 2. 2, A un (ii) Every element ofS9 can be written as a product of cycles of lengthsn1,...,n r such that n1 +...+nr = 9, whereni≥1 for alli. The order of any permutation of this cycle type is the least common multiple ofn1,...,n r. The maximum of l.c.m. is20 which is attained forr = 2, n1 = 4, n2 = 5. 4, B un (b) (i) Possible cycle types of elements ofS5 are 2, 3, 4, 5, (2,2), (2,3), where we omit cycles of length 1. The orders of elements of these cycle types are 2, 3, 4, 5, 2, 6, respectively. In particular,S5 does not contain elements of order|S5|= 120. An alternative way to see thatS5 is not cyclic is to note thatS5 is not commutative. Indeed,(12)(13)̸= (13)(12). 3, A un(ii) Write Mn = ( 1 n 0 1 ) . Then we haveMnMk = Mn+k for alln,k ∈Z. This implies (Mn)−1 = M−n. In particular, our group is{(M1)n|n∈Z}, therefore it is an infinite cyclic group. 3, A (iii) Letρbe the rotation through the angle2π/m. It preserves the regularm-gon. Moreover, every rotation preserving the regularm-gon isρi, wherei = 0, 1,...,m−1. Thus these rotations form a cyclic group of orderm with generatorρ. 1, A un (c) H is necessarily a subgroup ofG. Take anyh∈H and consider the sequenceh,h 2,h 3,.... By assumption, this sequence is contained inH. Moreover, h has finite order, sayn. Then hn =e, so thate∈H. Next, we haveh−1 =hn−1, soh−1∈H. ThusH contains e and is closed under taking inverses, and so is a subgroup ofG. 7, C (Total: 20 marks) Total A marks: 46 of 48 marks Total B marks: 34 of 30 marks Total C marks: 17 of 18 marks Total D marks: 23 of 24 marks