NYJC 2026 FM TP- Linear Algebra Set 3 (Solutions)

2026 J2 H2 FM Linear Algebra (Set 3 – Solutions) 1 In this question, V denotes the set of vectors of the form a b c d       , where a, b, c and d are real numbers. You may assume that V forms a linear space under the usual operations of vector addition and multiplication by a scalar. (a) Show that the subset of V for which 2a b d c+ − = does not form a linear space. [1] (b) Determine whether or not the subset of V for which both 42a b c d+ = + and 2a b c d− = + forms a linear space. [2] (c) Determine whether or not the subset of V for which 42a b c d+ = + or 2a b c d− = + forms a linear space. [2] (d) State the dimension of the subspace of V such that 3 2 0a b c d+ − + = and provide a basis for this linear space. [3] Solution: 2020/HCI Promo/Q8 (a) Let 2V be the given subset of V. 2 1 1 1 1 V      as 21 1 1 1+ − = . However, 2 1 12 1 1 V      as 22 2 2 2 2+ − =  . Since 2V is not closed under scalar multiplication, 2V is not a linear space. (b) Let 3V be the given subset of V. Clearly, 3 0 0 0 0 V      as ( ) ( ) ( )0 4 0 2 0 0 and 0 2 0 0 0+ = + − = + . Consider 12 12 3 12 12 , aa bb Vcc dd                       , and 12,kk   Since ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 1 1 2 2 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 2 2 1 1 2 2 4 4 4 22 2 k a k a k b k b k a k b k a k b k c k d k c k d k c k c k d k d + + + = + + + = + + + = + + + and ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 1 1 2 2 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 2 2 1 1 2 2 2 2 2k a k a k b k b k a k b k a k b k c k d k c k d k c k c k d k d + − + = − + − = + + + = + + + Intersection of subspaces are subspaces.

Thus 12 12 1 2 3 12 12 + aa bbk k Vcc dd                        3V is closed under addition and scalar multiplication (c) Consider 1 1 2 1      which satisfies 42a b c d+ = + but not 2a b c d− = + . Consider 1 0 1 0      which satisfies 2a b c d− = + but not 42a b c d+ = + . However 1 1 2 1 0 1 2 1 3 1 0 1            +=                  does not satisfy both 42a b c d+ = + and 2a b c d− = + as ( ) ( )2 4 1 2 3 1+  + and ( ) ( )2 2 1 3 1−  + . Thus subset is not a linear space as it is not closed under addition. A union of subspaces may not be a subspace. (d) 3 2 0 3 2a b c d c a b d+ − + =  = + + 1 0 0 0 1 0 3 2 1 0 0 1 a b a b dc d                = + +                        ( )1dim 3V= 1 0 0 0 1 0Basis , , 3 2 1 0 0 1            =                  since the 3 vectors are clearly linearly independent. 2 2018/NJC/Prelim An orthogonal matrix Q is one such that T ,=Q Q I where I is the identity matrix. (a) Find the possible determinants of an orthogonal matrix. [3] (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) T T T 2 det det det det det det det det det 1 det 1 = = = = = = Q Q I Q Q I Q Q I Q Q I Q Q

For any vectors 3,, uv  it can be shown that T .=u v u v (b) Suppose 1 2 3, andv v v are the column vectors (in that order) of a 3  3 orthogonal matrix Q with real entries. Use the above result to show that 1 2 3, andv v v are unit vectors that are pairwise perpendicular. [4] (b) ( ) 1 1 1 1 2 1 3 T 2 1 2 3 2 1 2 2 2 3 3 3 1 3 2 3 3 1 0 0 0 1 0 0 0 1 T T T T T T T T T T T T         = = =            v v v v v v v Q Q v v v v v v v v v v v v v v v v v Consider T ijvv , for i, j = 1, 2, 3. If ,ij= 2 1 1 1 1 T ii ii i i = = = = vv vv v v If ,ij 0 0 T ij ij ij = = ⊥ vv vv vv Alternatively: Note that ii=Qe v for i = 1, 2, 3, where  1 2 3,,e e e are the standard basis vectors for 3.  For i, j = 1, 2, 3, such that ,ij For i, j = 1, 2, 3, such that ,ij= ( ) ( ) T T TT TT (since ) 0 i j i j ij ij ij ij = = = == =  = v v v v Qe Qe e Q Qe e e Q Q I ee ( ) ( ) 2 T T TT TT (since ) 1 i i i i i ii ii ii =  = = = == = v v v v v Qe Qe e Q Qe e e Q Q I Suppose 1 2 3 1 2 2 1 1 12 , 1 and 2 ,3 3 32 2 1 −           = = − = −                v v v and the linear transformation T is defined as follows: ( ) 1 2 3 T     = u if and only if 1 1 2 2 3 3 ,  = + +u v v v where 1 2 3, and    are real constants. (c) The column vectors of an orthogonal matrix are called orthonormal vectors. Given that 12,vv and 3v as defined above are orthonormal vectors, find the matrix A representing the linear transformation T. [2] (c) ( ) 11 1 1 2 2 3 3 1 2 3 2 2 33               = + + = =           u v v v v v v Q

Thus 1 1T 2 3 1 2 2 1 2 1 23 2 2 1    −        = = = − −       −    Q u Q u u i.e. ( ) 1 2 2 1T 2 1 2 .3 2 2 1  = − − − uu Hence, 1 2 2 1 2 1 23 2 2 1  = − − − A Alternatively: (d) Hence find the subspace of 3  whose range space under T is given by : , . p q p q pq    +  [3] (d) Let ( ) 1 2 3 T p q pq           ==       +    u

( ) ( ) 1 1 2 2 3 3 1 2 3 1 3 2 3 () 1 2 2 2 1 1 1 12 2 1 23 3 3 32 1 2 1 10 01 11 p q p q pq pq pq   = + + = + + + = + + +    −                   = + − + − + −                                      = + −          u v v v v v v v v v v Required subspace :, p q p q pq  = −  +  3 Let V be the vector space of all real polynomials of degree at most n, and let :D V V→ be the linear transformation representing differentiation, that is, d() d pDp x= . (a) Explain why D is a linear transformation. [1] (b) State bases for the kernel and range of D. [2] (c) Give a reason why the set of real polynomials of degree n is not a vector space. [1] ACJC JC2 Prelim 9649/2019/02/Q1 (a) For any two polynomials p and q in V and any two real numbers  and  , d( ) ( ) d dd dd ( ) ( ) D p q p q x pq xx D p D q       + = + =+ =+ Hence D is a linear transformation. (b) Need d( ) 0 d pDp x== . Thus pk=  . A basis for the kernel of D is { 1 } . Let 1 1 1 0 ....nn nnp a x a x a x a − −= + + + + ( ) 12 11 d 1 ....d nn nn p na x n a x ax −− −= + − + + A basis for the range of D is   12, , ..., , 1nnx x x−− (c) The set of real polynomials of degree n is not a vector space as it does not contain a zero element. (It must have nax term, 0a ). OR If p and q are polynomials with the same term nax as the term in the highest power, then pq− would be a polynomial of degree 1n− and does not belong to the set. Hence it is not closed under addition.

4 Do not use a calculator in answering this question. The linear transformation T, from 22 ,→  defined by T : where , where , , and are constants .x x a b a b c dy y c d       =           AA  It is given that the transformation under T has a line 1L of invariant points. (a) Explain why A has an eigenvalue equal to 1. [1] (b) Hence show that 1ad a d bc+ = + + . [2] It is given that 341 435 =  − A . (c) Find an eigenvector 1e that corresponds to eigenvalue 1 of A and the Cartesian equation of 1.L [2] It is given that the transformation under T has an invariant line 2.L (d) By finding the other eigenvalue and its corresponding eigenvector 2e of ,A give the Cartesian equation of 2.L [3] (e) Show that 1L and 2L are perpendicular. [1] (f) By considering your answer to part (d) and (e), and the image of  +12ee when it is transformed under T, where ,,   describe the geometrical transformation T ex