F07 - Matrices and Linear Spaces - Tutorial Set 2 (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 2 Solutions) Page 1 of 7 National Junior College 2016 – 2017 H2 Further Mathematics Topic F7: Matrices and Linear Spaces (Tutorial Set 2 Solutions) Basic Mastery Questions 1 (i) No. 0 1x S  , 0 1x S  , but    0 1 0 1 0 0x x x S      . (A1 fails) (ii) Yes. (To verify non-empty, A1 and A6 as T is a subset of 3 ) (iii) Yes. (To verify all the 10 axioms) (iv) Yes. (To verify all the axioms) (v) No.  1 1, 0, 0  0, A10 fails unless  1, 0, 0 is the zero vector.  1 2, 0, 0  0 , A10 fail unless  2, 0, 0 is the zero vector. However, there is only one zero vector in the vector space, so it is not a vector space. (vi) Yes. (To verify all the axioms) 2 (To verify non-emptiness, A1 and A6 as they are subsets of 3 ) (i)   1, 1, 2 . (ii)     2,1,0 , 3,0,1 . 3 (To verify non-emptiness, A1 and A6 as they are subsets of 2P ) (i)   2,1 2x x (ii)   2 2x x  4 (i), (ii), (iv) and (iv): Yes (iii) No.  1,1,0 the set but    2 1,1,0 2, 2,0 the set  as    2 2 2 5 0 2 1    . (v) No.  0,1,1 the set but    2 0,1,1 0, 2, 2 the set  . (vii) No. An empty set is not a vector space. (viii) No.  1,0,1 the set ,  1,0, 1 the set  but      1,0,1 1,0, 1 2,0,0 the set    . 5 To show linearly independent, let        1,0, 1 0,2,0 1,1,1 0,0,0a b c     . 1 0 1 0 1 0 1 0 0 2 1 0 0 2 1 0 1 0 1 0 1 0 1 0 a b c a a b c b a b c c                                      . Since the determinant of the coef ficient matrix is 4, the coefficient matrix is invertible and there is exactly one solution to the system which is the trivial solution. Thus, the set is linearly independent. To show spanning, let        1,0, 1 0, 2,0 1,1,1 , ,p q r x y z    . 1 0 1 1 0 1 0 2 1 0 2 1 1 0 1 1 0 1 p q r x p x p q r y q y p q r z r z                                      . Since the determinant of the coefficient matrix is 4, the coefficient matrix is invertible and there is exactly one solution to the system for any vector  , ,x y z in 3 . Thus, the set spans 3 . (i) Yes (ii) No. It is not linearly independent as      0,0,0 0 1,0, 1 0 0, 2,0   (iii) No. It is not linearly independent as        0, 2,0 0 1,0, 1 0 1,1,1 2 0,1,0    (iv) No.        0,0,1 0,2,0 1,1,1 , 2 ,k l l k l l    has no solution, it does not span 3 www.KiasuExamPaper.com 659

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 2 Solutions) Page 2 of 7 6 (a) Let  : ,U V W   v w v w . Since V0 and W0 , U  0 0 0 . Thus U is non-empty. 1 2, U u u , 1 2, V v v and 1 2, W w w s.t. 1 1 1 u v w , 2 2 2 u v w .        1 2 1 1 2 2 1 2 1 2        u u v w v w v v w w . Since 1 2 V v v and 1 2 W w w , 1 2 U u u . U is closed under addition. U u and k   . ,V W   v w s.t.  u v w .  k k k k   u v w v w . Since k Vv and k Ww , k Uu . U is closed under scalar multiplication . Therefore U is a subspace of n . (b) No. Yes. Yes. Yes. (Solution omitted) 7   ,0 :U x x   and   0, :W y y   are both subspaces of 2 . For    1,0 , 0,1 U W  ,      1,0 0,1 1,1 U W    , so U W is not a vector space. 8(a) (i) The column space of 1 1 1 0 0 2 2 0 1 0 1 1 0 0 1         is 3 . Performing row operations, 1 1 2 4 1 2 1 1 1 0 0 1 1 1 0 0 1 1 1 0 0 2 2 0 1 0 0 4 2 1 0 0 1 0 1 1 0 0 1 0 2 1 0 1 0 0 0 1                                . Thus a basis is       1, 2,1 , 1, 2,1 , 0,1,0 (1st, 2nd and 4th columns) (ii) The column space of 1 1 1 1 0 0 0 3 3 0 0 1 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 1             is 4 . Since the first 3 columns are linearly independent, the leading 1’s in a row -echelon form are in the 1 st, 2nd, 3rd and 7th columns. Thus, a basis is 1 3 1 3 1 0 0 0, , ,1 0 0 0 1 0 0 1                           . 8(b) (i) Consider the corresponding spanning set for 3 ,         1,1,0 , 2,0,1 , 0, 2,1 , 1,1,1 . The column space of 1 2 0 1 1 0 2 1 0 1 1 1         is 3 . Performing row operations, 1 2 0 1 1 2 0 1 1 2 0 1 1 2 0 1 1 0 2 1 0 2 2 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 2 1                                           . Thus, a basis for 3 is       1,1,0 , 2,0,1 , 0, 2,1 , correspondingly, a basis for 2P is   2 21 ,2 , 2x x x x   . (ii) 2 0x y z   , let y  and z  . Then 2x    . The vector space can be rewritten as 2 2 1 1 0 0 1                                                    , so a basis can be 2 1 1 , 0 0 1                        www.KiasuExamPaper.com 660

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 2 Solutions) Page 3 of 7 Practice Questions 9 The 2 2 zero matrix O satisfies  AO OA O , so WO . Thus W is non-empty. , W B C , AB BA and AC CA . We have          A B C AB AC BA CA B C A , so W B C . Thus W is closed under addition. W D and k   , AD DA . We have        k k k k  A D AD DA D A , so k WD . Thus W is closed under scalar multiplication. Therefore, W is a subspace of  2,2M  . 10 (a) Other than the two extreme cases wi th dimensions 0 and 2 respectively, the dimension of any subspace of 2 must be 1, i.e. there is only one vector, say u , in a basis. Thus the subspace is  ku , which is a line through the origin. (b) Other than the two extreme cases with dimensions 0 and 3 respectively, the dimension of any subspace of 3 must be 1 or 2. When there is only one vector in a basis of the subspace, the subspace is a line through the origin (following a similar argument from (a)). When there are exactly two vectors in a basis of the subspace, say u and v, the subspace is  k lu v . As u and v are non -zero and linearly independent, the subspace is a plane through the origin. 11 1 , 1 , 1 b a a b b a                                 must not be linearly independent. (If so, it will become a basis as the dimension of 3 is 3). This means, 1 0 1 0 1 0 b a x a y z b b a                                        has non-trivial solution, so we have 3 3 3 3 1 1 1 3 1 0 1 b a a b a b ab ab ab a ab b b a            . 12 The vector space concerned is the row space of the matrix 1 2 1 3 1 2 2 10 8 7 7 8           . 1 2 1 1 2 1 1 2 1 3 1 2 0 7 5 0 7 5 2 10 8 0 14 10 0 0 0 7 7 8 0 21 15 0 0 0                                         , so a basis is     1, 2, 1 , 0, 7,5  and the dimension is 2. (If we wish to preserve the vectors, we may consider the column space of the transpose). Let      1, 4, 1,2, 1 0, 7,5a      . Solve it, we have 2 171, , 7 7a    . www.KiasuExamPaper.com 661

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 2 Solutions) Page 4 of 7 13 1 0  . Consider the equation 1 2 2 3 3k k k  a b b 0 (*) Substituting 1 1 2 2 3 3    a b b b , we have       1 1 1 2 2 3 3 2 2 3 3 1 1 1 1 2 2 2 1 3 3 3 k k k k k k k k                 b b b b b 0 b b b 0 This equation has only trivial solution as 1 2 3, ,b b b are linearly independent. i.e. 1 1 1 2 2 1 3 3 0k k k k k       . When 1 0  , 1 0k  , the