NATIONAL UNIVERSITY OF SINGAPORE MA1101R

Transcription

1 Student Number: NATIONAL UNIVERSITY OF SINGAPORE - Linear Algebra I (Semester 2 : AY25/26) Time allowed : 2 hours INSTRUCTIONS TO CANDIDATES. Write down your matriculation/student number clearly in the space provided at the top of this page. This booklet (and only this booklet) will be collected at the end of the examination. 2. Please write your matriculation/student number only. Do not write your name.. This examination paper contains SIX questions and comprises FIFTEEN printed pages. 4. Answer ALL questions. 5. This is a CLOSED BOOK (with helpsheet) examination. 6. You are allowed to use two A4 size helpsheets. 7. You may use scientific calculators. However, you should lay out systematically the various steps in the calculations) Examiner s Use Only Questions Marks Total

2 PAGE 2 Question [ marks] Let S = {u, u 2, u } be a subset of R where u = (,, ), u 2 = (,, 2), u = (2,, ). (i) [2 marks] Show that S forms a basis for R. (ii) [2 marks] Let w = (7,, ). Find the coordinate vector of w with respect to S. (iii) [4 marks] Let T = {v, v 2, v } be another basis for R where v = (,, ), v 2 = (2, 2, 2), v = (, 2, ). Find the transition matrix from T to S. (iv) [2 marks] Suppose the coordinate vector of x with respect to T is (a, b, c). coordinate vector of x with respect to S. Find the (i) There are many ways to show that u, u 2, u is a basis. Here we use u, u 2, u to form a matrix A such that 2 det(a) = 2 = Hence A is invertible, which means u, u 2, u form a basis for R. (ii) We need to express (7,, ) = c u + c 2 u 2 + c u R 2+( )R R +( )R So we can solve for c =, c 2 =, c = R +R 2 Hence the coordinate vector (w) S = (c, c 2, c ) = (,, 4) (iii) To find the transition matrix P from T to S, express v, v 2, v in terms of the basis S: R 2+( )R R +R 2 R 2 2 +( )R R R 2 R +( )R So P = R 2+( )R R +( 2)R

3 PAGE Question Continue your working below. (iv) [x] S = P [x] T = a b = c a + 2b + c 5a 4a c Continue on page 4 and 5 if you need more writing space. 4

4 PAGE 4 Question 2 [ marks] 2 Let A =. (i) [2 marks] Find the eigenvalues for A. Justify your answer. (ii) [8 marks] Show that A is diagonalizable and find an invertible matrix P and a diagonal matrix D such that P AP. (i) The eigenvalues are the diagonal entries of this lower triangular matrix, namely,2 and. (Note: There is no need to solve the characteristic equation of A.) R (ii) I A = 2 +R R +R 2 R 4 +R R 2 R (I A)x = x = (, s + 2t, s, t) = s(,,, ) + t(, 2,, ). So a basis for the eigenspace for eigenvalue is (,,, ), (, 2,, ). 2I A = R R 4 (2I A)x = x = ( t, t, t, t) = t(,,, ). So a basis for the eigenspace for eigenvalue 2 is (,,, ). I A = 2 2 R 2 +( )R R +( )R 2 R 4 +( )R (I A)x = x = (,,, t) = t(,,, ) So a basis for the eigenspace for eigenvalue is (,,, ). R 2 +( )R R +( )R 2 5

5 PAGE 5 Question 2 Continue your working below. Since there are four linearly independent eigenvectors for this 4 4 matrix, the matrix A is diagonalizable. Let P = 2. Then P AP = D = 2. Continue on page 4 and 5 if you need more writing space. 6

6 PAGE 6 Question [ marks] Let V = span{(,,, ), (, 2,, ), (,, 2, 2)}. (i) [ marks] Using the Gram-Schmidt process, find an orthogonal basis for V. (ii) [2 marks] Using the result in (i), find the projection of (2, 2, 2, ) onto V. (iii) [2 marks] Extend your basis in (i) to an orthogonal basis for R 4. (iv) [ marks] Find aleast squares solution of the system Ax = b where 2 A = 2 2 and b = (i) Let u = (,,, ), u 2 = (, 2,, ) and u = (,, 2, 2). v = u = (,,, ), v 2 = u 2 u 2 v v = (, 2,, ) 6 (,,, ) = (,,, ), v v v = u u v v v v u v 2 v 2 v 2 v 2 = (,, 2, 2) (,,, ) (,,, ) = (,,, ). An orthogonal basis for V is {v, v 2, v }. (ii) The projection of v = (2, 2, 2, ) onto V is p = (v v ) (v v ) v + (v v 2) (v 2 v 2 ) v 2 + (v v ) (v v ) v = (,,, ) + 6 (,,, ) + (,,, ) = (2,,, 2). (iii) Note that v p = (,,, ) is orthogonal to each v i. Then is an orthogonal basis for R 4. {v, v 2, v, (,,, )} 7

7 PAGE 7 Question 6 (iv) A T Ax = A T v is 6 5 x = R 2+( 2)R R 9 2 +R 6 9 R R 2 R 2 2 R +R R 2 +( )R 2 2 So a least squares solution to Ax = v is 2 R +( )R 2 R +( 2)R Alternatively, the answer can be obtained by solving Ax = p where p is the projection in (ii) above.. Continue on page 4 and 5 if you need more writing space. 8

8 PAGE 8 Question 4 [ marks] Let A be a 2 matrix and B be a 2 matrix such that AB = (i) [ marks] Find a basis for the row space of AB and state the rank of AB. (ii) [2 marks] Show that (AB) 2 = 5AB. (iii) [2 marks] What is the rank of BA? Justify your answer. (iv) [ marks] Find BA. Show clearly how you derive your answer. (i) We check that R R 2 R R 7 7 5R R R 4R 2 25 So a basis for the row space is given by {(, 7, 7), (, 4, 5)} and rank(ab) = 2. (ii) One verifies that: 7 7 (AB) 2 = = 5AB (iii) rank(ba) rank(a(ba)b) = rank((ab) 2 ) = 2. Since BA is 2 2, so rank(ba) = 2. (iv) From (ii), (BA) = BABABA = B(AB) 2 A = B(5AB)A = 5(BA) 2. From (iii), we have BA is full rank and invertible. ( ) 5 It follows that BA = 5I =. 5 9

9 PAGE 9 Question 4 Continue on page 4 and 5 if you need more writing space.

10 PAGE Question 5 [ marks] (a) [4 marks] Let T : R R 2 be a linear transformation and u =, u 2 =, u = a basis for R. Suppose T (u ) = ( ) 2, T (u 2 ) = ( ), T (u 6 ) = ( ) 2. 2 Write down the standard matrix of T and find Ker(T ) explicitly. Observe that: e = = (u 2 + u 2 u ). So T (e ) = (T (u 2 ) + T (u 2 ) T (u ) = e 2 = = (u u u ). ( ) + ( ) + ( ) ( ) ( ) = ( ) = ( ). 2 ( ). 4 So T (e 2 ) = (T (u 2 2) + T (u ) T (u ) = e = = (u 2 + u u 2 ). ( ) ( ) ( ) ( ) 2 So T (e ) = (T (u 2 ) + T (u ) T (u 2 ) = + =. ( ) 2 2 Hence the standard matrix of T is given by ( ) 2 By solving x y = ( ), we get the general solution x = 2t, y = t, z = z So Ker(T ) = {(2t, t, ) t R}.

11 PAGE Question 5 (b) [6 marks] Let T : R n R n be a linear operator such that its standard matrix is diagonalisable. Prove that R(T ) = R(T T ) and Ker(T ) = Ker(T T ). Let A be the standard matrix for T. Then A has n linearly independent eigenvectors, say v,..., v n, associated to eigenvalues λ,..., λ n, respectively. Suppose that λ = = λ k =, and λ i if i > k. For each i, Av i = λv i, and thus A 2 v i = λ 2 v i ; so v,..., v n are the eigenvectors of A 2 associated to eigenvalues λ 2,..., λ 2 n. Note that λ 2 i = if i k and λ 2 i if i > k. Then Ker(T ) = nullspace of A = span{v,..., v k } = nullspace of A 2 = Ker(T T ). R(T T ) = column space of A 2 column space of A = R(T ), and dim R(T T ) = n dim Ker(T ) = n dim Ker(T T ) = dim R(T ). We conclude that R(T T ) = R(T ). Continue on page 4 and 5 if you need more writing space. 2

12 PAGE 2 Question 6 [ marks] (a) [4 marks] The nullspace of a 4 matrix A is given by span,. Determine whether each of the following is true or false: (i) The first two columns of A are linearly independent. (ii) The second and fourth columns of A are identical., being in the nullspace of A means the homogeneous system Ax = has a general solutions x = y = t z = s + t w = s which gives two equations x = and y z + w =. Since the nullity of A is 2, by dimension theorem, the rank of A is also 2. Hence the rref R of A is given by R =. We see that the first two columns of R are linearly independent. So the corresponding columns in A must also be linearly independent. Hence (i) is true. We also observe that the second and fourth columns of R are identical. So the corresponding columns in A must also be identical. Hence (ii) is true.

13 PAGE Question 6 (b) [6 marks] Let V = span{v, v 2, v, v 4 } be a vector space such that v i are unit vectors for all i and v i v j < if i j. (i) Show that no two vectors among {v, v 2, v, v 4 } are linearly dependent. (ii) Prove that dim V. Clearly v i. (i) Assume that v, v 2 are linearly dependent. Let v = cv 2. Then > v v 2 = c(v v ) = c. We would have > v v = c(v 2 v ) >, which is a contradiction. Hence, v and v 2 are linearly independent. Similarly for any other two vectors. (ii) Assume that v, v 2, v are linearly dependent. Then any vector in v, v 2, v is a linear combination of the other two. Write v = c v + c 2 v 2. > v v = c + c 2 (v v 2 ) and > v 2 v = c (v v 2 ) + c 2. If c >, then c 2 (v v 2 ) < c <, and thus c 2 >. By Cauhy-Schwarz inequality, < v v 2 v v 2 =. We would have c < c 2 ( v v 2 ) c 2 and c 2 < c ( v v 2 ) c which is a contradiction (based on our assumption that c > ). Therefore, c < and hence c 2 <. However, this would imply that v v 4 = c (v v 4 ) + c 2 (v 2 v 4 ) >, a contradiction. Hence, v, v 2 and v are linearly independent. So dim V. 4

14 PAGE 4 (Additional working spaces for ALL questions - indicate your question numbers clearly.) 5

15 PAGE 5 (Additional working spaces for ALL questions - indicate your question numbers clearly.) [END OF PAPER]