B a qq t for some matrix q with orthonormal columns u

This preview shows page 27 - 29 out of 56 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Nature of Mathematics
The document you are viewing contains questions related to this textbook.
Chapter 4 / Exercise 51
Nature of Mathematics
Smith
Expert Verified
b. A = QQ T , for some matrix Q with orthonormal columns u 1 , . . . , u m . Note that Q T Q = I m , since the ij th entry of Q T Q is u i · u j . Then A 2 = QQ T QQ T = Q ( Q T Q ) Q T = QI m Q T = QQ T = A . 43. Examine how A acts on u , and on a vector v orthogonal to u : Au = (2 uu T - I 3 ) u = 2 uu T u - u = u, since u T u = u · u = u 2 = 1 . Av = (2 uu T - I 3 ) v = 2 uu T v - v = - v, since u T v = u · v = 0 . Since A leaves the vectors in L = span( u ) unchanged and reverses the vectors in V = L , it represents the reflection about L . Note that B = - A , so that B reverses the vectors in L and leaves the vectors in V unchanged; that is, B represents the reflection about V . 44. Note that A T is an m × n matrix. By Facts 3.3.7 and 5.3.9c we have dim(ker( A T )) = n - rank( A T ) = n - rank( A ). By Fact 3.3.6, dim(im( A )) = rank( A ), so that dim(im( A )) + dim(ker( A T )) = n . 45. Note that A T is an m × n matrix. By Facts 3.3.7 and 5.3.9c, we have dim(ker( A )) = m - rank( A ) and dim(ker( A T )) = n - rank( A T ) = n - rank( A ), so that dim(ker( A )) = dim(ker( A T )) if (and only if) A is a square matrix. 260
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Nature of Mathematics
The document you are viewing contains questions related to this textbook.
Chapter 4 / Exercise 51
Nature of Mathematics
Smith
Expert Verified
ISM: Linear Algebra Section 5.3 46. By Fact 5.2.2, the columns u 1 , . . . , u m of Q are orthonormal. Therefore, Q T Q = I m , since the ij th entry of Q T Q is u i · u j . If we multiply the equation M = QR by Q T from the left then Q T M = Q T QR = R , as claimed. 47. By Fact 5.2.2, the columns u 1 , . . . , u m of Q are orthonormal. Therefore, Q T Q = I m , since the ij th entry of Q T Q is u i · u j . By Fact 5.3.9a, we now have A T A = ( QR ) T QR = R T Q T QR = R T R . 48. As suggested, we consider the QR factorization A T = PR of A T , where P is orthogonal and R is upper triangular with positive diagonal entries. By Fact 5.3.9a, A = ( PR ) T = R T P T . Note that L = R T is lower triangular and Q = P T is orthogonal. 49. Yes! By Exercise 5.2.45, we can write A T = PL , where P is orthogonal and L is lower triangular. By Fact 5.3.9a, A = ( PL ) T = L T P T . Note that R = L T is upper triangular, and Q = P T is orthogonal (by Exercise 11). 50. a. If an n × n matrix A is orthogonal and upper triangular, then A - 1 is both lower triangular (since A - 1 = A T ) and upper triangular (being the inverse of an upper triangular matrix; compare with Exercise 2.3.35c). Therefore, A - 1 = A T is a diagonal matrix, and so is A itself. Since A is orthogonal with positive diagonal entries, all the diagonal entries must be 1, so that A = I n . b. Using the terminology suggested in the hint, we observe that 51. a. Using the terminology suggested in the hint, we observe that I m = Q T 1 Q 1 = ( Q 2 S ) T Q 2 S = S T Q T 2 Q 2 S = S T S , so that S is orthogonal, by Fact 5.3.7. b. Using the terminology suggested in the hint, we observe that R 2 R - 1 1 is both orthogonal (let S = R 2 R - 1 1 in part a) and upper triangular, with positive diagonal entries. By Exercise 50a, we have R 2 R - 1 1 = I m , so that R 1 = R 2 . Then Q 1 = Q 2 R 2 R - 1 1 = Q 2 , as claimed.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture