Linear Algebra with Applications (3rd Edition)

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In conclusion: The kernel of L consists of all symmetric matrices, and the image consists of all skew-symmetric matrices. 61. Note that the first three matrices of the given basis B are symmetric, so that L ( A ) = A A T = 0, and the coordinate vector [ L ( A )] B is 0 for all three of them. The last matrix of the basis is skew-symmetric, so that L ( A ) = 2 A , and [ L ( A )] B = 2 e 4 . Using Fact 4.3.2, we find that the B -matrix of L is 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 . 63. By Exercise 2.4.62b, the given LDU factorization of A is unique. By Fact 5.3.9a, A = A T = ( LDU ) T = U T D T L T = U T DL T is another way to write the LDU factorization of A (since U T is lower triangular and L T is upper triangular). By the uniqueness of the LDU factorization, we have U = L T (and L = U T ), as claimed. 65. Write 10 A = a b c d ; it is required that a, b, c and d be integers. Now A = a 10 b 10 c 10 d 10 must be an orthogonal matrix, implying that ( a 10 ) 2 +( c 10 ) 2 = 1, or a 2 + c 2 = 100. Checking 135
Chapter 5 SSM: Linear Algebra the squares of all integers from 1 to 9, we see that there are only two ways to write 100 as a sum of two positive perfect squares: 100 = 36+64 = 64+36. Since a and c are required to be positive, we have either a = 6 and c = 8 or a = 8 and c = 6. In each case we have two options for the second column of A , namely, the two unit vectors perpendicular to the first column vector. Thus we end up with four solutions: A = . 6 . 8 . 8 . 6 , . 6 . 8 . 8 . 6 , . 8 . 6 . 6 . 8 or . 8 . 6 . 6 . 8 . 67. a. We need to show that A T Ac = A T x , or, equivalently, that A T ( x Ac ) = 0. But A T ( x Ac ) = A T ( x c 1 v 1 − · · · − c m v m ) is the vector whose i th component is ( v i ) T ( x c 1 v 1 − · · · − c m v m ) = v i · ( x c 1 v 1 − · · · − c m v m ), which we know to be zero. b. The system A T Ac = A T x has a unique solution c for a given x , since c is the coordinate vector of proj V x with respect to the basis v 1 , . . . , v m . Thus the coefficient matrix A T A must be invertible, so that we can solve for c and write c = ( A T A ) 1 A T x . Then proj V x = c 1 v 1 + · · · + c m v m = Ac = A ( A T A ) 1 A T x . 5.4 1. A basis of ker( A T ) is 3 2 . (See Figure 5.8.) 3. We will first show that the vectors v 1 , . . . , v p , w 1 , . . . , w q span R n . Any vector v in R n can be written as v = v + v , where v is in V and v is in V (by definition of a projection, Fact 5.1.4). Now v is a linear combination of v 1 , . . . , v p , and v is a linear combination of w 1 , . . . , w q , showing that the vectors v 1 , . . . , v p , w 1 , . . . , w q span R n . Note that p + q = n , by Fact 5.1.8c; therefore, the vectors v 1 , . . . , v p , w 1 , . . . , w q form a basis of R n , by Fact 3.3.4d. 5. V = ker( A ), where A = 1 1 1 1 1 2 5 4 . Then V = (ker A ) = im( A T ), by Exercise 4. 136
SSM: Linear Algebra Section 5.4 Figure 5.8: for Problem 5.4.1 . The two columns of A T form a basis of V : 1 1 1 1 , 1 2 5 4 7. im( A ) and ker( A ) are orthogonal complements by Fact 5.4.1: (im A ) = ker( A T ) = ker( A ) 9. See Figure 5.9. x 0 is the shortest of all the vectors in S . (See Figure 5.9.) 11. a. Note that L + ( y ) = A T ( AA T ) 1 y ; indeed, this vector is in im( A T ) = (ker A ) , and it is a solution of L ( x ) = Ax = y .

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