Solucionario-de-algebra-lineal-Kolman-octava-edicion.pdf

B show that the sum and difference of two lower

Info icon This preview shows pages 13–16. Sign up to view the full content.

View Full Document Right Arrow Icon
(b) Show that the sum and difference of two lower triangular matrices is lower triangular. (c) Show that if a matrix is upper and lower triangular, then it is a diagonal matrix. Solution. (a) As A above, let B = [ b ij ] be also an upper triangular matrix, S = A + B = [ s ij ] be the sum and D = A - B = [ d ij ] be the difference of these matrices. Then, for every i > j we have s ij = a ij + b ij = 0 + 0 = 0 respectively, d ij = a ij - b ij = 0 - 0 = 0. Thus the sum and difference of two upper triangular matrices is upper triangular. Example. 1 2 3 0 1 2 0 0 1 + 3 2 1 0 3 2 0 0 3 = 4 4 4 0 4 4 0 0 4 , sum of two upper triangular matrices, which is also upper triangular; 1 0 0 2 1 0 3 2 1 - 1 0 0 1 1 0 1 1 1 = 0 0 0 1 0 0 2 1 0 , difference of two lower trian- gular matrices, which is also lower triangular. (b) Similar. (c) If a matrix is upper and lower triangular, then the entries above the main diagonal and the entries below the main diagonal are zero. Hence all the entries off the main diagonal are zero and the matrix is diagonal. T.6. (a) Show that if A is an upper triangular matrix, then A T is lower triangular. (b) Show that if A is an lower triangular matrix, then A T is upper triangular. Solution. (a) By the definition of the transpose if A T = [ a T ij ], and A is an upper triangular matrix, a ij = 0 for every i > j and so a T ji = a ij = 0.
Image of page 13

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
13 Hence A T is lower triangular. (b) Similar. Page 37-38. T.4. Show that the product of two diagonal matrices is a diagonal matrix. Solution. Just verify that a 11 0 ... 0 0 a 22 ... 0 ... ... ... ... 0 0 ... a nn b 11 0 ... 0 0 b 22 ... 0 ... ... ... ... 0 0 ... b nn = a 11 b 11 0 ... 0 0 a 22 b 22 ... 0 ... ... ... ... 0 0 ... a nn b nn . T.5. Show that the product of two scalar matrices is a scalar matrix. Solution. Just verify that a 0 ... 0 0 a ... 0 ... ... ... ... 0 0 ... a b 0 ... 0 0 b ... 0 ... ... ... ... 0 0 ... b = ab 0 ... 0 0 ab ... 0 ... ... ... ... 0 0 ... ab . Short solution. Notice that any scalar matrix has the form a.I n = a 0 ... 0 0 a ... 0 ... ... ... ... 0 0 ... a . Then, obviously ( a.I n )( b.I n ) = ( ab ) .I n shows that prod- ucts of scalar matrices are scalar matrices. T.6. (a) Show that the product of two upper triangular matrices is an upper triangular matrix.
Image of page 14
14 CHAPTER 1. MATRICES (b) Show that the product of two lower triangular matrices is a lower triangular matrix. Sketched solution. (a) A direct computation shows that the product of two upper triangular matrices a 11 a 12 ... a 1 n 0 a 22 ... a 2 n ... ... ... ... 0 0 ... a nn b 11 b 12 ... b 1 n 0 b 22 ... b 2 n ... ... ... ... 0 0 ... b nn is upper triangular too. Indeed, this is a 11 b 11 a 11 b 11 + a 12 b 22 ... a 11 b 1 n + a 12 b 2 n + ... + a 1 n b nn 0 a 22 b 22 ... a 22 b 2 n + a 23 b 3 n + ... + a 2 n b nn ... ... ... ... 0 0 ... a nn b nn , an upper triangular matrix. Complete solution. Let P = [ p ij ] = AB be the product of two upper triangular matrices. If i > j then p ij = n k =1 a ik b kj = i - 1 k =1 a ik b kj + n k = i a ik b kj = ( a i 1 b 1 j + ... + a i,i - 1 b i - 1 ,j ) + ( a ii b ij + ... + a in b nj ). Since both matrices A and B are upper triangular, in the first sum the a ’s are zero and in the second sum the b ’s are zero. Hence p ij = 0 and P is upper triangular too.
Image of page 15

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 16
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern