Assignment 6 Solution

Assignment 6 Solution - MATH 223 Linear Algebra Winter 2008...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 6 1. Suppose that A is an m × n matrix and B is an n × m matrix. [Hence, both AB and BA are defined and square, but not necessarily of the same size.] Show that tr ( AB ) = tr ( BA ). If C and D are square matrices of the same size which are similar, show that tr ( C ) = tr ( D ). [ tr ( C ) is the trace of C , that is, the sum of the entries on the diagonal.] Solution: If the ( j, k )-entry of A is a j,k and the ( k, j )-entry of B is b k,j , then the ( j, j )-entry of AB is n k =1 a j,k b k,j . The trace of AB is then m j =1 n k =1 a j,k b k,j . Similarly, the ( k, k )-entry of BA is m j =1 b k,j a k,j and the trace of BA is n k =1 m j =1 b k,j a j,k . Multiplication in the field is commutative, so these two sums are the same (in a different order). That is, tr ( AB ) = tr ( BA ). Now if C and D are similar, say D = P - 1 CP . Taking A = CP and B = P - 1 , AB = C and BA = D . By the last paragraph, C and D have the same trace. 2. For each of the following matrices A j ( j = 1 , 2 , 3) over the real numbers, (a) find all the eigenvalues of A j ; (b) find a basis for each eigenspace; (c) find the minimal polynomial min A j and the characteristic polynomial χ A j ; (d) find — if possible — an invertible matrix P j such that P - 1 j A j P j is invertible. If this is not possible, explain why. (e) Find A 11 1 . (2 11 = 2048 and 3 11 = 177147.) A 1 = - 7 - 5 5 4 2 - 1 - 6 - 6 7 , A 2 = 3 - 2 3 10 - 3 5 5 - 4 7 , A 3 = 6 4 - 1 4 6 - 1 - 4 - 4 3 . Solution: We start with A 1 for some reason. det ( λI - A 1 ) = det λ + 7 5 - 5 - 4 λ - 2 1 6 6 λ - 7 = ( λ +7) det λ - 2 1 6 λ - 7 - 5 det - 4 1 6 λ - 7 - 5 det - 4 λ - 2 6 6 = ( λ + 7)[( λ - 2)( λ - 7) - 6] - 5[ - 4( λ - 7) - 6] - 5[ - 24 - 6( λ - 2)] = ( λ + 7)( λ 2 - 9 λ + 8) - 5( - 4 λ + 22) - 5( - 6 λ - 12) = λ 3 - 2 λ 2 - 5 λ + 6 . 1
Image of page 1

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

View Full Document Right Arrow Icon
[I won’t present the calculation of the characteristic polynomial in such detail in future. Note that tr ( A 1 ) = 2, as it should be.] It’s easy to see that 1 is a root of this polynomial and λ 3 - 2 λ 2 - 5 λ + 6 = ( λ - 1)( λ 2 - λ - 6) = ( λ - 1)( λ + 2)( λ - 3). So we have three distinct eigenvalues 1, -2 and 3. We must have that min A 1 = χ A 1 = ( λ - 1)( λ + 2)( λ - 3). 1 I - A 1 = 8 5 - 5 - 4 - 1 1 6 6 - 6 , which row-reduces to 1 0 0 0 1 - 1 0 0 0 . A basis for the eigenspace is 0 1 1 . - 2 I - A 1 = 5 5 - 5 - 4 - 4 1 6 6 - 9 , which row-reduces to 1 1 0 0 0 1 0 0 0 . A basis for this eigenspace is - 1 1 0 . 3 I - A 1 = 10 5 - 5 - 4 1 1 6 6 - 4 , which row-reduces to 1 0 - 1 3 0 1 - 1 3 0 0 0 . A basis for this eigenspace is 1 1 3 . With P 1 = 0 - 1 1 1 1 1 1 0 3 , we have P - 1 1 A 1 P 1 = D = 1 0 0 0 - 2 0 0 0 3 . Now A 1 = P 1 DP - 1 1 , so A 11 1 = P 1 D 11 P - 1 1 = 0 - 1 1 1 1 1 1 0 3 1 0 0 0 - 2048 0 0 0 177147 3 3 - 2 - 2 - 1 1 - 1 - 1 1 = - 181243 - 179195 179195 - 173048 - 175096 175097 - 531438 - 531438 531439 .
Image of page 2
Image of page 3
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