Assignment 6 Solution

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

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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

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[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 .
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