This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 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 ( I A 1 ) = det + 7 5 5 4  2 1 6 6  7 = ( +7) det  2 1 6  7  5 det 4 1 6  7  5 det 4  2 6 6 = ( + 7)[(  2)(  7) 6] 5[ 4(  7) 6] 5[ 24 6(  2)] = ( + 7)( 2 9 + 8) 5( 4 + 22) 5( 6  12) = 3 2 2 5 + 6 . 1 [I wont present the calculation of the characteristic polynomial in such detail in future. Note that tr ( A 1 ) = 2, as it should be.] Its easy to see that 1 is a root of this polynomial and 3 2 2 5 +6 = (  1)( 2  6) = (  1)( + 2)(  3). So we have three distinct eigenvalues 1, 2 and 3. We must have that min A 1 = A 1 = (  1)( + 2)(  3). 1 I A 1 = 8 5 5 4 1 1 6 6 6 , which rowreduces to 1 0 0 1 1 0 0 . A basis for the eigenspace is 1 1 . 2 I A 1 = 5 5 5 4 4 1 6 6 9 , which rowreduces to 1 1 0 0 0 1 0 0 0 . A basis for this eigenspace is  1 1 ....
View
Full
Document
This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.
 Fall '07
 Loveys
 Linear Algebra, Algebra

Click to edit the document details