Question

Problem 11 Let A= 3 -1 0 0 16 -5 0 0 16 0 -2 15 -3 -15 2 8 a) [3 pts) Compute the characteristic polynomial of A and find its roots. b) (4 pts] For each eigenvalue of A find a basis for the corresponding eigenspace. c) [3 pts] Determine if A is defective. Justify your answer. d) [6 pts) If A is defective, determine the defective eigenvalue or eigenvalues, and find a Jordan chain (or set of Jordan chains) in the corresponding generalized eigenspace that provides a canonical basis for that space.

Answer 1

3 AS 16 16 0 -1 -5 15 0 0 -3 -15 0 2 charateristics botynomial of A =) (A-d Il=0 3-d -1 t6 16 - 5-d 15 0 -3-d -15 O 2 8-d -3-3-3 +70+6=0 -> (+1)(d+1) (ded 5d +6)=0 2 (d+12 (d-2 (d-3)=0 d=-1, -1, d=2, d=3 eigen values - -1,-1, 2, 3 eigen vectors (A-d, I), = 0 O 4 - 16 -4 Vi=0 15 -15 o on solving U- DO

For da=2 (A-da Iva=0 16 16 7 15 02-0 0 -5 2 -15 6 solving 02- 3 For da=3 (A-da I) v 3 = O 16 colo 16 -8 0 IS -15 3=0 -6 o . on sokuing 03 5 lerno ) basis tor eigenspale 13) For Eda Four Edza

© Yes. matrix A is detective Because geometric mak tiblicity of eigen vakue dizal is But algebric muktiblicita 18 6.m. of (di= -1) A.m of (d. = -1) A is not matrix diagonakizable. d-Ear dim (Ed.) = 1 A.M. Of(d, = -1) @ Detective eigenvalue - -1 generalized eigen vectos for eigen value (di=) (A-d, I) W- & 16 =) f 16 o -2 15 -15 2 0 "13] 0 241 : انع solving 263 xu 23 = 0 => 213=0 and 24 =o and 24+4x2 = -1 24 = -1-4X2 Wg ON wa= JORDAN chain &i=1-4