Question

0 ſi 1 19. (5 points) Find the eigenvalues and eigenvectors of A= 0 2 2 Lo 03

Answer 1

19 The given matrin is Az UIO 03 0 2 2 xet Then be the eigen value of A. Characteristic equation det (A -> I 3+3) = 0 07 1-) 1 0 0 2-2 2 20 o 3-7 or, 27 (1-1) 2 20 371 2 or, 20 03, 20 08, » [62-») (3-2) –0x2 (1-2) (2-2) (3-2) 21,2,3 Therefore eigen value of a dazu dz=2, 23=3. (9) be the eigen rector Corresponding to the eigen-value C21 (4-31 13x3) X1 = 0 are det X, Then

1 1-21 O 2-21 2 03-)) 3) 0-13 07, 0 0 1 2 2 H 2 Ooo 0 0o, 0 I 4+2ZI 24 D ) 42o, 7120. ** (17) () (0) to aiel neu The eigen vector corresponding is (5). det be the eigen velfor corresponding to 22=2 then (A-2733) X2 20 X22 J 22 10 or, 0 0 0 re IL 82 o 2 n 0 2) - the 2220 2 n 2 J2, Z220.

X2= (0) (n veltos corres- Therefore eigen -ponding to A2=2. is (). Let X₂= be the eigen L3 Z3 veetor Then corresponding to 23=3. (4- 3 13x3) X320 00, -2 lo -12 :) (2)-( =) -283 + 7320 23722320 J3 22x3 2232 22 2 283.ie Ez ²83 !. X3 nz 2 ang 73 23 sez Laz 2 1 Therefore ei gen vector corres- -ponding to a3 a3 is (2) The eigen value of a are dal, 72²2, 73²3 and corresponding eigen vector are (5), (6). (1) respectively

20 since, eigenvalue of A are 4121, 8222, 70323 corresponding eigen rector and are (:) (6) (4) respectively We choose pi 0 1 2 Now AP K I 01 0 22 0 0 3 2 2 1 2 3 2 6 00- 0 3 a o 1 2 0 0 0 3 2 where P.D Da diag {1,2,3}. Therefore AP a PD os, plap PlAP = D. where pa 1 1 1 and 2 Ο Ο 0 20 03 A is diagonali zable matrix.