Question

Show that any two eigenvectors of the symmetric matrix corresponding to distinct eigenvalues are orthogonal. -1 0-1 0-1 0 -107 Find the characteristic polynomial of A. far - 41 - Find the eigenvalues of A. (Enter your answers from smallest to largest.) (11, 12, 13) = Find the general form for every eigenvector corresponding to 11. (Uses as your parameter.) X1 = Find the general form for every eigenvector corresponding to 12. (Use t as your parameter.) x2 = (0.t,0) Find the general form for every eigenvector corresponding to iz. (Use u as your parameter.) X3 = Find X1 X2: X1 X2 = 0 Find X1 X3 X1 X3 = 0 Find X2 X2 X2 X3 = 0 Determine whether the eigenvectors corresponding to distinct eigenvalues are orthogonal. (Select all that apply.) x, and x, are orthogonal. x, and xy are orthogonal. *2 and xy are orthogonal,

Answer 1

solution: The given matrix is, -| A -1 o 0 7 polynomial of a "". The characteristic is 1 2 1 - Al Al = 0 => 2+1 o 0 o 2+) o 2-7 0 = (a +1) (*+) (-->) + [-(+)) = -> (2+1) [(a +1) (1 - 1) - 1) = 0 Gati) (=62-3) = 0 The characteristic polynomial of 6+1) (2=67 -1) = 0 => A is

from () We get, 2+1 = 0 or ²-67-8=0 6 + √36+32 a -1 => a 2 3 I r17 The are , eigenvalues of a 3-17, -1, 3 + 117 X ef be. the eigenvector of the eigenvalue 21 = 3-117 Corresponding to Then AXI axi -) (A – 2,1) x1 = 0 117-4 0 — 124 0 /17 - 4 o X2 -1 0 17+4 xa -> (117 - 4) 24 – x3 (5174) X2 -24 + (377 +4) X3 = 0

which together implies, (577-4) (517-4) X2 X3 ro -C., =0 s X3 S ut 4 from (ii) (114 - 4) and from (1), x2 general form of the eigens vector X1 (s, o, s (877-4) (177-4)s The S 0 Now let X2 be the eigenveetor Corsesponding the eigenvalue Az = -1 to A X2 72 X2 = (A - 72 I ) X2 = 0

0 0 0 20-10) o EY -763 O -74 +883 - 0 =0 =) - 74 +8.0 74 o and is arbitrary. Cet H2=t X2 omo [10, +, 0) X₂ is the Now to cet the eigenveetor Corresponding 3+ /17 eigenvalue = 73 Then - 73 X3 =) (A-73 1) X3 =0

h-tis 24 h-tir- X2 :) o - X3 -57+4 => -13 tis o -> (-117-4)* (-57-4) 12 24 + (-17 +4) *3 = 0 which together implies, - 74 + (-NTF +4) 33 = 0 ENT7-4) * v) =0 ut 23 u n from (iv 2) 24 =(-NT2+4) u from (v) , LENT+4) u (-177 +4)6,0 22 O Ex :

N೮ಬ, X1 X2 = (s, 0, 5(7-4) (0,4,0) O +0+0 XI. X2 and, X, X3 tin 11 ( S, 0, -->)(4 3/517 4(+4), 0, u su (-1* +4) + su(173-4 su (-va +4 +VTA -4) 11 = XI. X3 To and X2 X3 (0, trg). (66175+), 0, ) O + 0 + 0 11 = 0 X2 X3 o

we can that say q xi and X2 are orthogonal. orthogonal. and X3 are q x, and X3 are 9 X2 and orthogonal.