Question

7. Orthogonally diagonalize the matrices by finding an orthogonal matrix Q and a diagonal matrix D such that QT AQ = D. 1 А 0 -1 0 0 -1 0 1 В = 2 0 0 1 0 1 0 0 0 0 1 0 1 0 0 2

Answer 1

Salution (i) A 0 -1 0 * To find eigen values of A, The charecterstic equation of A is given by IA-211 = 0 1-a O O ka = 0 -4 (1-7)(2-732_o] - { [O-E 1) (-))] =0 2.-233-2-1) = 0 (1-1) [1-23²-1) = Q-_) (1 +82 28-1) = 0 (-a) a 6-2) = 0 2= 1,0,2 * Let be the eigen vector corresponding to the eigen value a=0 then y Z SO [A-01) (I=

1 to 1 y - о 0 1 X = z U =0 OC -z =0 y -36 += If we take Z=1 then x = 2 and y=0 Son is an Nero eigen vector for 250. * het Norg be the eigen vector of a corresponding to the eigen value a= 2, then (A– 11] [2] SO 1-1 -1 ie o Neco =0 b 1-2 0 0 1-1 =0 роз too son ZO=OC ie > -ZCO and -20 CO x g= is an eigen vector for a= 2. O * Eigen vector corresponding to a = 2. [A-2013 0

1-2 0 -1 o 1-2 so D Z 1 1 1 o 86N co » Z = -2 -X-Z 50 y=0 -06-Z = 0 Take x = 1, then z=-1, y=0 . is an eigen vector four a=2. Eigen vectors of A are given by = [:] V = li Vz SOI since LV,V>= [ ol] 0 <V2, Viz? = [o io] 1 L", z>= 1 [ = 1-LDO O Hence reg. in are orthogonal. het @= ez - in one cli) ==(!) Luty with a + [:] - viacos con los 1/52 ez = U 1111 -452 "Q=[ę e ez il. Q is orthogonal matrix. ご and QTAQ = 11 D 2 O 2

(ii) 2 B o ܝܕ 0 2 Charecterstic equation is 2-2 1 1-2 O 0 1-2 D 2-4 L-a 0 0 1-d (2-2) L-a 1 o 1-2 SO 0 2-2 0 O -0 (2-1) [(-a)?[-2]] - (-1) (-1) [- (-1)] 2–13? (2-1)2-(1-1,= 0 L-a) (2-1)-1] =0 (2-22² (4+ 40-1)=0 92 + 3) =0 Q-13? (1-1) 6-3) = 0 (2-1)3 6-3) = 0 2² (2-a) ² G² la = 1, 1, 1,3 het a X2 be the eigen vector corresponding to a=2, then XA 2-1 0 O 1-1 o 2-1 o 8 8 88? =0 o 0 304 0 241

6 6 D M X2 х X4 >) 3G+ oct = 0 ic. xq = -X .: Rank of the coefficient matrix is iie. nullity is 3. Hence there are 3 eigen vectors a=1. (i) take x = 1 = x = -1 and X=0=X3 then - X2 Xz xq. and 2-1, 0-0 (ii) take og zo = Xx=0 then 02 09 = Ouro X4 and 062=0, K3=1 (11) take OG = 0 =) 569 = 0 24 then = OG 063 34 O .:. For i=1, vectors are eigen V = 12 - - L OOO Dz

Let ୪ ୪୪ be eigen vector corresponding to a=3. xal 2-3 0 1 t 1-3 O 35 X2 33 xa -O O O 1-3 0 -- 2-3 0 -2 O 5854 O -2 O O 0 0 = ) -34 + = - 2x2=0 - 2063 = 0 x - 39=0 = %=o=og and Xa=09 ni is an eigen vector for a=3. Jeg oco V 0-00 omto V2 = EL -oop . .. U <UL, V >= [do (100-1] 13 <ur, ung > = [10 100][i]

<u, 19.= [oo 1 0] 1) =1-1-D <u, VG >= (loo Hence us, s, uz, va are orthogonal. 1 e,= - Šż = -1500 L-Y2 13 ir *** | ez = 2-00 vo 11 Vizul 0 1152 eq = wall :@=[e e ę eq 1 R- Oost- goo is the orthogonal matrix h Ja ܝܙ o such that QT B Q = 0 0 o - 0 0 3