Question

The columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that A = QR. gallon | A nos A vw ok o on (83) O A. R= OB. R=

Answer 1

-2 3 -2 -2 Solution: -2 3 Here A = -2 -2 4 -2 -2 q1 = a1 = -2 -2 4 4i||- y(- 2)² + 5² + ( - 2)² + 4² = /49 = 7 r11 = 91 | |ai || "12 = 91 a, = -2 4 3 92 = az - l12 ' 9 = 42| - V52 + 2² + 0² + o² = v29 = 5.3852 r22 =

0.9285 2 0.3714 1 92 = 92 = | |4: || 5.3852 2 0.9285 0.3714 e- [4, 4:] - 4 ru "12 R = 0 5.3852 o r22 checking Q x R = A? 2 0.9285 -2 3 0.3714 5 7 Qx R = 0 5.3852 -2 -2 -1 -2 3 5 7 and A = -2 -2 4 4 NIN

Find QR Decomposition (Householder Method)... -2 3 -2 -2 Solution: -2 3 Here A= -2 -2 4 4 -2 3 A, = -2 -2 4 4 -2 a1 = -2 4 4|| - V- 2)* + 5² + ( - 2)² + 4² = v49 = 7 -2 v, = a, - sign (a,,1) ||a, | |e, = %3D -2 -2 100 0 4 v v 0 100 [ 55 -2 4 ] = н -1-2. v v1 0 010 70 31 8 -2 35 35 0 0 1 0 8 19 7 35 35 7. 4 -7 -7 -2 3 4 5 7 0 -3 н, А,- х 2 7 35 35 31 8 -2 -2 0 -4 4 8 19 7 7 35 35 Now removing 1st row and 1 st column, we get -3 A, = 2 -4

-3 az = -4 | |4: a, ||- V- 3)2 + 2² + ( - 4)² = vz9 = 5.3852 -3 -8.3852 V2 = aq - sign(a,1) ||a1||e1 = + 5.3852 x 2 2 -4 -4 100 -8.3852 -0.5571 0.3714 -0.7428 v1 v H2I-2. 2 [ -8.3852 2 -4 ] = 0 10 0.3714 0.9114 0.1772 %3D 90.311 v V1 -4 -0.7428 0.1772 0.6457 -0.5571 0.3714 -0.7428 -3 5.3852 H2 A2 = 0.3714 0.9114 0.1772 -0.7428 0.1772 0,6457 -4 Since, H,HA = R -2 3 -7 -7 2 2 0 -0.5571 0.3714 -0.7428 0 5.3852 = R НанА- 0 0.3714 0.9114 0.1772 2 31 -2 -2 35 35 0 -0.7428 0.1772 0.6457 4 4 19 7 35 35 Also A = H,H,R and A - QR, . Q = H,H, 0.9285 -0.1061 0.2122 4 0.3714 0.2653 -0.5306 L. 0 -0.5571 0.3714 -0.7428 Q = H,H2= 0 0.3714 0.9114 0.1772 31 0.9539 0.0923 35 35 0 -0.7428 0.1772 0.6457 4 19 0.0923 0.8154 7. 7 35 35 checking Q x R = A? 0.9285 -0.1061 0.2122 -7 -7 -2 3 0.3714 0.2653 -0.5306 0 5.3852 Q xR = -2 -2 0.9539 0.0923 4 4 4 0.0923 0.8154 L.