Question

NEED HELP !! LINEAR ALGEBRA

Problem 5.3.11 142 Diagonalize the matrix A3 4 0 For this problem, we will go through the steps of computing the characteristic polynomial (by definition the characteristic polynonial is defined by det(4 followed by computing the eigenvectors. From there the diagonalization will be computed. XI)), port synpy as sp rl1,4,-2 Input the raws #Enter space or camias separated integers for your problem r2: 3,40 r1, ..1, 4, -2· e@peran {type: "string. r2-3, 4, e aparan Itype:"string" r3. "-3, 1, 3. #eparan {type :"string") 31 "3,1,3 rs- [r1, r2, 3 results matrix_int_syp(rs) p.pprint(results) 1 4 -2 Problem 4.7.19 Let P 3 -5 0 and vi 5 and g a. Find a basis u,2, u for R3 such that P is the change-of-coordinates matrix from u, u2, us) to fvi,vz,v,] b. Find a basis u1,2,for Rsuch that P is the change-of-coordinates matrix from vI, V2, U31to w,W2,wsh Recall from the definition of change-of-coordinate matrix, that defining P to be the change of coordinate vectors from U to V means that the column vectors of P are the coordinates of U when Vis the set of basis vectors. That said, if we have the coordinates in V (as defined by P, then l น1 иг из-M ข2 tly I * P for the first question. For the second question, we are claiming that [vi 23ww2 P, so we use the inverse of P import sympy as sp rl: 1,2,-1 #input the rows #Enter space or comas separated integers for your problem 21 3,-5,0 r1: .1, 2, -1. #eparan {type : "string") r2-.-3, -5, e' #@param (type : "string") r3.4, 6, 1. ส@paran {type. "string" } 3: 4, 6,1 rs [r1, r2, r3] results- matrix int_syp(rs) sp.pprint (results) 3 -5 6 Problem 5.3.18 7 -164 Diagonalize the matrix A 6 13 2 12 16 1 This will be worked with the diagonalize function. nport synpy as sp li -7,-16, 4 tInput the rows #Enter space or comas separated integers for your proble" P21 6, 13,-2 r1--7,-16, 4psram ftype:"string ran (type r3 12, 16, 1' agporan (type: "string 3 12,16,1 rs- [rl, r2, r31 results natrix int syp(rs) sp.pprint(results) tRenenber to execute this cell 7-16 4 6 13-2 12 16 1 mport numpy as np inport synpy as sp #Start of student change area 8-np.array(-, -16, 41) np.array(s, 2-np.anray([12, 16, 1 End ot student change area P, D A.disgon lize) rint( "P" sp.pprint(P) print("D") p.pprint(o) print Verification) sp.pprintAP PD) sp.ppintAP D p.zeros (A.shape[o]))

Answer 1

Solution: A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such thatA- PDP HereA=| 340 Find eigenvalues of the matrixA (-1-A) 4-2 3 (4 - 2) 00 3 13 -) ·(-1-1)((4-1) × (3-A)-0x1)-4(3 x (3-A)-Ox3) + (-2)(3 × 1-(4-1) x 3) = 0 (-1-A)(12-72+2-)-49-3A)-2(-9+3A) = 0 12-562-)-(06-12)--18+)-0 2 3

(a 2)-0 or (i- 3)-0 or (a 5-0 The eigenvalues of the matrix A are given by - 2, 3,5, 1. Eigenvectors forA- -2 -2 2. Eigenvectors for 3 3. Eigenvectors for 5

The eigenvectors compose the columns of matrix P -2 The diagonal matrix D is composed of the eigenvalues 2 0 0 0 0 5 Now find p-1

4 24 35 24 -2 Adj(P) = Adj| 1

-(1 x 11x 1) x1) (-2) x 1 (-2) x 1 (-2) x 1 33 -()

11 7 15 24 3 8 5 11 1 8 24 5 15 5 Now, P Aj(P) 5 11 1 8 24 4 35 24 5 15 5 3 11 6 7 35 35

3 116 7 35 35 PDP-1 Now verify that A 2 0 0 x0 30 0 0 5 X2+1X0+1X 4+0+0 0 +00+0+ -2-0-0 0- +0 0-0-5 2+0 0 0+3+0 0+0+5

4 2 3 5 3 116 4 7 35 35 2 3 5 11 385 9 ㄨㄧㄧㄧㄨㄧ-+-X- 35 8 53 7 35 8 5 3 7 11 98 6 9 8 35 8 5 35 8 5 35 35 12 5 44 315 24 3 5 7 35 5 7355 7 1522 9 45 12 9 15 7 35 57355 7 1522 24 45 12 24 15 7 3557 355 7

-14-2 -14-2 P.D P-13 4 0 3 1 3 -14-2 And A-3 4 0

Part(b):

7 16 4 Here A=| 6 13-2 12 16 1 Find eigenvalues of the matrix A (7-2) 16 4 (13-А) -2 |=0 12 16 (I-i) (7.วั(13-λ) x(1-A)-(-2) × 16) _ 16(6 x (1-A)-(-2) x 12) + 4(6 x 16-(13-A)×12) = 0 . (7-A)((13-141+ λ2)-(-32)) . 16((6-6)-(-24)) + 4(96-(156-12ג0-(( .: (7-A)(45-141+ λ2)-16(30-6) +4(-60+ 120-0 . (315-1431+2u2-λ3)-(480-96) + (-240+480-0 λ3-21A2-1+405-0 Roots can be found using newton raphson method Newton Raphson method for x. 21x2-x+ 405-0 x-20.04159

x3 -21x2-x + 405 Now, using long division-20041590.95841x-20.20797 Newton Raphson method for x2 -095841x-20.20797- X--4.04159 x2 -0.95841x -20.20797 Now, using long division+4 04150x-5 Now, x-5-0 The eigenvalues of the matrix A are given by -20.04159, 4.04159, 5 1 . Eigenvectors for λ 20.04 159 0.92013 Vi=| 0.5 2. Eigenvectors for λ-_ 4.04159 1.0868 V2 = | 0.5 3. Eigenvectors for 5

0.8 30.35 2. The eigenvectors compose the columns of matrix P 0.92013 -1.0868 0.8 ·P=| 0.5 0.5.0.35 1. The diagonal matrix D is composed of the eigenvalues 20.04159 00 D- 04.04159 0 0 3. Now find P- 0.92013 -1.0868 0.8 P0.5 0.5 -0.35 09201 x0.5 -035 1.0868 x 0.5 -0.35 1+0.8 x [0.5 05 0.9201 x (0.5 × 1-(-0.35) x 1) + 1.0868 x (0.5x 1-(-0.35) × 1) + 0.8 x (0.5 x 1-0.5x 1) 0.9201x (0.5 0.35)+1.0868x (0.5 0.35)+0.8 x (0.5-0.5) 0.9201 x (0.85) + 1.0868 x (0.85) + 0.8 x (0) -0.78210.9238 0

1.7059 0.92013 -1.0868 0.8 Adj(P) = Adj 0.5 0.5-0.35 0.5 0.5 0.5 -0.35 0.5 -0.35 0.92013 -1.0868 -1.0868 0.8 0.92013 0.8 1.0868 0.8 0.92013 0.8 0.92013 -1.0868 0.5 0.5 0.35 0.5 0.35 0.5 +(0.5 x 1-(-0.35) x 1) (0.5 x 1 - (- 0.35) x 1) +(0.5 x 1-0.5 x 1) -1.0868) x 1-0.8 x 1) +(0.9201 x 1-0.8x 1) -(0.9201 x 1-(-1.0868) x 1) -((-1.0868) х (-0.35)-0.8 х 0.5) -(0.9201 х (-0.35)-0.8 х 0.5) +(0.9201 х 0.5-(-1.0868) х 0.5) +(0.5 +0.35) (0.5 0.35 +(0.5 - 0.5) = | -(-1.0868-0.8) -(0.9201-0.8) -(0.9201 + 1.0868) +(0.3804 -0.4) -(-0.322 -0.4) +(0.4601 + 0.5434) 0.85 -0.85 0 =| 1.8868 0.1201-2.0069 0.0196 0.722 1.0035

0.85 1.8868 -0.01962 0.85 0.12013 0.72205 0 2.00693 1.00347 Now,PAd(P) 0.85 1.8868 -0.01962 - x0.85 0.12013 0.72205 1.7059 0 2.00693 1.00347 0.49827 1.10605 -0.0115 =1-0.49827 0.07042 0.42327 0 1.17647 0.58824 0.49827 1.10605 -0.0115 .. P-1=1-0.49827 0.07042 0.42327 0 1.17647 0.58824 Now verify that A - PDP-1 0.92013 -1.0868 0.8 20.0415900 0.5 0.5 -0.35 x04.04159 0

.9201 x 20.0416-1.0868 x 0+0.8 x0 0.9201 x 0-1.0868 x-4.0416+0.8x0 0.9201 x 0-1.0868 x0+0.8 x5 0.5 x 20.0416 0.5 x 0-0.35 x0 0.5 x 0+0.5x -4.0416-0.35 x0 x0+0.5x0-0.35 x5 1x0+1x -4.04161x0 1 × 20.0416 + 1 x 0 1 0 18.4409 0+0 0+4.39240 0+0+4 10.0208 + 0 + 0 0-2.0208 + 0 0 + 0-1.75 20.0416+0+0 0-4.0416+0 0+0 5 18.44093 4.3924 4 10.0208 -2.0208 -1.75 20.04159 -4.04159 18.44093 4.39244 0.49827 1.10605 0.0115 (PxD)×(P-1):| ー 10.0208-2.0208-1.75 -0.49827 0.07042 0.42327 20.04159 -4.041595 01.17647 0.58824 18.4409 x 0.4983+4.3924 x - 0.49834x0 18.4409 x 1.1064.3924 x 0.0704+4x - 1.1765 18.4409 x- 0.01154.3924 x 0.42334 x 0.5882 10.0208 x 0.4983-2.0208 х-0.4983-1.75 × 0 10.0208 x 1.106-2.0208 0.0704-1.75 ×-1.1765 10.0208 ×-0.01 15-2.0208 × 0.4233-1.75 x 0.5882 20.0416 x 0.4983. 4.04 16 х -0.4983 + 5 x 0 20.04 16 × 1.106-4.0416 x 0.0704 + 5 x-1.1765 20.0416 x -00115-4.0416 × 0.4233 + 5 x 0.5882 9.1886-2.1886+0 20.3966 0.3093-4.7059 -0.2121 1.85922.3529 .9931 1.0069+0 11.0835 -0.1423 2.0588 -0.1152-0.8553-1.0294 99862 2.0138 +0 22.167-0.2846- 5.8824 0.2305 1.7107 2.9412 7 16 4 ー613-2 12 16 1

7 16 4 P.D.P-1-6 13 -2 12 16 1 7 16 4 And A6 13 -2 12 16 1