Key to MATLAB Exercise 4 School of Mathematical Sciences Xiamen Univer http:/edjpkc.xmu.ed 00 Jordan 400 A is similar to B because JordanA is equal to JordanB >>VA, DA=eig(A) 00.0000 0 0 0000-1.0000 0 DA= 200 0 020 >>VB, DBFeig(B) VB 00.0000 0 00.3162 1.00001.0000 200 020 >> rank( VA)rank(VB) C, 2, 4 are the eigenvalues of A and B, and the rank of va equals that of VB, which means that a has the same jordan normal form as b. therefore a is similar to b >> A=rand(4 4) P.poly(A) 1.0000-1.54300.0898-0.03100.1878 It means that the characteristic polynomial is x-1.543x+0.0898x-003 1x+0.1878 14313 -0.2427+0.40lll -0.2427-040llil The eigenvalues (or characteristic roots ) of A are 1. 4313, 0.5970.-02427+0.401li and Key to Ex4-2Key to MATLAB Exercise 4  School of Mathematical Sciences Xiamen University  http://gdjpkc.xmu.edu.cn  Key to Ex4­2  1  0  0  3  ­6  ­3  JordanB =  4  0  0  0  2  1  0  0  2  A is similar to B because JordanA is equal to JordanB.  Or >> [VA, DA]=eig(A) VA =  0  0.0000  0  0  0  1.0000  1.0000  ­1.0000  0  DA =  2  0  0  0  2  0  0  0  4  >> [VB, DB]=eig(B) VB =  0  0.0000 0  0  0 0.3162  1.0000  1.0000 0.9487  DB =  2  0  0  0  2  0  0  0  4  >> rank(VA)==rank(VB) ans =  1  2, 2, 4 are the eigenvalues of A and B, and the rank of VA equals that of VB, which means that  A has the same Jordan normal form as B. Therefore A is similar to B.  3． >> A=rand(4);  >> P=poly(A) P =  1.0000  ­1.5430  0.0898  ­0.0310  0.1878  It means that the characteristic polynomial is 4 3 2 x -1.543x + 0.0898x - 0.031x + 0.1878 >> roots(P) ans =  1.4313  0.5970  ­0.2427 + 0.4011i  ­0.2427 ­ 0.4011i  The eigenvalues  (or characteristic roots ) of A are 1.4313,  0.5970.  ­0.2427+0.4011i  and