Question

[ 5 2 31 8. (9 points) Let M = 3 8 3 . ( 2 0 4] -11 | 3 | a. Show that uj = 0 , 12 = -3 , uz = 6 are eigenvectors of M, and 1 1 1 determine the corresponding eigenvalues. b. Using your answer to part (a) what is det(M)? c. Using your answer to part (a), what is the characteristic polynomial of M? d. Using your answer to part (a), is M diagonalizable? Why or why not? L2 [1]

Answer 1

, 8) Ler M=1 15 2 37 3 8 3 and u = 1 Kay Criven three veclorsu, olju, -3 we have to check if they are cigenvectors of M or not and have to find the coresponding eigenvawes (if they are eigenvectors) - We know that if we multiply the matrix with eigenvedon and obtain a multiple of that vector This is based on the defination of an eigenvectors the vecto ne is an eigenvection cone sponding to the eigenvalue VER and mama Aif Avenu . In the given problem, if we can show that Mu, = true; M4 = 1242 and M43 -13 43 Then we can say that u1,ų, and uz are eigenvectors of M corresponding to the eigenvalues X., X2 and its respectively.

Now, lets check ; 5 5 2 31-1 Mul, = | 3 8 30 1 2 0 4] L [putting the values of 'n and up . T.27 - I-5+0+ 31 1 -3+0+ 3 1 - 2 +0+4 ] szü; Since 1,55 1:] = dou, with X=2.. Hence, Muli is a scalan multiple of up. So us is an eigenvectos, corresponding to the eigenvale Aqaun ی § 8 31-3 20 AL { 5-6+6 3-24 + 6 2-0 +8 i = -15 - 5. 1-3 [ 2 ] MN , : = 5 U2 = Az U₂ with 2 = 5. since muz is nothing but a scalan multiple of Uz. So, uz is an eigenvector corresponding to the eigenvalue 12=51 of the matrix. M. .. T5 2 37T 37 And, Mu, = 3 8 3 1 2 0 4 T 15+12+37 | 9 +48 + 3 = L6+0+4J [ 301 60 [ 31 101 = 10 uz

- 13u3 with 13-10 Muz gives us a scalon multiple of Uz. 6o uz is an eigenvector of matrix m corresponding to the eigenvalue [23-101 Thus, ui ,Uz and Uz are eigenvectors of M (Praved) : And the corresponding eigenvalues aire. 2; 5. 'and 10 respectively not using our answer to part (a) , we have to determine der (M) 1. We know that the determinant of a manix Ais equal to the product of its eigenvalues Xis. from part (as we came to know that the matrix M has three eigen values as 11=2, 12 = 5 and 13-10. Thus, det (M) = xilas (product of eigenvalues) = 2*5*10 = 100 . Hencee det (M) = 100) Henee Key we have to determine the characteristic polynomial of M - using our answer to part (a). If we considen an nan marix A. The characteristic polynomial of A denoted by Pa (x) is the polynomial defined by : PA (t) = det (a I - A). Cukere I is the nwn identity - matrix and we also know that it has the eigenvawes as roots. So, if the matrix A has a distinct eigenvalues Xos X2,---, an then the characteristic polynomial of A is given by ; Pa (x) = (x-1))(x-x2)(x-43) ------(x~ An)

Now, since from part (a) , we came to know that the matrix M hous three distinct eigenawen as x1=2, 12=5 and 13 = 10 ,thus the characteristic polynomial of M is given by ; | PM(x) = (x-2)(x-5) (x-10) . [d> we ( we know that an non matrix A is diagonalizable if it has n distinct eigenvalues. from the given problem i we see that M is a 3*3 marrix and from past (as we obtained M has three distinct eigenvalus ie, 2, 5 and 10. Hence M is diagonalizable