2. The union of the zero vector with the set of all eigenvectors with an eigenva
ID: 3135898 • Letter: 2
Question
2. The union of the zero vector with the set of all eigenvectors with an eigenvalue is called the eigenspace associated with . The dimension of the eigenspace associated with , or equivalently the maximum number of linearly independent eigenvectors associated with , is referred to as the eigenvalue's geometric multiplicity. (a) Let A and B be similar matrices. Prove that the geometric multiplicities of the eigenvalues of A and B are the same. (b) Are the following matrices similar? 1-4 -2 A:=10 1 0 0 4 3 3 0 0 B:= 1011 0 0 1Explanation / Answer
2.(a). If A and B are n×n matrices, these are called similar if there exists an invertible matrix P such that B = P-1AP.This implies that A = PBP-1. Now, if Av = v, then PBP-1 v = v or, BP-1 v = P-1v . Thus, if v is an eigenvector of A, with eigenvalue , then P-1v is an eigenvector of B with the same eigenvalue. This means that every eigenvalue of A is an eigenvalue of B and vice-versa. Hence, A and B have the same eigenvalues. Further, v and P-1v are the same vector, written in different coordinate systems. Therefor, A and B have the same eigenvalues, the same eigenvectors and the same geometric multiplicities.
(b). The characteristic equationofA is det(A-I3) =0 or, 3 -52+7-3 = 0 or, (-1)2(-3)= 0. Thus, the eigenvalues of A are = 3 and 2 = 3 =1. The eigenvectors of A associated with its eigenvalue 1 are solutions to the equation (A-I3) X = 0. To solve this equation, we have to reduce the matrix A-I3 to its REEF which is
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Now, if X = (x,y,z)T, then the equation (A-I3) X = 0 is equivalent to y+z/2 = 0 or, y = -z/2. Then X = (x,-z/2,z)T = x (1,0,0)T+z/2(0,-1,2)T. Hence, the eigenvectors of A associated with its eigenvalue 1 are v2 =(1,0,0)Tand v3 = (0,-1,2)T. Similarly, the eigenvector of A associated with its eigenvalue 3 is (-1,0,1)T. Since A has 3 distinct ( and linearly independent) eigenvectors, it is diagonalizable.
Further, the characteristic polynomial of B is det(B-I3) =0 or, 3 -52+7-3 = 0 or, (-1)2(-3)= 0. Thus, the eigenvalues of A are = 3 and 2 = 3 =1. The eigenvectors of B associated with its eigenvalue 3 and 1 are (0,1,0)T and (1,0,0)T respectively. Since B does not have 3 distinct eigenvectors, it is not diagonalizable.
A and B have the same rank (3), same determinant (3), same trace(5), same characteristic polynomial, but A is diagonalizable while B is not diagonalizable. Hence A and B are not similar.
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