[1 0-27 0 3 0 2 0 5 has a single real eigenvalue λ = 3 with algebraic multiplicity three. (a) Find a basis for the associated eigenspace. Basis = {}. (b) Is the matrix A defective? DA. A is defective because the geometric multiplicity of the eigenvalue is less than the algebraic multiplicity The matrix A= B. A is not defective because the eigenvectors are linearly independent OC. A is not defective because the eigenvalue has algebraic multiplicity three OD. A is defective because it has only one eigenvalue

[1 0-27 0 3 0 2 0 5 has a single real eigenvalue λ = 3 with algebraic multiplicity three. (a) Find a basis for the associated eigenspace. Basis = {}. (b) Is the matrix A defective? DA. A is defective because the geometric multiplicity of the eigenvalue is less than the algebraic multiplicity The matrix A= B. A is not defective because the eigenvectors are linearly independent OC. A is not defective because the eigenvalue has algebraic multiplicity three OD. A is defective because it has only one eigenvalue

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.3: Eigenvalues And Eigenvectors Of N X N Matrices

Problem 9EQ

See similar textbooks