The matrix A = 0 -87 -8 2 0 8 has a single real eigenvalue X (a) Find a basis for the associated eigenspace. Го Basis = { }. -4 4 = 4 with algebraic multiplicity three. (b) Is the matrix A defective? DA. A is defective because it has only one eigenvalue foctive because the geometric multiplicity of the eigenvalue is less than the algebraic multiplici
The matrix A = 0 -87 -8 2 0 8 has a single real eigenvalue X (a) Find a basis for the associated eigenspace. Го Basis = { }. -4 4 = 4 with algebraic multiplicity three. (b) Is the matrix A defective? DA. A is defective because it has only one eigenvalue foctive because the geometric multiplicity of the eigenvalue is less than the algebraic multiplici
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Transcribed Image Text:The matrix A=
0 0 -87
-4 4 -8
0
2
8
has a single real eigenvalue A = 4 with algebraic multiplicity three.
(a) Find a basis for the associated eigenspace.
Basis = {
(b) Is the matrix A defective?
DA. A is defective because it has only one eigenvalue
OB. A is defective because the geometric multiplicity of the eigenvalue is less than the algebraic multiplicity
Oc. A is not defective because the eigenvalue has algebraic multiplicity three
OD. A is not defective because the eigenvectors are linearly independent
Expert Solution
Step 1
Introduction:
The matrices that have an eigenvalue decomposition are those that are not faulty. Theorem. If and only if an IRmm decomposition has an eigenvalue, it is not faulty. In light of this, diagonalizable is a different word meaning non-defective.
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