A = dz dt [333] -3 5 AZ =

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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How do we calculate eigenvectors for repeated eigenvalues and how is it different from how we calculate eigenvectors when we have two roots produced from the characteristic polynomial? For example, in the picture, we have a coefficient matrix for a system of linear differential equations. We calculate that the repeated eigenvalue is 2, and we would solve the equation (A minus lambda) times Z is equal to zero as shown in the other picture. This should give us the vector [1, 1] for our first eigenvector. But how would we use this to solve for the second eigenvector to get two linearly independent solutions?

A
ZP
=
=
AZ
dt
det(A – XI) =
= (x - 2)²
-
1 3
-3 5
2
Transcribed Image Text:A ZP = = AZ dt det(A – XI) = = (x - 2)² - 1 3 -3 5 2
AZ = XZ
(A-X) Z=0
Transcribed Image Text:AZ = XZ (A-X) Z=0
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