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Let's now find the corresponding eigenvectors: Plugging in a₁: Plugging in a2: 0 1 (9) 0 - 0 - λ 1 0 α β 0 2 λβ = 2λα = β = 2a λα = 2λβ = a = 2β 2λγ = y = 1 2λγ = λβ = λα = β = a λα = λβ = a =β 2λγ = λγ =1=0, = |αι) = = 102) λβ = αα λα = αβ 2λγ = = αγ -= ( (0)

Let's now find the corresponding eigenvectors: Plugging in a₁: Plugging in a2: 0 1 (9) 0 - 0 - λ 1 0 α β 0 2 λβ = 2λα = β = 2a λα = 2λβ = a = 2β 2λγ = y = 1 2λγ = λβ = λα = β = a λα = λβ = a =β 2λγ = λγ =1=0, = |αι) = = 102) λβ = αα λα = αβ 2λγ = = αγ -= ( (0)

Advanced Engineering Mathematics
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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I'm having a hard time understanding how they're getting these vectors from the matrix. I follow up to plugging in the eigenvalues and solving the variables, then I get lost for like the first one beta = alpha = 0.

When I solve and plug in beta = 2*alpha = 2 * (2*beta) -> beta = 4beta, so is it zero because thats the only number that makes this equality?

So like it's not so much algebra after solving as looking at the equality and making sense of it?

Let's now find the corresponding eigenvectors: Plugging in a₁: Plugging in ag: 6 : 90--0--- λ 1 α 0 0 2 Ο - λβ = 2λα → β = 2a λα = 2λβ → a = 2β } → |ai) 2λγ = 2λγ = y=1 λβ = λα = β = a - λα λβ → a = β 2λγ = λγ = y = 0, = → 102) = λβ = αα λα αβ αγ = -) (3) 12
Transcribed Image Text:Let's now find the corresponding eigenvectors: Plugging in a₁: Plugging in ag: 6 : 90--0--- λ 1 α 0 0 2 Ο - λβ = 2λα → β = 2a λα = 2λβ → a = 2β } → |ai) 2λγ = 2λγ = y=1 λβ = λα = β = a - λα λβ → a = β 2λγ = λγ = y = 0, = → 102) = λβ = αα λα αβ αγ = -) (3) 12
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