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anonymous
 5 years ago
How do you solve for eigenvectors? I get the eigenvalues. Here is the problem.
[2 0 0
1 1 2
1 0 1]
The characteristic polynomial is x^3+2x^2+x2 (obviously is lambda but I am using x for now)
eigen values are x1= 1
x2= 1
x3= 2
How would you do the eigenvectors?
anonymous
 5 years ago
How do you solve for eigenvectors? I get the eigenvalues. Here is the problem. [2 0 0 1 1 2 1 0 1] The characteristic polynomial is x^3+2x^2+x2 (obviously is lambda but I am using x for now) eigen values are x1= 1 x2= 1 x3= 2 How would you do the eigenvectors?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Replace your eigenvalues, one at a time, for lambda in your (AxI) matrix where x is lambda. So you'll have 3 separate matrix equations, (AxI) v = 0, to solve for three different eigenvectors. You're almost there, just that one last step!
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