anonymous
  • anonymous
Can someone help me with eigenvalues and eigenvectors?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
I need help with problem number 6
anonymous
  • anonymous
Use: \[det(A-\lambda I) = 0\] to find the eigenvalues. For two by two matrices I think you will get two values of lambra (as a quadratic in lambda will come out).

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anonymous
  • anonymous
I basically need help with part C
anonymous
  • anonymous
Lambda=0 and 4
anonymous
  • anonymous
Is that correct?
anonymous
  • anonymous
You have to find a vector X such that \[ AX=\lambda X \]
TuringTest
  • TuringTest
\[\det(A-I\lambda)=\det\left[\begin{matrix}2-\lambda & 2 \\ 2 & 2-\lambda\end{matrix}\right]=4-2\lambda+\lambda^2-4\]sorry gotta eat!
anonymous
  • anonymous
LOL Go Eat:D
anonymous
  • anonymous
For the corresponding eigenvectors, I think it's a case of finding X in \[A-\lambda I = X\] when you know the two values of lambda...
anonymous
  • anonymous
Turing its the opposite
TuringTest
  • TuringTest
you have I*Iambda-A I'm guessing, it's the same it is the same
anonymous
  • anonymous
oh sorry \[(A-\lambda I)X = 0\] (the ero matrix)
anonymous
  • anonymous
zero*
anonymous
  • anonymous
ohhh ya that is the one i have in my textbook
anonymous
  • anonymous
Check out this example :http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Worked_example
anonymous
  • anonymous
Thanks I figured it out :D
mathmate
  • mathmate
For \( \lambda=0 \) reduced matrix is \[\left[\begin{matrix}1 & 1 \\ 0 & 0\end{matrix}\right]\]which gives an eigenvector of (1,-1). For \( \lambda=4 \), the reduced matrix is\[\left[\begin{matrix}-1 & 1 \\ 0 & 0\end{matrix}\right]\]whose eigenvector is (1,1).
anonymous
  • anonymous
Thanks Guys :D I really appreciate ur help
mathmate
  • mathmate
You're welcome! :)

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