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beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.1Let's call our matrix A. Take the matrix \[A  \lambda I\] I.e, the matrix A with lambda subtracted from the diagonal entries. Solve the equation \[det (A  \lambda I) = 0\] This should be a polinomial equation on lambda. The roots of the equation are the eigenvalues of A.

RyanL.
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah the characteristic equation I saw that on my book but in order to set this up you need to have the matrix set up as up triangle matrix but I also read on my book that the values on the main diagonal for a triangle matrix are the eigenvalues so why bother finding the determinant if we can simply row reduce to get a triangle matrix.

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.1"Yeah the characteristic equation I saw that on my book but in order to set this up you need to have the matrix set up as up triangle matrix" I don't think you do, it works with any matrix. Now if n is big you might not want to calculate the determinant and you might want to use a specialized algorithm. You'll need to ask someone else about those :) But this is fine for n=2 or n=3.

DanielxAK
 2 years ago
Best ResponseYou've already chosen the best response.0You've both hinted at the underlying concept. It's called Eigenvalue Decomposition. Wikipedia does a decent job explaining it: http://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix However, this isn't very easy to put into practice. Once n gets sufficiently large (doesn't take long), you'll need to implement computational methods that will give you an approximate answer (or converges to the true eigenvalues) to solve anything in an efficient manner.
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