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Eigenvectors

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Answers to Common Questions
･ 1 Learn and understand the definition of an "eigenvector." It is found for an n x n square matrix A and... ･ 2 Find the eigenvalues of the matrix by using the characteristic equation det (A -- LI) = 0. "Det" stands... ...
http://www.ehow.com/how_5206393_calculate-eigenvectors...   See entire page »
That's good. We want a matrix filled with 1s for the eigenvectors corresponding to -1. What you do is then use gaussian elimination to get from this matrix: 1 1 1 1 1 1 1 1 1 to this matrix: 1 1 1 0 0 0 0 0 0 which is just done by subtrac...
http://answers.yahoo.com/question/index?qid=2008111017...   See entire page »
Given a matrix A, an eigenvector of A, say x, is a vector that, when multiplied to A, will give a scalar multiple of x. i.e Ax=[lambda]x where lambda is any number from the set of real numbers, and is called an eigenvalue. (Please note that...
http://wiki.answers.com/Q/What_is_eigen_values_and_eig...   See entire page »
Answers to Other Common Questions
eicantanswer eigenvalues and eigenvectors come together as a pair. If you take a matrix and right-multiply it by an eigenvector, you get an answer that is the same as the eigenvector multiplied by a scalar - that scalar is an eigenvalue. So...
http://answers.yahoo.com/question/index?qid=20090805163...
No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
http://wiki.answers.com/Q/Do_similar_matrices_have_the_...
Mathematically, two different kinds of eigenvectors need to be distinguished: left eigenvectors and right eigenvectors. However, for many problems in physics and engineering, it is sufficient to consider only right eigenvectors. The term "...
http://uk.answers.yahoo.com/question/index?qid=20090211...
No, it's not. Maybe you have the right idea, but what you've written down doesn't make a lot of sense. The nullspace is {x : Ax = 0}. Can you write down what the set of eigenvectors corresponding to zero is?
http://www.physicsforums.com/showthread.php?t=186999
The eigenvalues you've found are correct, although I think you should not round them off. Using the quadratic formula you can easily find the eigenvalues to be 1+sqrt(2) and 1-sqrt(2). I think you can still find the approximate eigenvectors...
http://www.vbforums.com/showthread.php?t=526183
the general case is much more interesting: suppose are eigenvectors with corresponding pairwise distinct eigenvalues then are linearly independent: the proof is by induction over : there's nothing to prove for n = 1. so suppose the claim is...
http://www.mathhelpforum.com/math-help/linear-abstract-...
I've used both the standard netlib lapack and blas, as well as the Intel Math Kernel Library. The Intel one is faster, but might not be free. You may be able to get a free version for educational use, but I don't recall for sure. The netlib...
http://forums.opensuse.org/programming-scripting/423899...