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Eigenvalue and Eigenvector |
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| Sep8-03, 04:49 AM | #1 |
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Eigenvalue and Eigenvector
can any one explain the the real meaning and purpose of eigen vlaue and eigen vectors..
[:)] |
| Sep8-03, 05:59 AM | #2 |
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You know how easy it is to work with diagonal matrices, right?
Consider the fact that (nearly) every square matrix can, after a suitable change of basis, be written as a diagonal matrix whose entries are simply its eigenvalues. So in one sense, using eigenvalues and eigenvectors lets you treat (nearly) any matrix similarly to a diagonal matrix, making the work easier. |
| Sep8-03, 06:48 AM | #3 |
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This works providing the matrix has unique eigenvalues, right? Do we need a full set of linearly independent eigenvectors?
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| Sep8-03, 06:53 AM | #4 |
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Eigenvalue and Eigenvector
Actually, in order for a matrix to be diagonalizable, it is NOT necessary that all the eigenvalues be unique. It IS necessary that all the eigenvectors be independent- that is that there exist a basis for the vector space consisting of eigenvectors.
Essentially, the eigenvectors are those vectors on which the linear tranformation acts like simple scalar multiplication. |
| Sep8-03, 07:49 AM | #5 |
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If you have some square matrix then a non-zero vector x in R^n is an eigenvector of A if Ax is a scalar multiple of x. This scalar is called an eigenvalue of A.
Q) So if you have some vector then scalar multiples of it only 'stretches' or 'compresses' it by a factor of your eigenvalue? Explain. And how can we use determinants in finding eigenvalues of a given matrix? (off-topic) Has this got anything to do with the Kronecker Delta in Tensor Calculus? |
| Sep8-03, 06:26 PM | #6 |
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When calculating the eigenvalues {λn} of a matrix A, you have to solve the equation: det(A-λI)=0. If we rewrite that in terms of matrix elements (IOW, with indices) we can write: det(Aij-λIij), the identity matrix Iij is none other than the Kronecker delta, δij. |
| Sep9-03, 10:23 PM | #7 |
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finding eigen vector for the matrix A, will it give the orthogonal quantity of the matrix..
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| Jan29-05, 08:37 PM | #8 |
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Recognitions:
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an n by n matrix M is diagonalizable if and only if the space R^n has a basis of eigenvectors of M, if and only if the minimal polynomial P of M consists of a product of different linear factors, if and only if the characteristic polynomial Q splits into a product of linear factors, and for each root c of Q, the kernel of M-cId has dimension equal to the power with which the factor (X-c) occurs in Q.
see http://www.math.uga.edu/~roy/ |
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