Eigenvalue and Eigenvector

shankar

can any one explain the the real meaning and purpose of eigen vlaue and eigen vectors.. Hurkyl

Staff Emeritus
Gold Member
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.

Lonewolf

This works providing the matrix has unique eigenvalues, right? Do we need a full set of linearly independent eigenvectors?

HallsofIvy

Homework Helper
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.

Oxymoron

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?

Tom Mattson

Staff Emeritus
Gold Member
Originally posted by Oxymoron
(off-topic) Has this got anything to do with the Kronecker Delta in Tensor Calculus?
Yes, it does.

When calculating the eigenvalues {&lambda;n} of a matrix A, you have to solve the equation:

det(A-&lambda;I)=0.

If we rewrite that in terms of matrix elements (IOW, with indices) we can write:

det(Aij-&lambda;Iij),

the identity matrix Iij is none other than the Kronecker delta, &delta;ij.

shankar

finding eigen vector for the matrix A, will it give the orthogonal quantity of the matrix..

mathwonk

Homework Helper
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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