# 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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