# Eigenvalues & eigenvectors of N x N matrix?

1. Jul 20, 2008

### sapling_pk

How to get eigenvalues & eigenvectors of N x N matrix?
Please can anyone help me out i have searched a lot but not able to find the solution.

Regards

2. Jul 20, 2008

### Hurkyl

Staff Emeritus
Your textbook should present a complete algorithm for computing them; have you looked there? If you've already looked at it, then in what way are you having trouble using it?

3. Jul 20, 2008

### HallsofIvy

My goodness! This is one of the major problems of Linear Algebra and, indeed, of mathematics in general! Surely, as Hurkyl suggests, any textbook on Linear Algebra will devote one or more chapters to this!

This is much too general a question for a forum like this. Can you post specific problems?

4. Jul 20, 2008

### mathwonk

c is an eigenvalue of A iff A-c fails to be invertible iff det(A-c) = 0. so compute det(A-c) considering c as a variable and set this polynomial equal to zero. if c is a root of it, then compute a basis for the kernel of A-c by gaussian elimination.

doing this for all roots c of det(A-c) gives a maximal independent set of eigenvectors, hence basis of them if one exists.

5. Jul 23, 2008

### sapling_pk

well actually i want find eigenvalues of huge matrix i.e 12 x 70000 so hope you have understood my problem.
Regards

6. Jul 23, 2008

### sapling_pk

And not only eigenvalues but also the eigenvectors.Because i am implementing a face recognition algorithm if someone give me any idea with respect to programming that will be appreciated.Thanks

7. Jul 23, 2008

### mathwonk

apparently you knlow more than i do, but here is what my old linear aklgebra book says:

assuming your matrix A is diagonalizable, and the largest eigenvalue is unique and much larger than the other eigenvalues, then for any vector u which has a non zero coefficient with respect to the corresponding "largest" eigenvector, Au has a large component of that eigenvector.

then (Au.u)/(u.u) is an approximation to the dominant eigenvalue.

iterating A makes the dominance more pronounced, so (Au.u)/u.u) will hopefully converge to the dominant eigenvalue if we repeat the calculation with Au in place of u, and continue many times.

these are called rayleigh quotients.

8. Jul 23, 2008

### JPRitchie

9. Jul 23, 2008

### Manchot

Have you thought about using a standard eigensolver package, like LAPACK?