Eigenvector mystery

[JurgenRo]

(10-20-2020 12:08 AM)Valentin Albillo Wrote:  

(10-19-2020 09:42 PM)Thomas Okken Wrote:  Eigenvalues are values λ such that Av = λv, so you find them by solving Av − λv = 0, or equivalently, (A − λI)v = 0, where I is the identity matrix, or equivalently, |A − λI| = 0. That last form is also known as the characteristic equation of A, and, being a polynomial of the same degree as the dimension of A, you can find its solutions, and thus the eigenvalues of A, using a polynomial root finder.

Thomas is fully right.

You can find a program to compute the coefficients of the Characteristic Polynomial (its roots are the eigenvalues) for real or complex matrices in my article:

      HP Article VA047 - Boldly Going - Eigenvalues and Friends

Also, there are many solved examples in the article, to make the matter crystal-clear.

V.

Well, a few things should still be mentioned here:
1. for the characteristic polynomial p(x)=|A-xI|, the term |A-xI| denotes the determinant of (A-xI).
2. the algebraic multiplicity of an eigenvalue (i.e. the multiplicity as zero of p) is always >= geometric multiplicity (i.e. the dimension of the eigenspace belonging to the eigenvalue).
3. I am not sure if "Dimension of a Matrix" is a valid definition. It's simply the number of rows (or columns).