Re: Characteristic Polynomials and Eigenvalues

*To*: mathgroup at smc.vnet.net*Subject*: [mg19381] Re: [mg19364] Characteristic Polynomials and Eigenvalues*From*: "Mark E. Harder" <harderm at ucs.orst.edu>*Date*: Mon, 23 Aug 1999 13:57:04 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Manuel, If a matrix, A, has eigenvectors, x, and eigenvalues,L (scalars), corresponding to those eigenvectors, then, by definition, A x= Lx. So, then A x -Lx= (A -L I) x=0, where I = the Identity matrix. The theory of linear equations, part of Linear Algebra, then requires that Det[ A -L I ]=0 (Det is a Mathematica function for the determinant of a matrix), and this is where the characteristic polynomial comes in: the Determinant on the lhs of this equation is the characteristic polynomial, and so the equation says that the roots of this polynomial are the eigenvalues for A. (Use Mathematica to show this for your matrix) It really has nothing to do with straight lines or curves; this is algebra, not geometry (although there are geometric applications). To understand the theory of linear equations and the finer points of eigensystem theory, you really do have to understand some linear algebra, and I think you are going to have to read some book on the subject. There is more than one book that takes you through linear algebra with the help of Mathematica. See the books section of the Wolfram website. Also, other books not involving Mathematica per se are "Applied Linear Algebra", by Noble & Daniel, which is geared to applications, of course. And there is always the Scaum's Outline for linear algebra, with lots of worked & solved problems. Hope this helps. -mark -----Original Message----- From: MAvalosJr at aol.com <MAvalosJr at aol.com> To: mathgroup at smc.vnet.net Subject: [mg19381] [mg19364] Characteristic Polynomials and Eigenvalues >Gentlemen: > >I have been studying linear algebra and with the aid of several programs and >add- ons to Mathematica the task has been a piece of cake. However, the time >comes when suddenly "understanding" leers its ugly head. >Given the vectors {4,-6}, {3, -7}, the characteristic polynomial is x^2 + 3 x >-10. The eigenvalues are (-5,2), the eigenvectors are (2,3) and (3,1). My >question: >What does the characteristic polynomial (since it discribes a curve) have to >do with the vectors (which are straight lines)? Or for that matter, the >eigenvalues and eigenvectors -derived from the matrix or the polynomial have >to do with the vectors? >I plotted the polynomial but can't figure out what it has to do with the >vectors. > >Thanks for whatever >Manuel >