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        riddles >> medium >> linear algebra
        
(Message started by: MonicaMath on Oct 10^th, 2009, 3:58pm)

Title: linear algebra
Post by MonicaMath on Oct 10^th, 2009, 3:58pm

How I can prove that :
given an nXn matrix A, then:

degree( minimal polynomial of A) = dim(V) where

V = {p(A) ; p(x) a polynomial }

?

Title: Re: linear algebra
Post by Eigenray on Oct 10^th, 2009, 6:11pm

Let K be the relevant field, and consider the K-algebra map

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif : K[x] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif V

determined by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(x) = A. What are the image and kernel of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif? And what is the dimension of K[x]/(f), for a polynomial f?

Title: Re: linear algebra
Post by MonicaMath on Oct 11^th, 2009, 11:23am

Thank you for your replay...

actually I need to prove the result using linear algebra and matrices theory, not using Abstract Algebra groups and homomorphisms,,, if possible

Title: Re: linear algebra
Post by Obob on Oct 11^th, 2009, 12:04pm

It could certainly be argued that Eigenray's suggested proof just uses linear algebra and matrix theory, although it is perhaps stated in slightly more high-tech language.

Let f(x) be the minimal polynomial of A, and say it has degree n. Then you can show 1, A, A^2,..., A^(n-1) form a basis for V:

[hide]showing they are independent follows immediately from the definition of the minimal polynomial; on the other hand, to see they span V, if you are given any polynomial p then use the division algorithm for polynomials to write p = qf + r, where r has degree at most n-1. Then p(A) = r(A), so in fact p(A) lies in V.[/hide]