Author |
Topic: linear algebra (Read 783 times) |
|
MonicaMath
Newbie
Gender: 
Posts: 43
|
 |
linear algebra
« on: Oct 10th, 2009, 3:58pm » |
 Quote  Modify
|
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 }
?
|
|
 IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender: 
Posts: 1948
|
|
Re: linear algebra
« Reply #1 on: Oct 10th, 2009, 6:11pm » |
Quote Modify
|
Let K be the relevant field, and consider the K-algebra map
 : K[x]  V
determined by (x) = A. What are the image and kernel of ? And what is the dimension of K[x]/(f), for a polynomial f?
|
|
IP Logged |
|
|
|
MonicaMath
Newbie
Gender:
Posts: 43
|
|
Re: linear algebra
« Reply #2 on: Oct 11th, 2009, 11:23am » |
Quote Modify
|
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
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: linear algebra
« Reply #3 on: Oct 11th, 2009, 12:04pm » |
Quote Modify
|
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:
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print