Linear algebra find the minimal polynomial

catsarebad
Messages
69
Reaction score
0

Homework Statement


M: V -> V linear operator st M^2 + 1_v = 0
find the POSSIBILITIES for min. pol. of M^3+2M^2+M+3I_v


Homework Equations





The Attempt at a Solution



using M^2 = -1_v,
i rewrote the operator(?) as
M^3 + M + I_v

i don't know what to do. i guessed min poly to be like x^3-x-1
but what would be other possibilities. i seriously doubt the one i guessed is even correct.

i know that, if min poly is u(x) then u(M) = 0.
 
Physics news on Phys.org
catsarebad said:

Homework Statement


M: V -> V linear operator st M^2 + 1_v = 0
find the POSSIBILITIES for min. pol. of M^3+2M^2+M+3I_v


Homework Equations





The Attempt at a Solution



using M^2 = -1_v,
i rewrote the operator(?) as
M^3 + M + I_v

i don't know what to do. i guessed min poly to be like x^3-x-1
but what would be other possibilities. i seriously doubt the one i guessed is even correct.

i know that, if min poly is u(x) then u(M) = 0.

M^3=(-M), isn't it?
 
got it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top