# Minimal polynomial

by vabamyyr
Tags: minimal, polynomial
 P: 66 I have matrix A $$\left(\begin{array}{ccc}6&2&-2\\-2&2&2\\2&2&2\end{array} \right)$$ Its characteristic polynomial is $$p(\lambda)=\lambda^3 - 10\lambda^2 + 32\lambda -32$$ Finding minimal polynomial i get: $$(I\lambda-A)^\vee=\left(\begin{array}{ccc}\lambda-6&-2&2\\2&\lambda-2&-2\\-2&-2&\lambda-2\end{array}\right)^\vee$$ I cant understand why this last result equals with $$\left(\begin{array}{ccc}\lambda^2-4\lambda&-2\lambda+8&2\lambda-8\\2\lambda-8&\lambda^2-8\lambda+16&2\lambda-8\\-2\lambda+8&2\lambda-8&\lambda^2-8\lambda+16\end{array} \right)$$ can someone explain?
 Sci Advisor HW Helper P: 9,398 would you mind retypsetting that so it is more legible?
 Sci Advisor HW Helper P: 9,398 Oh, and if you check that the char poly has 3 distinct roots it will be its minimal poly. other than that, it is of the form (x-a)^2(x-b) a different from b or (x-a)^3, check if those happen and if they do the minimal polys are divisors of these and you can check which, A satisfies.
HW Helper
P: 2,589

## Minimal polynomial

Also, you can observe right away that (0 1 0)T is an eigenvector with eigenvalue 2 (look at the middle column of A). So 2 is a root of your characteristic polynomial, which will make it easier for you to factor.
 P: 66 I guess im on my own again.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,708 Perhaps it would help if you explain what the "v" in $$(I\lambda-A)^\vee$$ means! In any case, it was explained to you that the minimal polynomial- the polynomial of smallest degree that A satisfies- is just the characteristic polynomial with duplicate factors removed. And it was pointed out that $\lambda- 2$ is a factor so it should be simple to determine all the factors. What more do you want?
HW Helper
P: 2,589
 Quote by vabamyyr I guess im on my own again.
No, we essentially answered your question for you. After having those questions answered, you went and edited your post and added some more stuff in (the stuff with the (A - xI)[sup]V[/sub]). If you edit a post, no one will notice any changes have been made unless they come and look at the thread i.e. when you edit it, the thread doesn't appear at the top of the "Linear & Abstract Algebra" forum like it would if you made a new post, and it doesn't appear when people search for "New Posts".

If you have a new question, make a new post or a new thread, don't edit an old post which every one thinks has already been answered. Moreover, it appears you only gave it less than 5 hours before complaining that "you're on your own again." Be more patient.
 P: 66 I have matrix A $$\left(\begin{array}{ccc}6&2&-2\\-2&2&2\\2&2&2\end{array} \right)$$ Its characteristic polynomial is $$p(\lambda)=\lambda^3 - 10\lambda^2 + 32\lambda -32$$ when i factor it i get $$p(\lambda)=(2-\lambda)(\lambda-4)^2$$ So removing the duplicate factor $$(\lambda-4)$$ i get that minimal polynomial is $$(\psi)=(2-\lambda)(\lambda -4)$$ and A really satisfies it. But lets assume I have matrix B $$\left(\begin{array}{ccc}6&2&2\\-2&2&0\\0&0&2\end{array} \right)$$ Its characteristic polynomial is the same as for A and following the same method as HallsOfIvy pointed out i get the same $$p(\psi)$$ for B as for A but when u put B into the equation you dont get 0 so this method doesnt work.