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Minimal polynomial

by vabamyyr
Tags: minimal, polynomial
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vabamyyr
#1
Feb23-06, 10:21 AM
P: 66
I have matrix A
[tex] \left(\begin{array}{ccc}6&2&-2\\-2&2&2\\2&2&2\end{array} \right) [/tex]

Its characteristic polynomial is
[tex]
p(\lambda)=\lambda^3 - 10\lambda^2 + 32\lambda -32
[/tex]

Finding minimal polynomial i get:
[tex] (I\lambda-A)^\vee=\left(\begin{array}{ccc}\lambda-6&-2&2\\2&\lambda-2&-2\\-2&-2&\lambda-2\end{array}\right)^\vee [/tex]

I cant understand why this last result equals with
[tex] \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) [/tex]
can someone explain?
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matt grime
#2
Feb23-06, 10:26 AM
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would you mind retypsetting that so it is more legible?
matt grime
#3
Feb23-06, 10:30 AM
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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.

AKG
#4
Feb23-06, 10:57 AM
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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.
vabamyyr
#5
Feb24-06, 04:04 AM
P: 66
I guess im on my own again.
HallsofIvy
#6
Feb24-06, 06:25 AM
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Thanks
PF Gold
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Perhaps it would help if you explain what the "v" in
[tex] (I\lambda-A)^\vee[/tex]
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 [itex]\lambda- 2[/itex] is a factor so it should be simple to determine all the factors. What more do you want?
AKG
#7
Feb24-06, 01:26 PM
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Quote 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.
matt grime
#8
Feb24-06, 01:42 PM
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Quote Quote by HallsofIvy
just the characteristic polynomial with duplicate factors removed.

That in my mind is a little misleading. It is the characteristic poly with 'unnecessary' duplicate factors removed in some sense.
vabamyyr
#9
Feb25-06, 07:13 AM
P: 66
I have matrix A
[tex] \left(\begin{array}{ccc}6&2&-2\\-2&2&2\\2&2&2\end{array} \right) [/tex]

Its characteristic polynomial is
[tex]
p(\lambda)=\lambda^3 - 10\lambda^2 + 32\lambda -32[/tex]

when i factor it i get [tex]p(\lambda)=(2-\lambda)(\lambda-4)^2 [/tex]
So removing the duplicate factor
[tex](\lambda-4)[/tex]
i get that minimal polynomial is
[tex](\psi)=(2-\lambda)(\lambda -4)[/tex]
and A really satisfies it.

But lets assume I have matrix B [tex]\left(\begin{array}{ccc}6&2&2\\-2&2&0\\0&0&2\end{array} \right)[/tex]

Its characteristic polynomial is the same as for A and following
the same method as HallsOfIvy pointed out i get the same
[tex]p(\psi)[/tex] for B as for A but when u put B into the equation
you dont get 0 so this method doesnt work.
matt grime
#10
Feb25-06, 08:05 AM
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Erm, no, as was pointed out you cannot *just* remove factors and assume that the minimal poly is the product of the distinct linear factors. All this tells you if indeed your manipulations are correct is that the minimal poly and the characterstic poly are the same. The method is correct, and has given you the minimal poly, it is just that you don't recognize this.

This was what I wished to clarify from HallsofIvy's post. The minimal poly is not the product of the distinct linear factors but the product of the distinct linear factors with the smallest multiplicity of each possible so that the matrix satisfies the polynomial.
mathwonk
#11
Feb25-06, 09:36 AM
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to illustrate matt's point about repeated factors in mminimal polynomials, consider the derivative operator D, acting on functions. on the space spanned by the functions e^x and xe^x, the minimal polynomial is (D-1)^2. while on the span of e^x, it is just D-1.


expressed in matrices, i.e. stripped of all meaning, for the ordered basis e^x, xe^x, we get a matrix of D with columns [ 1 0]^t and [1 1]^t,

where ^t means transpose.

the point is that the matrix is not disgonalizable, but is a sum of a diagonal matrix (namely the identity matrix), and a "nilpotent matrix", here the matrix which is all zeroes except for a 1 in the upper right corner.

It is the rpoesence of this nilpotent summand which prevents the minimal polynomial from having distinct irreducible factors.


or think about the map D, on the space of polynomials of degree at most n-1. The eigenvalues are all zero, and the characteristic polynomial is D^n, which is also the minimal polynomial.


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