Minimal Polynomial and Jordan Form

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Homework Help Overview

The discussion revolves around a 6x6 matrix A with a minimal polynomial of p(s) = s^3. Participants are exploring the implications of this polynomial on the characteristic polynomial, possible Jordan forms, and the rank of the matrix.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the minimal polynomial and the eigenvalues, questioning how to derive the characteristic polynomial from the given information. There is exploration of the possible sizes of Jordan blocks and their implications for the matrix structure.

Discussion Status

Some participants have provided insights into the eigenvalues and the structure of Jordan blocks, while others are questioning the validity of certain configurations and their corresponding minimal polynomials. The conversation is ongoing, with various interpretations being considered.

Contextual Notes

There is a focus on the constraints of the problem, particularly the requirement for the matrix to be 6x6 and the implications of the minimal polynomial on the eigenvalues and Jordan form. Participants are also grappling with the need to ensure that all eigenvalues are accounted for in the characteristic polynomial.

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Homework Statement


Suppose that A is a 6x6 matrix with real values and has a min. poly of p(s) = s^3.

a) Find the Characteristic polynomial of A
b) What are the possibilities for the Jordan form of A?
c) What are the possibilities of the rank of A?


Homework Equations



See below.

The Attempt at a Solution



a) I only see that 3 of the eigenvalues are zero, but don't know how to find the rest for the characterisitic polynomial

b) The Jordan blocks can be size 1,2, or 3 i.e. [L 1 0; 0 L 1; 0 0 L], [L 0 0; 0 L 0; 0 0 L], [L 1 0; 0 L 0; 0 0 L] where L are the eigenvalues from the min. poly. (equal to zero)

c) rank(A) + dim N(A) = n, where N(A) is the nullspace of A, and n = 6. Do I just need to find the nullspace of A (and if so, how?) or am I going down the wrong direction.
 
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aznkid310 said:

Homework Statement


Suppose that A is a 6x6 matrix with real values and has a min. poly of p(s) = s^3.

a) Find the Characteristic polynomial of A
b) What are the possibilities for the Jordan form of A?
c) What are the possibilities of the rank of A?

Homework Equations



See below.

The Attempt at a Solution



a) I only see that 3 of the eigenvalues are zero, but don't know how to find the rest for the characterisitic polynomial

It is a theorem that the roots of the minimal polynomial are exactly the eigenvalues. So what are all the eigenvalues?? Can you find the characteristic polynomial now??

b) The Jordan blocks can be size 1,2, or 3 i.e. [L 1 0; 0 L 1; 0 0 L], [L 0 0; 0 L 0; 0 0 L], [L 1 0; 0 L 0; 0 0 L] where L are the eigenvalues from the min. poly. (equal to zero)

Are you sure that second matrix can occur? What is its minimal polynomial??
Also, the matrices need to be 6x6.

c) rank(A) + dim N(A) = n, where N(A) is the nullspace of A, and n = 6. Do I just need to find the nullspace of A (and if so, how?) or am I going down the wrong direction.

You probably need to use the Jordan forms you found in (2). You can easily see the rank of those.
 
micromass said:
It is a theorem that the roots of the minimal polynomial are exactly the eigenvalues. So what are all the eigenvalues?? Can you find the characteristic polynomial now??

Are you sure that second matrix can occur? What is its minimal polynomial??
Also, the matrices need to be 6x6.

You probably need to use the Jordan forms you found in (2). You can easily see the rank of those.

If the min poly is p(s) = s^3, doesn't that mean that only three of the six are zero? How would I find the other 3 eigenvalues? I must not be understanding this correctly.

I understand that the matrix needs to be 6x6, but I only know 3 of the eigenvalues? I guess if I understand the first part of your reply, it'll answer this one.
 
No. All eigenvalues are roots of the minimal equations. Since s^3= 0 has only s= 0 as root, all six eigenvalues are 0.
 
Thanks HallsofIvy. So using this info, we can have possible jordan blocks of size 1,2, or 3. So the possible matrices are:

[0 1 0
0 0 1
0 0 0]

[0 0 0
0 0 0
0 0 0]

[0 1 0
0 0 0
0 0 0]

Where I can fill out the rest of the 6x6 with zeros if I wanted the original matrix. My question is, why can't the second form occur?
 
You have to write 6x6 - matrices...
 
Right, so for example:

[0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 1 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]
 
Are you sure these have minimal polynomial s^3?? Are you sure these are the only matrices??
 
Each of the three eigenvalues (zero) in the min. poly could have jordan blocks of size 1,2, or 3. So:

[0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 1 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0]

Or does the fact that the min. poly is s^3 indicate that we must have one 3x3 Jordan block, in which case we need to fill out the remaining matrix with either another 3x3, a 2x1 and a 1x1, or a 1x1 and 2x1:

[0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0]

[0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0]

[0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 1
0 0 0 0 0 0]
 
Last edited:

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