# Minimal Polynomial and Jordan Form

1. Dec 1, 2011

### aznkid310

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

See below.

3. The attempt at a solution

a) I only see that 3 of the eigenvalues are zero, but dont 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.

2. Dec 1, 2011

### micromass

Staff Emeritus
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.

3. Dec 1, 2011

### aznkid310

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.

4. Dec 2, 2011

### HallsofIvy

Staff Emeritus
No. All eigenvalues are roots of the minimal equations. Since $s^3= 0$ has only s= 0 as root, all six eigenvalues are 0.

5. Dec 2, 2011

### aznkid310

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 cant the second form occur?

6. Dec 2, 2011

### micromass

Staff Emeritus
You have to write 6x6 - matrices...

7. Dec 2, 2011

### aznkid310

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]

8. Dec 2, 2011

### micromass

Staff Emeritus
Are you sure these have minimal polynomial $s^3$?? Are you sure these are the only matrices??

9. Dec 2, 2011

### aznkid310

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: Dec 2, 2011