Jordan Form- Please confirm my solution

1. May 31, 2010

talolard

1. The problem statement, all variables and given/known data
edit, sorry its not jordan form, I misplaced words.
Hey,
Let A be an nxn matrix over some field F. $$A \neg 0$$ and $$A^{2009}+A^{2007}=0$$. Is a diagnizable?

Attempt at solution
we can notice that $$A^{2009}+A^{2007}=0 \iff A^{2007}(A^2+I)=0 \iff A^{2007}=0 or A^2=-I$$
If $$A^2 = -I$$ then a is a diagnol matrix with eigenvalues i and -i (i is defined as the sqaure root of -1 in the field. if the field does not have sucha number then A is not diagnizable).
If [/tex] A^2 \neq -I [/tex] then$$A^{2007}=0$$ but the A is nillpotent and so has less the n eigenvectors, so A is not diagnizable. q.e.d.
Is this correct?
Thanks
Tal
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: May 31, 2010
2. May 31, 2010

Staff: Mentor

You are omitting a possibility. For square matrices A and B, it is possible for AB = 0 where neither A nor B is the 0 matrix.

Here's a simple example.
$$\left[ \begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array} \right] \left[ \begin{array}{c c} 0 & 2 \\ 0 & 0 \end{array} \right] = \left[ \begin{array}{c c} 0 & 0 \\ 0 & 0 \end{array} \right]$$

3. May 31, 2010

talolard

Hi Mark,
As I see it, I am only talking about AA, and not Ab, two different matrices. if B= A^2006 abnd AB=-0 tehn AB=A*A*A&2006=A^2007=0 which still leaves A nilpotent.
If that is not the case then we need A^2=-I which still means that A is diagnozable iff sqrt(-1) exists in the field.
What am I missing?
Thanks
Tal

4. May 31, 2010

Staff: Mentor

You are talking about two different matrices: A2007 and A2 + I. You said that since A2007(A2 + I) = 0, then it must be that A2700 = 0 or A2 + I = 0.

What I said was that it is possible for the product of two matrices to be zero, where neither matrix is the zero matrix.

It is true that if A = 0 or B = 0, then AB = 0, but the converse is not necessarily true, as my example shows.

5. Jun 1, 2010

talolard

Ok, heres an improvment:
Assume A is diagnizable. Let v be an eigenvector of A with an eigenvalue, c, that is not 0. then we have
$$(A^{2009}+A^{2007})\vec(v) =A^{2009}\vec(v) + A^{2007}\vec(v) = c^{2009}\vec(v) + c^{2007}\vec(v)=0 \iff c=i or c=-i$$
If A is a 2x2 matrix over a field where i exists then the matrix is diagnizable. otherwise it i not.
if RkA> 3 then for A to be diagnizable it would need other eigenvalues but none would atisfy the condition.
Is that correct?
Thanks
Tal