Linear algebra: Jordan normal forms

[rbpl]

Suppose the characteristic polynomial of a matrix A is $$\lambda$$^3($$\lambda$$-1)($$\lambda$$-2). If the nullity of A is two, what are the possible Jordan normal forms of A up to conjugation?I think that an example of a matrix with such characteristic polynomial is:

0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 1 0
0 0 0 0 2

But then isn't the nullity 1 in this case?

Otherwise, the possibilities for the above matrix are:

0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 1 0
0 0 0 0 2

0 1 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
0 0 0 0 2

0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 1 0
0 0 0 0 2

Am I correct?
What should I do about the nullity which is 2?

 

[lippyka]

Yes, you are correct; the above matrices are all valid Jordan normal forms with the given characteristic polynomial. For the nullity of two, the only additional possibility is:0 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 10 0 0 0 2