# Jordan Forms

1. Dec 14, 2009

### chuckles1176

1. The problem statement, all variables and given/known data
find all Jordan forms of 8x8 matrices given the minimal polynomial x^2*(x-1)^3

2. Relevant equations

3. The attempt at a solution

The roots are clearly 0,1 and 0 has degree 2 while 1 has degree 3. The forms would be made up of the blocks [0,0;1,0] corresponding to 0 and [1,0,0;1,1,0;0,1,1] corresponding to 1.

So all possible 8x8 forms would be different combinations of the blocks such that they are always both included at least once and dimension 1 blocks being either of the roots such that the overall dimension of the blocks is 8.

-I am not convinced this is the solution because of the geometric multiplicities of 1 and 0 would affect the entries next to the main diagonals...I'm just not sure how if at all.

2. Dec 15, 2009

### HallsofIvy

Staff Emeritus
Every such 8 by 8 matrix will have 2 "0"s and 3 "1"s on the diagonal. The "Jordan" form may or may not have "1" above each number on the diagonal. For the "0"s, then, you can have either
$$\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}$$
or
$$\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}$$

For the "1"s there are 4 possiblilties.

3. Dec 16, 2009

### chuckles1176

sorry I should have made this a bit more clear in the question, but can there exist a Jordan block (given the minimal polynomial above) with 1's on the main diagonal of dimension 2 such that it satisfies the 8x8 dimension req? i.e. assuming we have the dim=3 blocks obtained from 1 and the dim=2 blocks obtained from 0, can we have a dim=2 block with 1's on the main diagonal, and an appropriate number of dim=1 blocks being either 0,1 to satisfy the 8x8 dim?

4. Dec 17, 2009

### HallsofIvy

Staff Emeritus
Yes, it is possible, that "1" be an eigenvalue of algebraic multiplicity 3 and geometric multiplicity 2 (or any positive integer less than or equal to 3). One possiblity would be
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\0 & 0 & 1\end{bmatrix}$$