# Finding Jordan Forms of 8x8 Matrices

• chuckles1176
In summary, to find all Jordan forms of 8x8 matrices given the minimal polynomial x^2*(x-1)^3, we would have to consider all possible combinations of 0 and 1 blocks such that there are always at least one of each and the total dimension is 8. The geometric multiplicities of 0 and 1 would affect the entries next to the main diagonals, but it is possible to have a Jordan block with 1's on the main diagonal of dimension 2 and still satisfy the 8x8 dimension requirement.
chuckles1176

## Homework Statement

find all Jordan forms of 8x8 matrices given the minimal polynomial x^2*(x-1)^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.

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.

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?

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
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\0 & 0 & 1\end{bmatrix}$$

## 1. What is a Jordan form of an 8x8 matrix?

A Jordan form of an 8x8 matrix is a special type of matrix that has a specific structure and is useful for solving certain types of problems in linear algebra. It is named after the mathematician Camille Jordan.

## 2. How do you find the Jordan form of an 8x8 matrix?

To find the Jordan form of an 8x8 matrix, you need to follow a specific set of steps. First, you need to find the eigenvalues of the matrix. Then, for each eigenvalue, you need to find the corresponding eigenvectors. Finally, you can form the Jordan matrix by placing the eigenvectors in a specific pattern.

## 3. Why is it important to find the Jordan form of an 8x8 matrix?

The Jordan form of an 8x8 matrix is important because it allows us to simplify complex matrices and make them easier to work with. It is also useful for solving problems related to diagonalization and finding the powers of a matrix.

## 4. Are there any limitations to finding the Jordan form of an 8x8 matrix?

Yes, there are some limitations to finding the Jordan form of an 8x8 matrix. One limitation is that not all matrices have a Jordan form. Another limitation is that the Jordan form is not unique, meaning there can be different Jordan matrices for the same matrix.

## 5. Can computer programs be used to find the Jordan form of an 8x8 matrix?

Yes, computer programs can be used to find the Jordan form of an 8x8 matrix. There are various software packages and online tools available that can efficiently calculate the Jordan form of a given matrix. However, it is still essential to understand the underlying concepts and steps involved in finding the Jordan form manually.

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