# Number of jordan blocks in Jordan decomposition

1. Dec 12, 2013

### jinawee

Given a matrix $$A$$. Is it possible to have a Jordan block form like:

$$\begin{pmatrix} \lambda & 1 & 0 & 0\\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1\\ 0 & 0 & 0 & \lambda\\ \end{pmatrix}$$

?

Last edited: Dec 12, 2013
2. Dec 12, 2013

### jgens

The matrix you named above has exactly that jordan canonical form. Sooooooo...yes.

3. Dec 12, 2013

### jinawee

Then I have another question.

The dimension of ker(A-λI) is the number of Jordan blocks associated to λ. And n, where n is given by dim(ker(A-λI)^n)=dim(ker(A-λI)^(n+1)) is the size of the largest Jordan block. The total size is dim(ker(A-λI).

So, how do we know which form is the correct one in the following case?

$$\begin{pmatrix} \lambda & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$$

or

$$\begin{pmatrix} \lambda & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$$

If there isn't enough information, what else do I need?

4. Dec 12, 2013

### jgens

I mean there are algorithms to turn any given matrix into its Jordan canonical form, so if you have a specific matrix, then there is always a way to tell the difference. To distinguish between the above cases you can always check that the matrices obtained by restricting the domain to a certain subspace have the desired block form. I have no idea if there are better ways of checking beyond these basic tricks, however, since I rarely use canonical forms.