So, I am currently reading about Jordan matrices and have understood that every matrix is similar to a Jordan matrix. I have been doing some calculations for simple matrices with distinct eigenvalues and everything has been going well. However, I have stumbled upon a problem which I can't figure out.
$A = \begin{bmatrix} 0 & 1 & 1\\ 1 & -1 & 1 \\ -1 & -1 & -2 \end{bmatrix}$
I want to find the Jordan normal form $J$ of $A$ with the corresponding transition matrix $S$.
The matrix seems to have three eigenvalues, all with value -1. Hence, I can't use theory about diagonalization to simply calculate the Jordan form. Could anyone explain where to begin?
Thanks!
Hint: The number of Jordan blocks corresponding to eigenvalue $\lambda$ is the geometric multiplicity of $\lambda$