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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!

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  • $\begingroup$ Beware, your first affirmation is false : every matrix in a field $K$ whose characteristic polynomial has all its roots in $K$ is similar to a Jordan form. In the case you are only dealing with $\mathbb{R}$ , this means that a matrix with coefficients in $\mathbb{R}$ is similar to a Jordan matrix in $\mathbb{R}$ if its eigenvalues are real. Otherwise, it is similar to a Jordan matrix only in $\mathbb{C}$ $\endgroup$ – Junkyards Oct 1 '17 at 15:59
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Hint: The number of Jordan blocks corresponding to eigenvalue $\lambda$ is the geometric multiplicity of $\lambda$

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  • $\begingroup$ Hey! I wrote in Matlab: null(A + eye(3)) and I get just a single vector, hence we should only have one Jordan block. Does this mean that the matrix $J$ consists of -1:s along the diagonal and two ones about the main diagonal? $\endgroup$ – Ludwwwig Oct 1 '17 at 15:58
  • $\begingroup$ Exactly, you have $2$ ones on the super diagonal. $\endgroup$ – Rab Oct 1 '17 at 16:26

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