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How to get the Additive Jordan Decomposition of a matrix from its Jordan Canonical Form?

I tried with few matrices and its Jordan Canonical Forms, but I could not understand. So please explain the relation.

Thanks in advance.

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  • $\begingroup$ I want to write a matrix A as A = $A_s + A_n$ where $A_s$ is semi simple and $A_n$ is nilpotent. I have found the Jordan Form (Say the matrix B) of A. Can I get Jordan Decomposition of A from B ? Sorry for not giving examples. $\endgroup$ – GA316 Oct 7 '13 at 16:40
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Let $A = S J S^{-1}$ be a Jordan decomposition. Note that $J$ has a form $J = D + N$, where $D$ is diagonal, and $N$ is strictly upper triangular with zeroes everywhere except some positions of the main superdiagonal, where it has ones. In other word,

\begin{align*} J &= \begin{bmatrix} J_{ij} \end{bmatrix}, \quad J_{ij} = \begin{cases} \text{$0$ or $1$}, & i = j-1, \\ \lambda_i, & i = j, \\ 0, & \text{otherwise}. \end{cases}, \\ D &= \operatorname{diag}(\lambda_1,\dots,\lambda_n), \\ N &= \begin{bmatrix} N_{ij} \end{bmatrix}, \quad N_{ij} = \begin{cases} \text{$0$ or $1$}, & i = j-1, \\ 0, & \text{otherwise}. \end{cases}. \end{align*}

So,

$$A = S J S^{-1} = S (D + N) S^{-1} = \underbrace{S D S^{-1}}_{A_s :=} + \underbrace{S N S^{-1}}_{A_n :=}.$$

Is this what you had in mind?

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  • $\begingroup$ exactly this is what I had in my mind. Thanks for your answer. $\endgroup$ – GA316 Oct 7 '13 at 18:02

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