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This may be an elementary question and I haven't been able to find if such a question has been asked on Math StackExchange but here it is:

suppose $M$ is an $n\times n$ matrix over $\mathbb{C}$ that is $GL_n(\mathbb{C})$-conjugate to $J$ where $J$ is a Jordan matrix with Jordan blocks $ J = \left[ \begin{array}{cccc} J_1 & 0 & \ldots & 0 \\ 0 & J_2 & \ldots & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & \ldots & J_k \\ \end{array} \right] $ where each $J_i$ is of rank $k_i$, all with the same eigenvalue $\lambda$; so $ J_i = \left[ \begin{array}{ccccc} \lambda & 1 & 0 & \ldots & 0 \\ 0 & \lambda & 1 & \ldots & 0 \\ 0 & 0 &\lambda & \ldots &\vdots \\ 0 & 0 & 0 & \ddots & 1 \\ 0 & 0 & 0 & \ldots & \lambda \\ \end{array} \right]. $ Then for any $ J' = \left[ \begin{array}{cccc} J_{\sigma(1)} & 0 & \ldots & 0 \\ 0 & J_{\sigma(2)} & \ldots & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & \ldots & J_{\sigma(k)} \\ \end{array} \right] $ where $\sigma:\{1,2,\ldots, k\}\rightarrow \{ 1,2,\ldots, k\}$ is injective, doesn't there exist $g\in GL_n(\mathbb{C})$ so that $gJg^{-1}=J'$? Basically, couldn't we always permute these Jordan blocks under the $GL_n(\mathbb{C})$-conjugation action?

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    Yes. ${}{}{}{}$2012-09-25

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Yes, we could. This is not specific to Jordan blocks. Quite generally, to any permutation $\pi$ there corresponds a permutation matrix $\Pi$ with $\Pi_{ij}=\delta_{i\pi(j)}$ with determinant $\pm1$. This permutation matrix permutes the rows of a matrix according to $\pi$ upon left-multiplication and the columns of a matrix according to $\pi^{-1}$ upon right-multiplication, so $\Pi J\Pi^{-1}$ permutes both the rows and the columns of $J$ according to $\pi$. Your rearrangement of the Jordan blocks is a specific case of such a permutation of the rows and columns of a matrix.

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    @math-visitor: You're welcome!2012-09-25