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Are matrices with the same Jordan blocks similar over $\Bbb C$? This seems to be the case, since say: $$\begin{array}{cc|c}0&1&0\\0&0&0\\\hline0&0&0 \end{array} \sim \begin{array}{c|cc}0&0&0\\\hline0&0&1\\0&0&0 \end{array}$$

by conjugation from $\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$.

"Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed;" -Wiki

That text is meant to simply mean that the Jordan form doesn't uniquely identify a matrix, without requiring some other rules, such as having eigenvalues listed in decreasing order from the top. But the matrices are still uniquely classified into a similarity class, and if they have the same blocks, they belong to the same class, right?

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Right, if they have the same blocks, then they are similar.

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