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?


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.