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After finishing some exercises, it seems that an $n\times n$ matrix $A$ will also have a $n\times n$ jordan form. Is my observation correct? If this is a common phenomenon, could you explain reasons behind it?

I guess it is about the dimension of eigenspaces, but I am not sure.

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    $\begingroup$ it depends on the space of matrices. Real valued matrices doesnt ever have Jordan normal formal forms, however we ever can define Jordan extended normal forms. Complex valued matrices ever have Jordan normal forms. This is because the domain of the matrix can be decomposed in generalized eigenspaces, wich are invariant under the action of the matrix. $\endgroup$ – Masacroso Feb 7 '18 at 3:29
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The Jordan form of a matrix is just the matrix transformed by a change of basis. They both represent the same linear map from an $n$-dimensional vector space to another, so they are of course both $n\times n$.

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