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We know that not all matrices can be diagonalized, but all matrices can be block diagonalized (with just one block) How can we find a similarity transformation leading to block diagonalization with the greatest possible number of blocks?

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Every matrix with elements in $\mathbb C$ has a Jordan Normal Form. The transform in the canonical basis will have blocks of sizes equal to the sizes of the generalized eigenspaces of the matrix.

The Jordan blocks have a very particular structure:

$$\left[\begin{array}{ccc}\lambda&1&0&\cdots&0\\0&\lambda&1&0&0\\0&\ddots&\ddots&\ddots&0\\0&0&0&\lambda&1\\0&0&0&0&\lambda\end{array}\right]$$

where the $\lambda$ is an eigenvalue for the matrix. It should be possible to prove that the block above can not be further reduced (although I have no proof ready in my magic pockets right now).


In the case you want real elements everywhere you can take a look here

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  • $\begingroup$ There’s a related block diagonalization for real matrices. $\endgroup$ – amd Apr 24 '18 at 18:46
  • $\begingroup$ @amd yep I have also answered on that one in one place or another. $\endgroup$ – mathreadler Apr 24 '18 at 18:54
  • $\begingroup$ I should have said that my matrix was real. The observation by @amd pointed me in the right direction. Thanks $\endgroup$ – Philip Roe Apr 24 '18 at 18:55
  • $\begingroup$ @PhilipRoe ok, you can see my other answer on the real Jordan form if it interests you. $\endgroup$ – mathreadler Apr 24 '18 at 18:57
  • $\begingroup$ To make things a little more concrete, all real matrices have diagonalizations that may be complex, and the jordan form of all real matrices is real. The distinction isn't between real and complex original matrices, but symmetric and nonsymmetric original matrices, where for symmetric matrices the diagonalization is also real (and is equal to the jordan form). $\endgroup$ – Y. S. Apr 24 '18 at 19:01

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