Prove that all normal matrices are semi-simple using Schur's decomposition












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Is there any elegant proof that shows that all normal matrices are semi-simple that comes from Schur's decomposition or its corrolaries? There is a proof that normal matrices are unitary diagonizable and then that diagonizable matrices are semi-simple, but it seems a little exhaustive to combine them both. Is there any better proof?










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    $begingroup$


    Is there any elegant proof that shows that all normal matrices are semi-simple that comes from Schur's decomposition or its corrolaries? There is a proof that normal matrices are unitary diagonizable and then that diagonizable matrices are semi-simple, but it seems a little exhaustive to combine them both. Is there any better proof?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Is there any elegant proof that shows that all normal matrices are semi-simple that comes from Schur's decomposition or its corrolaries? There is a proof that normal matrices are unitary diagonizable and then that diagonizable matrices are semi-simple, but it seems a little exhaustive to combine them both. Is there any better proof?










      share|cite|improve this question









      $endgroup$




      Is there any elegant proof that shows that all normal matrices are semi-simple that comes from Schur's decomposition or its corrolaries? There is a proof that normal matrices are unitary diagonizable and then that diagonizable matrices are semi-simple, but it seems a little exhaustive to combine them both. Is there any better proof?







      proof-writing diagonalization matrix-decomposition






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      asked Dec 17 '18 at 15:33









      ViniLLViniLL

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