Show that a is the length of the biggest Jordan chains corresponding to an eigenvalue.












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$begingroup$


Let V be a finite-dimensional vector space, T belong to L(V) and T has
an eigenvalue x with the corresponding space of generalized eigenvectors Ux. Let
a = min{k : (T-x)^k|Ux = 0}. Show that a is the length of the biggest Jordan
chain(s) corresponding to x. Can anyone help me? I have no clue where to start and how to proceed. Any help would be appreciated. Thanks in advance.










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$endgroup$

















    0












    $begingroup$


    Let V be a finite-dimensional vector space, T belong to L(V) and T has
    an eigenvalue x with the corresponding space of generalized eigenvectors Ux. Let
    a = min{k : (T-x)^k|Ux = 0}. Show that a is the length of the biggest Jordan
    chain(s) corresponding to x. Can anyone help me? I have no clue where to start and how to proceed. Any help would be appreciated. Thanks in advance.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let V be a finite-dimensional vector space, T belong to L(V) and T has
      an eigenvalue x with the corresponding space of generalized eigenvectors Ux. Let
      a = min{k : (T-x)^k|Ux = 0}. Show that a is the length of the biggest Jordan
      chain(s) corresponding to x. Can anyone help me? I have no clue where to start and how to proceed. Any help would be appreciated. Thanks in advance.










      share|cite|improve this question









      $endgroup$




      Let V be a finite-dimensional vector space, T belong to L(V) and T has
      an eigenvalue x with the corresponding space of generalized eigenvectors Ux. Let
      a = min{k : (T-x)^k|Ux = 0}. Show that a is the length of the biggest Jordan
      chain(s) corresponding to x. Can anyone help me? I have no clue where to start and how to proceed. Any help would be appreciated. Thanks in advance.







      linear-algebra generalizedeigenvector






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 4:57









      Ashis JanaAshis Jana

      122




      122






















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