Show that a is the length of the biggest Jordan chains corresponding to an eigenvalue.
$begingroup$
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.
linear-algebra generalizedeigenvector
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra generalizedeigenvector
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra generalizedeigenvector
$endgroup$
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
linear-algebra generalizedeigenvector
asked Dec 7 '18 at 4:57
Ashis JanaAshis Jana
22
22
add a comment |
add a comment |
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