Difference between revisions of "Diagonalizable"

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An operator ''T'' on a finite-dimensional vectorspace ''V'' is diagonalizeable if ''V'' has a basis of eigenvectors for ''T''.
 
An operator ''T'' on a finite-dimensional vectorspace ''V'' is diagonalizeable if ''V'' has a basis of eigenvectors for ''T''.
  
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[[Category: Mathematics: Linear Algebra]]
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[[Category:Mathematics]]

Revision as of 23:33, July 4, 2008

An operator T on a finite-dimensional vectorspace V is diagonalizeable if V has a basis of eigenvectors for T.