stable - Generalized eigenvalues April 23, 2007 Let V be a...

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Unformatted text preview: Generalized eigenvalues April 23, 2007 Let V be a vector space over F and let S be a linear operator on V . Recall that S is said to be nilpotent if there exists a positive integer m such that S m = 0. Theorem: Suppose that V is finite dimensional and S is a linear oper- ator V V . Then V admits a direct sum decomposition into S-invariant subspaces V = V n V s such that S V n is nilpotent and S V s is an isomorphism. Furthermore if T is any linear operator on V which commutes with S , then V n and V s are T- invariant. Proof: For each i , let K i be the kernel of S i and let R i be the image of S i . Then 0 = K K 1 K 2 K i V and V = R R 1 . Since V is finite dimensional, there exists an integer m such that K m = K m + k and R m = R m + k for all k 0. Let V n := K m and let V s := R m . Let us check that V n V s = 0. If v R m , v = S m ( x ) for some x V , and if also v K m , then S m ( v ) = 0. It follows that S 2...
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This note was uploaded on 12/05/2010 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.

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stable - Generalized eigenvalues April 23, 2007 Let V be a...

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