# stable - Generalized eigenvalues Let V be a vector space...

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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 0 K 1 K 2 · · · ⊆ K i ⊆ · · · V and V = R 0 R 1 · · · ⊇ 0 . 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 m ( x ) = 0, so x K 2 m . Since K 2 m = K m , in fact S m ( x ) = 0, i.e. , v = 0. Next we check that

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• Fall '08
• GUREVITCH

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