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Lemma 25 invariance each of the generalized

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Unformatted text preview: a linear operator under . Proof. Suppose . so that . Since and is invariant commute .□ Let and be vector spaces. The direct sum is the vector space with elements , where and and operations of vector addition and scalar multiplication defined by and , where also and and . For example, . Theorem 2.6 (Primary Decomposition). Let be a linear operator on a complex vector space with distinct eigenvalues and let be the generalized eigenspace of with eigenvalue . Then is the algebraic multiplicity of generalized eigenspaces, i.e. . and is the direct sum of the 5 Chapter 2 part B Proof. This is proved in Hirsch and Smale. Remark. We can choose a basis combined to obtain a basis for each eigenspace. By theorem 2.6, these can be for Warning. The labeling for generalized eigenvectors given above is Meiss’ notation. Note that the eigenvectors are relabeled to give the basis for . This keeps the notation simple but the labels must be interpreted correctly depending on context. Semisimple-Nilpotent Decomposition Shift notation from as linear operato...
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