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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| + | ==Denseness of diagonalizeable operators== | ||
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| + | The space of complex linear operators on <math>\mathbb{C}^n</math> may be identified with the vectorspace <math>M_n(\mathbb{C})</math> of nxn matrices with complex coefficients. As such, it inherits a natural structure as a topological space. | ||
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| + | Given this topology, the set of diagonalizeable maps is a dense subset of <math>M_n(\mathbb{C}^n)</math> | ||
[[Category:Mathematics]] | [[Category:Mathematics]] | ||
Revision as of 01:59, July 5, 2008
An operator T on a finite-dimensional vectorspace V is diagonalizeable if V has a basis of eigenvectors for T.
Denseness of diagonalizeable operators
The space of complex linear operators on <math>\mathbb{C}^n</math> may be identified with the vectorspace <math>M_n(\mathbb{C})</math> of nxn matrices with complex coefficients. As such, it inherits a natural structure as a topological space.
Given this topology, the set of diagonalizeable maps is a dense subset of <math>M_n(\mathbb{C}^n)</math>