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Mathematics-Online course: Linear Algebra - Normal Forms - Diagonalisation

Basis of Eigenvectors

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If there exists a basis of eigenvectors $ b_j$ associated with eigenvalues $ \lambda_j$ of a given matrix $ A$, then

$\displaystyle B^{-1} A B = \operatorname{diag}(\lambda_1,\ldots,\lambda_n), \quad B = (b_1,\ldots,b_n) \,. $

(Authors: App/Burkhardt/Höllig)

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  automatically generated 4/21/2005