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by srean
38 days ago
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KalMann is correct. Jordan canonical form decomposition is more general. Every matrix in an algebraically closed field will have such a decomposition. This is not true for spectral decomposition. Only diagonalizable matrices will have a spectral decomposition and they are a smaller subset. That said, Jordan form is uglier than spectral decomposition, to my taste that is. Spectral decomposition so beautiful and neat. |
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