question is as stated in the title. Is there also a way to prove it? Just curious, help will be greatly appreciated.
2026-03-29 04:41:36.1774759296
If A is diagonalizable, and λ is a eigenvalue of A, is (A-λI) diagonalizable as well?
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Choose a matrix $P$ such that $PAP^{-1}$ is diagonal. Then $$P(A - \lambda I)P^{-1} = PAP^{-1} - P(\lambda I)P^{-1} = PAP^{-1} - \lambda I$$ is a sum of diagonal matrices, thus also diagonal.