For a linear operator P on a finite-dimensional space V , the property P^2= P implies
that V = the direct sum of Null P and Range P . Prove that if P^2= P , then P is
diagonalizable and its eigenvalues can only be 0 or 1.
Eigenvalues
2013-08-06 9:04 pm
回答 (1)
2013-08-07 12:22 pm
✔ 最佳答案
The minimal polynomial for P is P^2 - P = P(P - I) = 0The matrix P is diagonalizable because the minimal polynomial factorize into distinct linear factors.
The eigenvalues can only be 0 or 1 because these are the only roots for the minimal polynomial.
See the wikipedia page Minimal polynomial for more details on it.
收錄日期: 2021-05-01 01:13:05
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