Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. **There will be no changes to other Yahoo properties or services, or your Yahoo account.** You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

## Trending News

# Eigenvalues

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.

### 1 Answer

- AndrewLv 68 years agoFavorite Answer
The minimal polynomial for P is P^2 - P = P(P - I) = 0

The 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.