匿名
匿名 發問於 科學及數學數學 · 7 年 前

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 個解答

評分
  • Andrew
    Lv 6
    7 年 前
    最佳解答

    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.

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