Proof that projection matrix P has lambda = 0 and lambda = 1

Assume that x is an eigenvector of P. Then P has atmost two eigenvalues (0 or 1).

    \[ \left\{\begin{matrix} P^{2}x = Px = \lambda x \\ P^{2}x = P\lambda x = \lambda^{2}x \end{matrix}\right. \Rightarrow \lambda x = \lambda^{2} x \Rightarrow \lambda = 0 \vee \lambda = 1 \]

  • Prove that there exists an eigenvector of P with eigenvalue equals 1:
    • We have P^{2} = P * P = P, therefore we can pick any column vectors of P and it will be an eigenvector of P with eigenvalue equals 1.
  • Prove that there exists an eigenvector of P with eigenvalue equals 0:
    • If the definition of projection matrix from WolframMathWorld is correct then we can let P = I and as the nullspace of I contains only one zero vector (and the fact that the eigenvector must be non-zero), we cannot find an eigenvector of I with eigenvalue equals 0. If P \in \mathbb{R}^{n\times n} and \text{rank}(P) < n then the nullspace of P contains eigenvectors of P.
    • This seems to be a mistake of Gilbert Strang in Introduction to Linear Algebra, 4th edition, page 286, sentence: “Projections have \lambda = 0 and 1“.

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