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

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

- Prove that there exists an eigenvector of with eigenvalue equals 1:
- We have , therefore we can pick any column vectors of and it will be an eigenvector of with eigenvalue equals 1.

- Prove that there exists an eigenvector of with eigenvalue equals 0:
- If the definition of projection matrix from WolframMathWorld is correct then we can let and as the nullspace of contains only one zero vector (and the fact that the eigenvector must be non-zero), we cannot find an eigenvector of with eigenvalue equals 0. If and then the nullspace of contains eigenvectors of .
- This seems to be a mistake of Gilbert Strang in Introduction to Linear Algebra, 4th edition, page 286, sentence: “Projections have and “.