# Find the eigenvalues of a projection operator

A projection operator $P$ is defined as $P^2$=$P$. Use this definition to find the eigenvalues of this operator.

In this question is it necessary to define what the projection operator is? And won't the eigenvalue just be zero?

Let $\lambda$ be an eigenvalue of $P$ for the eigenvector $v$. You have $\lambda^2 v = P^2 v = P v = \lambda v$. Because $v \neq 0$ it must be $\lambda^2 = \lambda$. The solutions of the last equation are $\lambda_1 = 0$ and $\lambda_2 = 1$. Those are the only possible eigenvalues the projection might have...
If $\mathbf x$ is an eigenvector of $P$ with eigenvalue $\lambda$, then $P(\mathbf x) = \lambda \mathbf x$. If $P^2 = P(P) = P$, then $P(P(\mathbf x)) = P(\lambda \mathbf x) = \lambda^2 \mathbf x$, but this must also equal $\lambda \mathbf x$. Therefore the eigenvalues of $P$ can only be members of your base field such that $\lambda^2 = \lambda$.
Assuming your base field is $\Bbb R$, there's one more value that satisfies this equation besides $\lambda =0$. What is it?
The eigenvalues are $0$ and $1$. Indeed, we know that for a projector $P$ defined on a vector space $E$, we have $$E= \ker P\oplus \operatorname{im}P=\ker P\oplus \ker(I-P)$$ $\ker P$ is the eigenspace associated with the eigenvalue $0$, $\ker(I- P)$ the eigenspace associated with the eigenvalue $1$. As $E$ is the direct sum of these eigenspaces, we have all eigenspaces, and all eigenvalues.