# 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...

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.

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?

Geometrically, eigenvalues are the scaling factors by which particular vectors are scaled when multiplied by the respective matrix. Since the projection matrix projects a vector to its column space (or you can think of some nonzero subspace $$W$$), one such type of vector is the vector that is already living in that subspace. (Projecting a vector to its plane/subspace does not change it right?) However, when doing so, the length doesn't change, so one eigenvalue must be $$1$$.

What about the vector that is perpendicular to $$W$$? Where will it go after the projection? Yes, it goes to the origin and hence the second eigenvalue must be $$0$$.

Although this approach does not prove anything, it gives a nice neat geometric intuition.