If $P$ is the matrix that projects $R^n$ onto a subspace $S$, explain why every vector in $S$ is an eigenvector, and so is every vector in $S^{\perp}$. What are the eigenvalues (Note the connection to $P^2 = P$, which means that $λ^2 = λ$ .)
I feel like I'm missing something from the question and therefore have no idea where to begin.
The only thing I can think of to get started is that if $S$ is a proper subspace of $R^n$ then, by the Invertible Matrix Theorem, 0 is an eigenvalue which also holds since $0^2=0$ from the given hint.
This is Review Question 5.6 Linear Algebra and Its Applications Fourth Edition by Gilbert Strang
As someone else stated, this works for an orthogonal projection. Think about it geometrically. Let $\bf{v}\in S$. Then the projection of $\bf{v}$ onto $S$ is just $\bf{v}$ itself. Thus $P\bf{v}$ $=\bf{v}$ $=1\bf{v}$, so $\bf{v}$ is an eigenvector with corresponding eigenvalue $\lambda =1$. Similarly, if $\bf{u}\in S^\bot$, then the projection of $\bf{u}$ onto $S$ is $0$. So $P\bf{u}$ $=\bf{0}$ $=0\bf{u}$, so $\bf{u}$ is an eigenvector with corresponding eigenvalue $\lambda =0$.