Let $V$ be an inner product space over a field $\mathbb{K}$, and let $W \neq V$ be a subspace in $V$ of finite dimension $> 0$.
Let
$P_W : V \to W$
be the orthogonal projection on $W$.(1) Show that $1$ is an eigenvalue for $P_W$, and the corresponding eigenspace equals $W$ .
(2) Show that $0$ is an eigenvalue for $P_W$, and the corresponding eigenspace is $W^{\perp}$.
I'm really stuck here.
I know that $P_W(v)=\lambda v$.
and the eigen space: $E_{P_W}(\lambda)=\{v \in V | P_W(v)=\lambda \cdot v\}=\operatorname{ker}\left(P_W-\lambda \cdot \operatorname{Id}_{V}\right)$
But how do I relate this to the orthogonal projection. I mean I hardly have any information about the two vector spaces, their basis, the inner product, any matrix representations, etc.
Best Answer
I'll answer the part regarding finding the eigenvalues, as it looks like the other part regarding the eigenspaces has been answered:
If $P_W$ is an orthogonal projection, then in particular, $P_W^2=P_W.$ So, if $P_Wv=\lambda v,$ then $P_W^2v=\lambda v.$ and $P_W^2v=\lambda^2 v.$ That is, $\lambda =\lambda ^2,$ so $\lambda=0$ or $\lambda=1.$