Eigenvalue and space for orthogonal projection

Question

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 $W^{\perp}$.

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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.

Answer 1

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 $\lambda=1.$

Answer 2

Orthogonal projection means that for $v\in V$, we can write $v=v_1+v_2$ with $v_1\in W$ and $v_2\in W^\perp$ and that then $P_W(v)=v_1$.

The eigenspace for eigenvalue $1$ consists of those vectors $v\in V$ for which $P_W(v)=v$. According to the above decomposition, this is equivalent to $v=v_1+v_2=v_1$, i.e., $v\in W$.

The eigenspace for eigenvalue $0$ consists of those vectors $v\in V$ for which $P_W(v)=0$, in other words, $v_1=0$, i.e., $v=v_2\in W^\perp$.