# Eigenvalue and space for orthogonal projection

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.

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 $$\lambda=1.$$

• That makes sense. But why is $P_W^2=P_W$? Does that go for $P_W^n=P_W$ - I can't find anything about that in my text book.
– mhj
Jun 3, 2019 at 20:18
• It's just one of the main properties (sometimes a definition) on a projection. Here's a link that might help: math.stackexchange.com/questions/1574758/…
– cmk
Jun 3, 2019 at 20:20
• I see it now. I think it makes perfectly intuitive sense, when thinking on vectors. If you project v down on W, you get p. If you try to project p on W again, of course you will get the same vector.
– mhj
Jun 3, 2019 at 20:24
• And the reason it also gives $\lambda^2v$ is because of: $\left(P_{W} \circ P_{W}\right) v=P_{\mathbf{w}}\left(P_{\mathbf{w}}(v)\right)=P_{\mathbf{W}}(\lambda v)==\lambda P_{W}(v)=\lambda \lambda V=\lambda^{2} v$
– mhj
Jun 3, 2019 at 20:42
• That is correct!
– cmk
Jun 3, 2019 at 20:43

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., $$v=v_2\in W^\perp$$.