Eigenvalues and eigenvectors for orthogonal projection I've been self-teaching myself linear algebra using Treil's Linear Algebra Done Wrong and I'm currently stumped on a problem and not sure how to start it. Here is the problem:

If someone could give me some hints on starting it, it would be greatly appreciated
 A: Hint: An orthogonal projection is what we call idempotent, meaning that applying the map twice to a vector is the same as applying the map once. Thus if we have an eigenvector $\vec{v}$ with 
$$
P\vec{v}=\lambda\vec{v},
$$
then we have both
$$
P^2\vec{v}=P\lambda\vec{v}=\lambda^2\vec{v}
$$
and 
$$
P^2\vec{v}=P\vec{v}=\lambda\vec{v}.
$$
Thus $\lambda^2=\lambda$, so $\lambda=0$ or $\lambda=1$. The case $\lambda=1$ is the case when the orthogonal projection leaves a vector unchanged, which occurs exactly when the vector lies in the space we are projecting onto. What could $\lambda=0$ represent? 
A: We have $V=E\oplus E^\perp$, so every vector $x$ in $V$ can we written uniquely as $x=y+z$ with $y\in E$ and $z\in E^\perp$. Letting $T$ be the orthogonal projection, we have that, if $\lambda \neq 0$:
$$T(x)=\lambda x \iff x=\lambda x + 0 \iff (\lambda -1)x=0$$ 
Added: By definition of orthogonal projection: $x=Tx + y$, with $Tx\in E$ and $y \in E^\perp$, and this is the unique way of writing $x$ as a sum of vectors in $E$ and $E^\perp$. Now, if $\lambda \neq 0$, since $\lambda x \in E$ it follows $x\in E$, and by uniqueness $x$ has to be written as $x=(\text{something in $E$}) + (\text{something in $E^\perp$})$. Since $x$ itself is in $E$, by uniqueness the only way to do this is $x=x+0$. The orthogonal proyection is thus $x=Tx=\lambda x$
Handling the cases $\lambda =0$ and $\lambda \neq 0$ separately might be useful (it requires an analogous reasoning).
