Eigenvalues and eigenvectors of a reflection about a plane Consider the linear transformation $T: \mathbb{R^3 \rightarrow R^3}$ given by the reflection about the plane $\textit{P}:x+2y-z=0$. In other words, $T(\textbf{v})=\textbf{v}-2\text{proj}_n\textbf{v}$, where $\textbf{n}$ is any normal vector for $\textit{P}$ and $v\in \mathbb{R^3}$.
a) Find the standard matrix $A= [T]$ for $T$
That was easy. I used the standard basis vectors $e_1$ and projected them on the normal as follows:
$T(\textbf{$e_1$})=\textbf{$e_1$}-2\text{proj}_n\textbf{$e_1$}$
$T(\textbf{$e_2$})=\textbf{$e_2$}-2\text{proj}_n\textbf{$e_2$}$
$T(\textbf{$e_3$})=\textbf{$e_3$}-2\text{proj}_n\textbf{$e_3$}$
I obtained the vectors $T(\textbf{$e_1$})$,$T(\textbf{$e_2$})$,$T(\textbf{$e_3$})$ which I then wrote as columns. Thus, 
$A=[T]=$$
\left(
\begin{array}{ccc}
\frac{2}{3} & \frac{-2}{3} & \frac{1}{3} \\ 
\frac{-2}{3} & \frac{-1}{3} & \frac{2}{3}\\ 
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\ 
    \end{array}
\right)
$
b) If $\textbf{n}$ is any normal to $\textit{P}$, find $T(n)$ use the result to find an eigenvalue of $A$
Well if I use:
$T(\textbf{$n$})=\textbf{$n$}-2\text{proj}_n\textbf{$n$}$, I get the same normal but with opposite sign (i.e. $T(\textbf{$n$})$=$-\textbf{n}$) . How is that supposed to help me find the eigen value of $A$ though?
I mean I shouldn't be finding the characteristic polynomial and such. I feel like there should be a way to approach this with only this bit of information.
c) If $\textbf{v}$ is any vector in the plane $P$, find $T(v)$ using the fact that $T$ is a reflection about $P$. Use the result to find another eigenvalue of $A$.
What does this mean by "using the fact that $T$ is a reflection about $P$" ? How is that supposed to help me find another eigen value?
d) Find eigenspaces corresponding to eigenvalues obtained in $(b)$ and $(c)$ using the fact that $T$ is refection about $P$.
Same confusion. What does it mean by "using the fact that $T$ is a reflection about $P$" ? How is that supposed to help me find the eigenspaces?
 A: You could find the eigenvalues and eigenvectors algebraically, i.e. calculate the eigenvalues as the roots of the characteristic polynomial and solve a linear, homogeneous system per eigenvalue to find the corresponding eigenvector(s). Here however, they want you to use the geometrical interpretation of the reflection to find them by reasoning.
The basic question you should ask yourself is: 
When reflecting about a plane, which vectors are mapped to a vector parallel with the vector you reflected; i.e. to a scalar multiple of that vector?
You should try to picture this geometrically. The different sub-questions  guide you through this.

b) If $\textbf{n}$ is any normal to $\textit{P}$, find $T(n)$ use the result to find an eigenvalue of $A$

Take any normal to $P$ and find its reflection: you expect to find the opposite vector as a result, right? This means any normal $\bf n$ is reflected to the vector $\bf -n$. So any normal $\bf n$ is an eigenvector with eigenvalue...
You can try it out with a few normals such as $(1,2,-1)$ or $(-2,-4,2)$.

c) If $\textbf{v}$ is any vector in the plane $P$, find $T(v)$ using the fact that $T$ is a reflection about $P$. Use the result to find another eigenvalue of $A$.

If you take a vector lying in the plane, what do you expect will happen when you calculate its reflection with respect to that plane? It remains unchanged! This means any vector $\bf p$ in the plane is reflected to the vector $\bf p$ itself, so any vector in the plane is an eigenvector with eigenvalue... 

d) Find eigenspaces corresponding to eigenvalues obtained in $(b)$ and $(c)$ using the fact that $T$ is refection about $P$.

With the reasoning in parts (b) and (c), you should ask yourself how many eigenvectors you can find in that situation:


*

*a normal to the plane is not unique, but all normals are... so you only have 1 linearly indepedent one;

*how many  linearly independent vectors lying in the plane can you find?

