b) I mean intuitively, I feel like the eigen value should just be -1. I am just not sure how to reason it out... Is it because that an eigen value shuld satisfy $\textbf{A}v=\lambda v$ ? So if I want the two normals to be equal to each other, $\lambda$ must equal 1? c) Well I feel like I would get another vector with opposite coordinates but I am not sure...

@FutureMathperson You are right: you expect to find $-1$ in that case because the vector is mapped to its opposite vector. An eigenvector satisfies $\textbf{A}v=\lambda v$ and through reasoning, you find that for a normal $\bf n$, you have $T(\bf n)=- \bf n$ so... ($v = \bf n$ and $\lambda = -1$).

I see you corrected the matrix so I'll delete that part of my answer.

Yes sorry. I forgot to mention that part. Wolfram alpha tells me that the other two eigenvectors are both equal to 1. I'm trying to reason out why though? I feel like reflecting a vector would reverse the sign somehow...

Have you read through my answer? Most vectors you reflect are not mapped to a vector parallel to the initial vector, but to some other vector. However, there are a few special cases: consider vectors normal to the plane (part b) and vectors lying in the plane (part c). After reflection, they end up being b) flipped, c) unchanged - you see?

@FutureMathperson I added a bit more detail in my answer.

Ohh! I see. So the eigen value would just be 1 because any point on the plane would end up mapping to itself so to find a scalar multiple, the eigen value would simply be 1.

@FutureMathperson Indeed, for part c! So the only vectors which are mapped to scalar multiples (parallel to itself) after reflection about a plane are normal vectors ($\lambda = -1$; part b) and vectors in the plane ($\lambda = 1$; part c).

For part d, there is only 1 possible normal to a plane so there is only 1 linearly vector one there. Since I am spanning $\mathbb{R^3}$, would I have 3 linearly independent vectors in total?

Yes: 1 for the normal and since a plane is two-dimensional, 2 in the plane; makes 3 in total.

How would that help with the eigenspace though? As far as I know, the eigenspace is space generated by the eigenvectors corresponding to the same eigenvalue. Wouldn't the eigen space just be $\text{span{-1,1,1}}$?

Yeah the conversation was becoming long.

Hi there. The eigenspace corresponding to an eigenvalue is the set of all eigenvectors with that eigenvalue. Now parts b and c should've led you to finding out that the only two eigenvalues are -1 (corresponding eigenvectors are normal to the plane) and 1 (corresponding eigenvectors are in the plane).

You give the eigenspaces as subsets spanned by linearly independent eigenvectors; for b (normal vectors, eigenvalue -1) this is the set spanned by 1 normal; take e.g. (1,2,-1) - do you agree?

Yes I agree.

Then for the eigenspace corresponding to eigenvalue 1, you expect the space to be two-dimensional (since a plane is two-dimensional).

That means you need to find two linearly independent vectors lying in the plane. Once you have found two, the subset spanned by these two form the eigenspace

It's it 3 dimensional since I am working with 3 variables?

Also for the normal, I get the opposite signs from you, i.e. (-1,-2,1). Is it because the eigen value is -1?

'The' normal is not unique, any (non-zero) scalar multiply of (1,2,-1) is a normal to the plane, so (-1,-2,1) and (5,10,-5) are all normals.

Note that (1,2,-1) is mapped to (-1,-2,1) and vice versa :-)

you can pick any to span the eigenspace

Yes that makes sense. Thank you very much!

Alright, you're welcome!