Future Math person

Yes that makes sense. Thank you very much!

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

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

Yes I agree.

Yeah the conversation was becoming long.

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 $\text{span{-1,1,1}}$?

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?

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.

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...

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...

I got it. Never mind.

I am not sure how I would apply it here... I can see the result directly but I am not sure how to reach that point.

Well it's basically how I would use the remainder theorem really.