Why are the eigenvalues for reflection across a line in $R^3$ 1,-1,-1? I don't fully understand why this is the case. You have a reflection along the line in question which would be itself and subsequently be eigenvalue 1. Tracking so far. Then you'd have some other vector perpendicular to the line in $R^3$ and you reflect it and you get -1. Is the third -1 because you have two vectors forming a plane orthogonal to the line and those two vectors are a basis for that plane? If so, why are they both negative? Why are there necessarily two? Is this just a convention thing?
Reference video (moved to the time of the subject in question): https://youtu.be/oUWRut4ssDQ?t=527
 A: Your hunches are on the right track and it sounds like you should just review some definitions.
When you reflect across a line, any vector on the line (a subspace of dimension 1) is unchanged (i.e. multiplied by 1), and any vector in the plane perpendicular to the line (a subspace of dimension 2) is turned into its opposite (i.e. multiplied by -1).
In general:
An eigenvector of a transformation $T$ is a nonzero vector $v$ with the property that $Tv$ is a scalar multiple of $v$, i.e. $Tv=\lambda v$ for some scalar $\lambda$. We call $\lambda$ the eigenvalue associated with $v$. For a particular eigenvalue, the corresponding eigenvectors (with 0 included) form a subspace. We call such a subspace an eigenspace.
If our overall vector space is finite-dimensional (like $\Bbb R^3$), then the equation $\det(T-\lambda I)=0$ (which expresses the fact that $Tv=\lambda v$ for some nonzero vector $v$ - see this video for an explanation) gives us a polynomial equation whose roots are the eigenvalues of $T$. It turns out that the multiplicity of such a root is equal to the dimension of the corresponding eigenspace. So when we list eigenvalues with repetition, the multiplicity corresponds to that of these polynomial roots as well as to the dimension of the eigenspaces.
A: A geometric intuition: Pick an orthonormal basis containing a vector on your line as one of the generators. It's clear that this line is preserved, so it has eigenvalue 1, by definition of your reflection. For the rest, notice that that orthogonal projection onto the complementary plane takes your reflection in a line down to a reflection through the origin in this plane. This is the map $(x,y) \mapsto (-x,-y)$, which inverts both of the other two basis vectors in our chosen basis. Thus the other two eigenvalues are both $-1$.
This argument extends to the planar symmetry case, which in the comments you are confused about, by swapping the role of the line and the plane. Now it's the plane which is fixed, so you have $2$ eigenvectors with eigenvalue $1$, given by an orthonormal basis for that plane. The orthogonal line to this plane is reflected, by definition of your map, so it's eigenvalue must be $-1$.
