saying that x_{1}is an eigenvector of T with eigenvalue 1 means that Tx_{1}= x_{1}. verifying this is just a routine computation:

now, for any real number λ, we have: T(λx_{1}) = λ(Tx_{1}) (since T is linear)

= λx_{1}, so any vector λx_{1}is also an eigenvector for T (with eigenvalue 1).

this means that the line L = {λx_{1}: λ in R} is an invariant subspace of T: T(L) ⊂ L (in fact, since T has non-zero determinant, T(L) = L).

geometrically, this means that T does not change the line L at all, in other words (since we know that T^{2}= I) T reflects the plane through the line L (we know it is a reflection and not a rotation of 180 degrees since det(T) = -1).

personally, i find the use of the trigonometry somewhat distracting. the underlying algebraic fact is this: if R is a matrix with det(R) = 1, and R^{n}= I, then R is a rotation "1/n-th of the way around a circle". of course, to generalize this entirely, one must allow n to be any real number greater than 1 (which is where the trigonometry comes in, since it becomes difficult to speak of non-integral powers of a matrix). but the important fact is that given a rotation matrix R, then there is a corresponding reflection matrix T, where T = RC, where C is the matrix:

geometrically, this means that for any rotation of the plane, if we pick the line left invariant by the associated reflection T, as the "new y-axis", and the line perpendicular to that line as the "new x-axis", then in these "new coordinates" T is just reflection about the y-axis, and in these "new coordinates" the matrix for T is simply C. so C is in some sense "the canonical reflection", since any other reflection T is just C followed by a rotation R.

in geometry, the points of the plane are all "uniform", the only "distinguished" point is the origin. the "coordinates" we introduce, in order to be able to use linear algebra, are somewhat arbitrarily chosen: we "pick" a direction and call it the x-axis, and then call the perpendicular direction the y-axis. this choice of axes is somewhat arbitrary, the geometry holds true no matter how we "label" our points.

now orthogonality, or "right-angled-ness" is a geometric property. it shouldn't depend on the coordinate system we are using. but to define it in the context of vector spaces (so as to define linear transformations that PRESERVE it), we need a way of defining orthogonality on vector spaces. this is normally done by defining an inner product (the USUAL inner product is the "dot product" in euclidean n-space). then T preserves orthogonality if Tu.Tv = u.v. note that we get something "for free" here, if we define the length of the vector u as:

|u| = √(u.u),

then if T preserves orthogonality: |Tu| = √(Tu.Tu) = √(u.u) = |u|, so T preserves lengths as well.

and we can go even further: we can define the DISTANCE between the vectors u and v as: |u - v|. then we have:

|Tu - Tv| = |T(u - v)| = √(T(u - v).T(u - v)) = √((u - v).(u - v)) = |u - v|, so T preserves distances as well.

in the 2x2 case (the euclidean plane) if T =

[a b]

[c d]

then T(x,y) = (ax + by,cx + dy). if T preserves the dot product, then a^{2}+c^{2}= 1, and b^{2}+d^{2}= 1, and ac+bd = 0

now consider the transpose of T:

[a c]

[b d].

note that TT^{T}=

[a^{2}+b^{2}ac+bd]

[ac+bd c^{2}+d^{2}] = I

hence det(TT^{T}) = det(T)det(T^{T}) = det(T)^{2}= 1, so evidently det(T) is a square root of 1, so either 1 or -1. on the other hand, if det(T) = 1 or -1, running the argument in reverse shows T preserves inner products.