Reflection about a k dimensional Euclidian subspace of $\mathbb{E}_{n}$ My question:
When is a such a reflection orientation preserved? When is  orientation not preserved?
 A: Answer: Orientation is preserved when $n-k$ is even, and it is not preserved when $n-k$ is odd.
To show this, rotate the environment so the $k$-dimensional subspace aligns with the first $k$ axes. Choose a figure of the $n$ axis vectors, so they are orthogonal to each other.
The first $k$ vectors align with the $k$-dimensional subspace, so they are unchanged under the reflection. The remaining $n-k$ vectors are orthogonal to the subspace, so each one is mapped to its negative when reflected.
Orientation is preserved if the determinant formed by the original $n$ vectors (as rows or as columns) has the same sign as that of the $n$ image vectors. In our case, $n-k$ vectors have been replaced with their negatives and the remainder are unchanged. Replacing one vector with its negative in a determinant replaces the determinant with $-1$ times the determinant. Therefore, replacing $n-k$ vectors with their negatives changes the determinant with
$$(-1)^{n-k}$$
times the original determinant. The sign changed if and only if $n-k$ was negative, and the conclusion follows.

Examples confirm this. In 2D space, reflection in a line changes orientation while reflection in a point does not. In 3D space, reflection in a plane or point changes orientation while reflection in a line does not.
A: An algebraic view of Rory's answer: if $B_{(n-k)\times n}$ is an orthonormal basis for the orthogonal complement of your subspace, then the reflection operator $R$ is given by
$$R = I - 2B^TB.$$
Then
$$\det R = \det[I_{n\times n} - 2B^TB] = \det[I_{(n-k)\times(n-k)} - 2BB^T] = \det[-I_{(n-k)\times(n-k)}] = (-1)^{n-k}$$
by the Sylvester determinant theorem, so the reflection is orientation-preserving if and only if $n-k$ is even.
