Notation: if v is a vector, then v* is its conjugate transpose and vv* is an outer product. If Q is a matrix, Q* is its conjugate transpose. C^m is the vector space over the complex numbers. The conjugate transpose is just the regular transpose ^T in R^m.
Say you have a vector v in C^m, then P_1 = (vv*)/(v*v) is the rank one projector onto the line: span(v). P_2 = (I - P_1) is the rank m - 1 complementary projector onto the hyperplane: ⊥span(v). Intuitively, P_1 isolates the component that exists in span(v), while the transform P_2 = (I - P_2) "shaves off" that component. If v has unit length, then P_1 = vv*, since the inner product v*v = 1. You can project onto more than just hyperplanes and lines. If you have an m by n matrix Q (n ≤ m) with pairwise orthonormal columns, then QQ* projects onto the space spanned by those columns. Notice the similarity between vv* and QQ*. Also, notice that QQ* = I if Q is a full basis for C^m.
What if you want to reflect a vector across a hyperplane ⊥span(v)? This is just a matter of inverting the component in span(v): P3 = (I - 2(P_1)). Here's an easy example of this in 3d space: imagine if the hyper plane is the xy-plane (z=0, x and y are free): the line orthogonal to this space is span(e3). The projector onto the xy-plane is the (I - [0, 0, e3]) = [e1, e2, 0]: which is just the identity matrix, except the last column is zero. (I - 2[0, 0, e3]) is just the identity matrix except the last entry in the diagonal negates the z component.
What are the eigenvalues of a transform that reflects a vector in C^m across a hyperplane? The components of the vector in ⊥span(v) hyperplane are untouched, while the component of the vector in span(v) is negated. Hence, the eigenvalues of reflector are all one except for a single negative one. The determinant is the product of all of these eigenvalues: negative one. Hence, the transform is unitary and the singular values are all 1. This is the solution to excercise 10.1 in Numerical Linear Algebra by Trefethen and Bau.