Theorem. An affine map from N-dimensional Cartesian space to itself is an affine transformation if and only if the associated N \times N matrix is an invertible matrix. In this case the inverse map is an affine transformation whose associated matrix is the inverse of the matrix associated with the original affine transformation.
Proof segments for the preceding theorem:
If f(x) = A x + b and A is an invertible matrix, then the map f is inverted by
If f(x) = A x + b is a bijective affine map, then A must be an invertible matrix.
Theorem. If p_{0}, p_{1}, ... p_{n} are barycentrically independent points of n-dimensional Euclidean space, i.e., none is a barycentric combination of the others, and q_{0}, q_{1}, ..., q_{n} are any points, then there is one and only one affine map f on n-dimensional space for which f(p_{0}) = q_{0}, f(p_{1}) = q_{1}, ..., f(p_{n}) = q_{n}. This map is an affine transformation if and only if q_{0}, q_{1}, ..., q_{n} are barycentrically independent.
Show that an affine transformation of the plane carries a parallelogram to a parallelogram.
Show that there is one and only one affine transformation of the plane carrying a given parallelogram to another given parallelogram in a given vertex-matching way.
Show that any affine transformation of the plane carries the point where the diagonals of a given parallelogram meet to the point where the diagonals of the image parallelogram meet.
Definition. If p_{0}, p_{1}, p_{2}, and p_{3} are barycentrically independent, then the tetrahedron with vertices p_{0}, p_{1}, p_{2}, and p_{3} is the set of all barycentric combinations of those vertices. Show that an affine transformation of 3-dimensional space must always carry a tetrahedron to a tetrahedron.