Recall from the study of linear algebra that if f is a linear map from R^{n} to itself and v =
| { | v_{1}, ..., v_{n} | } |
is a linear basis of R^{n}, then the matrix of f with respect to the basis v is the n \times n matrix M whose j-th column, for 1 <= j <= n, is the column of coordinates of f(v_{j}) relative to v, i.e.,
| f(v_{j}) = SUM_{i = 1}^{n}[ M_{ij} v_{i} ], 1 <= j <= n . |
Definition. If p =
| { | p_{0}, ..., p_{n} | } |
is an affine basis of R^{n} and f is an affine map from R^{n} to itself then the affine matrix of f with respect to the affine basis p is the (n + 1) \times (n + 1) matrix M whose j-th column, for 0 <= j <= n, is the column of barycentric coordinates of f(p_{j}) relative to p, i.e.,
| f(p_{j}) = SUM_{i = 0}^{n}[ M_{ij} p_{i} ] with SUM_{i = 0}^{n}[ M_{ij} ] = 1 , 0 <= j <= n . |
Proposition. If p is a point of R^{n} having barycentric coordinates (x_{0}, ..., x_{n}) relative to the affine basis p and if f is an affine map having matrix M relative to p, then f(p) is the point of R^{n} having barycentric coordinates (y_{0}, ..., y_{n}) relative to p where the vectors x and y, when regarded as columns, are related by the formula y = Mx.
Proof. Because f preserves barycentric combinations and p = x_{0} p_{0} + ... x_{n} p_{n} with x_{0} + ... + x_{n} = 1, it follows that
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One needs to check that the last line is indeed a barycentric combination of the p_{i}, i.e., that y_{0} + ... + y_{n} = 1. This follows from the fact that y is the x-barycentric combination of the (weight 1) columns of M
Show that the map phi:R^{2} --> R^{3} given by (x_{1}, x_{2}) -> (x_{1}, x_{2}, 1 - x_{1} - x_{2}) is an affine map.
Conclude from the first exercise that if tau is translation of R^{2} by the vector a = (a_{1}, a_{2}), then phi(tau(x)) = phi(x) + ã where ã is the weight 0 triple (a_{1}, a_{2}, a_{3}) with a_{3} = - a_{1} - a_{2}.
(Continuing) Find the affine matrix of the translation tau.
Find the affine matrix of the half turn of R^{2} about the point c, i.e., the affine transformation x -> 2 c - x.
Show that if M is the affine matrix of the affine transformation f(x) = U x + v of R^{2}, then det M = det U.