A translation of R^{n} is a function
of the form f(x) = x + a, where a is a fixed vector and x is a variable point. In fact, the f of this last formula is called ``translation by the vector a'' and is sometimes denoted by T_{a}.
Two subsets G and H of R^{n} are called parallel if one is a translate of the other, i.e., if there is a vector a such that T_{a}(G) = H. In this situation one sometimes writes H = G + a.
This general notion of parallelism for sets reduces in the special case where the sets are lines to the usual notion of parallelism for lines as the following proposition shows.
Proposition. If L_{1} is the line given parametrically by phi_{1}(t) = p_{1} + t v_{1} and L_{2} likewise the line given by phi_{2}(t) = p_{2} + t v_{2}, where p_{1} and p_{2} are given points and v_{1} and v_{2} are given non-zero vectors, then L_{1} and L_{2}, considered as subsets of R^{n}, are parallel if and only if there is a scalar c such that v_{2} = c v_{1}.
Proof. If v_{2} = c v_{1}, then
and L_{2} = T_{a}(L_{1}) where a = p_{2} - p_{1}. (Note that the scalar c must be non-zero since v_{2} is non-zero and, therefore, that phi_{1}(ct) parameterizes the same line as phi_{1}(t). ) Conversely, assume that L_{2} = T_{a}(L_{1}). Since p_{2} = phi_{2}(0) is a point of L_{2}, it must be of the form T_{a}(q) for some point q of L_{1}. Likewise, p_{2} + v_{2} = phi_{2}(1) is a point of L_{2}, and it must be of the form T_{a}(r) for some point r of L_{1}. Suppose that q = phi_{1}(u) and r = phi_{1}(u'). Then one has
and we see that v_{2} is a scalar multiple of v_{1}.
Proposition. If G, H, and J are subsets of R^{n} with G parallel to H and H parallel to J, then G is parallel to J.
Proof. If H = T_{a}(G) and J = T_{b}(H), then J = T_{b}(T_{a}(G)). Since the result of following T_{a} with T_{b} is the translation T_{a + b}, i.e., T_{b}(T_{a}(x)) = T_{a + b}(x) for any point x, one sees that J = (T_{b} \circ T_{a})(G) = T_{a + b}(G).
Proposition. Any two non-empty fibers of a linear function are parallel subsets of the function's domain. In fact, any given non-empty fiber of a linear function is the translate of the function's kernel by any member of the given non-empty fiber.
Proof. If M is an m \times n matrix, then f(x) = Mx is a linear function from R^{n} to R^{m}, and every linear function from R^{n} to R^{m} has this form. One proceeds by showing that every non-empty fiber of f is a translate of the kernel of f. Suppose that f^{-1}(b) is a non-empty fiber of f. Let a be any given point in f^{-1}(b). Then f(a) = b. The issue now is to establish that f^{-1}(b) = T_{a}(Ker f). This is a statement that two sets are equal. To prove that two sets are equal one may argue that each is a subset of the other. Suppose that x is given in T_{a}(Ker f). Then x = z + a where z is in Ker f and, therefore, f(z) = 0. Hence,
so x is in f^{-1}(b), and, therefore, T_{a}(Ker f) is seen to be a subset of f^{-1}(b). For the reverse inclusion suppose now that x is given in f^{-1}(b). Let w = x -a. Then
Hence, w is in Ker f, and since x = a + w = T_{a}(w), one sees that x is in T_{a}(Ker f). Thus, f^{-1}(b) is seen to be a subset of T_{a}(Ker f), and these two sets are equal since each is a subset of the other.
Review the daily handouts looking for statements that you do not understand, that you find wrong, or that you find lacking adequate justification in the context of the course so far. (The midterm test will be Wednesday, March 17.)
Web versions of these handouts are available at the URL
http://math.albany.edu:8000/math/pers/hammond/course/mat220/assgt/ .