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Example text

F (en )]. 7) Proposition: The map f from IFn to the vector space X is linear if and only if f = [f (e1 ), f (e2 ), . . , f (en )]. In other words, L(IFn , X) = {[v1 , v2 , . . , vn ] : v1 , v2 , . . , vn ∈ X} (≃ X n ). 3) the map at : IRn → IR : x → a1 x1 + a2 x2 + · · · + an xn = [a1 , . . , an ]x, and, in this case, at ej = aj , all j. 8) at = [a1 , . . , an ] = [a]t . Notation: I follow MATLAB notation. , [V, W ] denotes the column map in which ﬁrst all the columns of V are used and then all the columns of W .

5 3 0 15jan03 2 −1 1 1 3 2 5 1 0 = id3 ? c 2002 Carl de Boor The pigeonhole principle for square matrices 41 The pigeonhole principle for square matrices We are ready for a discussion of our basic problem, namely solving A? = y, in case A ∈ IFm×n , hence y ∈ IFm . When is A 1-1, onto, invertible? We answer all these questions by applying elimination to the augmented matrix [A, y]. 5)Corollary, all its columns must be bound. , m ≥ n. Further, if m = n, then, by the time we reach the last column of [A, y], there is no row left to bind it.

Vn ]a = v1 a1 + v2 a2 + · · · + vn an . Thus, IFm×n = L(IFn , IFm ). Thus, a matrix V ∈ IFm×n is associated with two rather diﬀerent maps: (i) since it is an assignment with domain m × n and values in IF, we could think of it as a map on m × n to IF; (ii) since it is the n-list of its columns, we can think of it as the linear map from IFn to IFm that carries the n-vector a to the m-vector V a = v1 a1 + v2 a2 + · · · + vn an . From now on, we will stick to the second interpretation when we talk about the domain, the range, or the target, of a matrix.