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By de Boor C.

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Example text

For this, consider the matrix QW , with the columns of W := [w1 , . . , wk ] the so-called Newton polynomials wj : t → 15jan03 h

Vm 15jan03 c 2002 Carl de Boor 28 2. Vector spaces and linear maps are elements of ran A, then there must be a sequence w1 , w2 , . . , wm in X with Awj = vj , all j. Since X is a vector space, it contains W a for arbitrary a ∈ IFm , therefore the corresponding linear combination V a = [Aw1 , Aw2 , . . , Awm ]a = (AW )a = A(W a) must be in ran A. In other words, if V ⊂ ran A, then ran V ⊂ ran A. Hence, if we wonder whether A is onto, and we happen to know an onto column map [v1 , v2 , . . , vm ] = V ∈ L(IFm , Y ), then we only have to check that the finitely many columns, v1 , v2 , .

18 Here are three questions that can be settled without doing any arithmetic. Please do so. (i) Can both of the following equalities be right? , numbers α, β for which e1 = (1, 3)α + (2, 5)β), given that AV := −5 3 2 −1 1 3 2 5 = id2 ? (iii) How does one conclude at a glance that the following equation must be wrong? −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?

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