By Hefferon J.
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Symbolic Reasoning allied to dull A lgebra. the manager examples of such structures are Hamilton sQ uaternions, Grassmann sC alculus of Extension and Boole sS ymbolic common sense. Such algebras have an intrinsic price for separate certain examine; they are also important of a comparative examine, for the sake of the sunshine thereby thrown at the common thought of symbolic reasoning, and on algebraic symbolism specifically.
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The number of columns of a matrix of coefficients A of a linear system equals the number n of unknowns. A linear system with at least one solution has at most one solution if and only if the space of solutions of the associated homogeneous system has dimension zero (recall: in the ‘General = Particular + Homogeneous’ equation v = p + h, provided that such a p exists, the solution v is unique if and only if the vector h is unique, namely h = 0). But that means, by the theorem, that n = r. 27 There is little danger of their being equal since the row space is a set of row vectors while the column space is a set of columns (unless the matrix is 1×1, in which case the two spaces must be equal).
The calculation for span is also easy; for any x, y ∈ R4 , we have that c1 ·(cos θ−sin θ)+c2 ·(2 cos θ+3 sin θ) = x cos θ + y sin θ gives that c2 = x/5 + y/5 and that c1 = 3x/5 − 2y/5, and so the span is the entire space. 25 Yes. Linear independence and span are unchanged by reordering. 26 No linearly independent set contains a zero vector. 28 Each forms a linearly independent set if v is ommitted. To preserve linear independence, we must expand the span of each. That is, we must determine the span of each (leaving v out), and then pick a v lying outside of that span.
B) We must show that if a subset of the domain is convex then its image, as a subset of the range, is also convex. Suppose that C ⊆ Rn is convex and consider its image h(C). 1]} is a subset of h(C). Fix any member tˆ· d1 +(1− tˆ)· d2 of that line segment. Because the endpoints of are in the image of C, there are members of C that map to them, say h(c1 ) = d1 and h(c2 ) = d2 . Now, where tˆ is the scalar that is fixed in the first sentence of this paragraph, observe that h(tˆ· c1 + (1 − tˆ) · c2 ) = tˆ· h(c1 ) + (1 − tˆ) · h(c2 ) = tˆ · d1 + (1 − tˆ) · d2 Thus, any member of is a member of h(C), and so h(C) is convex.