By Hefferon J.

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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.