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# Now 0 sv2 forces s 0 since v w i1 i2 i3 combinatorial

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Unformatted text preview: ince v w I.1, I.2, I.3 Combinatorial Structures is 2 nonzero vector, and so has nonzero length. Since w = sv, we have w = 0v = 0. a 14 I.4, I.5, I.6 Unlabelled structures FS: Part A.1, A.2 3 21 II.1, II.2, II.3 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 26 9 Nov 2 Comtet74 Handout #1 (self study) Labelled structures I (c). Yes. YouII.5, II.6 set u = (1, 0, 0)Labelled structures II , and w = (0, 0, 1). These are distinct and none is can , v = (0, 1, 0) 4 28 II.4, the zero vector, and Combinatorial u · vCombinatorial + (0)(1) + (0)(0) = 0, u · w = (1)(0) + (0)(0) + (0)(1) = 0, note that = (1)(0) 5 Asst #1 Due Parameters and vOctw = (0)(0) + parameters (0)(1) = 0. So they are orthogonal. · 5 III.1, III.2...
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