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Unformatted text preview: Notes on linear algebra (Monday 17¢h October, 2016, 23:10) page 202 (here, we have renamed the indices a and o as x and y). (The reason for rewriting D this way was to get rid of the letters a and o; indeed, we are going to reuse these letters for vectors.) Proof that D contains the zero vector: Recall that the zero vector of IR3 is 03x1 2 (0,0,0)T. This vector 03x1 has the form (a, 0,2a —|— o)T for some 1a,?) E IR (namely, for a = 0 and o = 0). Thus, it belongs to D (by the definition of D). In other words, D contains the zero vector. Proof that D is closed under addition: Let a E D and w E D. We must prove that o + w E D. We have n E D = {(x, 0,231: + y)T | any E R} (by (157)). In other words, n has the form (x, 0,23: -l— y)T for some x,y E IR. Fix two such x,y, and denote them by acmyv. (I do not want to simply call them x,y, because I will introduce two other such 3:, y shortly.) Thus, xv, ya are real numbers and satisfy a = (xv, 0,235., + yv)T. We have to E D = {(x,0,2;r +y)T | x,y E IR} (by (157)). In other words, to has the form (x, 0,22: + y)T for some any E R. Fix two such x,y, and denote them by xw, yw. Thus, xw, yw are real numbers and satisfy to = (xw, 0, wa + yw)T. From I) = (xv,0,2xz, + yU)T and w = (xw,0,2xw + yw)T, we obtain '0 + w = (x... (1.2.x, + yaT + (am. 0.22:.“ + wa r x 'T' ...
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