MATA22Y 2018, Tutorial #1, Lucas Ashbury-Bridgwood. oce hours: Mondays 1011, IC 404. assignments & sol posted Fridays on Quercus/modules

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1 MATA22Y 2018, Tutorial #1, Lucas Ashbury-Bridgwood 1 today tutorial info A1 2 tutorial info (0:100:20) Lucas Ashbury-Bridgwood lucas.ashbury.bridgwood@mail.utoronto.ca math graduate student studying functional analysis, e.g. LOT of linear algebra!) C -algebras (uses A oce hours: Mondays 1011, IC 404 assignments & sol posted Fridays on Quercus/modules quizzes (5) based on assignments, lecture, readings every other week weeks 3, 5, 7, 9, 11 so rst quiz next week! 50 min long written second half of tutorial assignments not graded, but you need to do them for the quizzes/tests in tutorials, I will help with the assignments, and we will do quizzes when they occur optional tutorial website: lucasbridgwood.wordpress.com/mata22/ post tutorial notes and possibly other stu 3 A1 3.1 linear combination solving (0:200:30) (1.1 #25) Find c such that [13, 15] is a linear combination of [1, 5] and [3, c] This means solving [13, 15] = r [1, 5] + t [3, c] for r, t Doing this shows what assumptions we need to make on c for it to work, giving us our c 1

2 13 = r + 3t and 15 = 5r + ct This implies r = 13 3t (*) This implies 15 = 5 (13 3t) + ct = 65 15t + ct This implies 80 = (c 15) t This is solvable, and hence t, and therefore r by (*), is obtained, provided c 15 QUESTION: Which is the answer? All c 15, or c = 15? All c 15, because if c 15, we can solve for r, t, therefore showing the required linear combination 3.2 True/false (0:300:40) (1.1 #39) (f) The span of any two nonzero vectors in R 2 is R 2 This means if v, w 0, then always span (v, w) = {rv + tw : r, t R} = R 2 False QUESTION: What is a counter-example? v = w = [0, 1], then span (v, w) = spanv = {[0, r] : r R} R 2 NOTE: Being able to come up with counter-examples is essential! (g) If v, w 0 in R 2 with v w (not parallel), then span (v, w) = R 2 True Proof at end if time (or on notes posted on tutorial website) (h) The span of any three nonzero, nonparallel vectors in R 3 is R 3 False QUESTION: What is a counter-example? span ([1, 0, 0], [0, 1, 0], [1, 1, 0]) = {[r, t, 0] : r, t R} R 3, and they are mutually nonparallel 2

3 3.3 Finding vector perpendicular to given two (0:400:50) (1.2 #16) Given u = [ 1, 3, 4], v = [2, 1, 1], nd a nonzero vector perpendicular to both u, v Need to nd r, s, t with [r, s, t] u, [r, s, t] v = 0 r + 3s + 4t = 0 2r + s t = 0 Two equations with 3 variables, can't solve (1 too many variables) Instead, set one variable to anything nonzero and solve two equations in two variables Set t = 1 (doesn't have to be 1, anything nonzero, and could use r or s too) Then r + 3s = 4 and 2r + s = 1 This implies r = 4 + 3s This implies 1 = 2 (4 + 3s) + s = 8 + 7s = s = 1 = r = 1 So [1, 1, 1] is such a vector QUESTION: Why can't set set t = 0? (we said anything nonzero) The solution will always be zero, and we need a nonzero vector e.g. 2 ( r + 3s = 0) + (2r + s = 0) gives 6s = 0 = s = 0 = r = Magnitude, dot product proof (0:501:00) (1.2 #43) Prove v + w, v w are perpendicular i v = w v + w v w (v + w) (v w) = 0 v v v w + w v w w = 0 v 2 v w + v w w 2 = 0 v 2 w 2 = 0 v 2 = w 2 v = w QED QUESTION: Why does v 2 = w 2 = v = w instead of v = ± w? v, w 0 always by norm properties, so taking square roots we know there is no negatives 3

4 3.5 BREAK (1:00-1:10) 3.6 Right triangle midpoint proof (1:101:20) (1.2 #46) Prove that the midpoint of the hypotenuse of a right triangle is equidistance from the three vertices DRAW PICTURE By denition of being the midpoint, x = 1 (v + w) = y 2 So x = y z = y + v = 1v 1w + v = 1 (v w) So z = x (and hence we are done) i 1 v w = 1 v + w 2 2 But v ± w 2 = (v ± w) (v ± w) = v v ± 2v w + w w = v 2 + w 2 since v w by assumption of right triangle Hence v + w = v 2 + w 2 = v w, and we are done QED 3.7 Quadrilateral to parallelogram (1:201:35) (A1 #5) Show the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram DRAW PICTURE To show the gure is a parallelogram, it is equivalent to show that the opposite sides have equal magitude, i.e. u 1 = u 3 and u 2 = u 4 Consider showing u 1 = u 3 By 1.2 #43, u 1 = u 3 (u 1 + u 3 ) (u 1 u 3 ) = 0 Using the midpoints, u 1 + u 3 = ( 1 v v ( 2 2) + 1 v v ) 2 4 = 1 (v v 2 + v 3 + v 4 ) But the sum of the v j starts and ends at the same place, so the sum is 0 4

5 Hence u 1 + u 3 = = 0 = (u 1 + u 3 ) (u 1 u 3 ) = 0, completing this case PRACTICE: Show u 2 = u 4 (same argument) As noted this completes the proof QED 3.8 Rhombus midpoint intersection (1:351:50) (A1 #6) Show that the diagonals of a rhombus bisect each other DRAW PICTURE v = ry + tx for some r, t > 0 Hence v = r (v w) + t (v + w) = (r + t) v + (t r) w This implies (1 r t) v = (t r) w If t r = 0, i.e. t = r, then (1 2r) v = 0 Since v 0, then 1 2r = 0, i.e. t = r = 1/2 This proves the claim If t r 0, then we can divide to get 1 r tv = w t r This means that v w QUESTION: Is this possible? No, clearly a rhombus can't be made where all the sides are parallel This shows that it must be true that t r = 0, where we know the claim holds QED NOTE: Getting better at solving problems like these is just practice 3.9 Extra proof of span in R 2 (1:502) If v, w 0 in R 2 are not parallel, then span (v, w) = R 2 Must show that given [x, y] R 2, we can solve [x, y] = rv + tw for r, t Let v = [, v 2 ], w = [w 1, w 2 ] So x = r + w 1 t and y = v 2 r + w 2 t Since v 0, one of, v 2 is nonzero 5

6 Without loss of generality, assume 0 (the argument is similar if v 2 0) Then x = r + w 1 t = x w 1t = r Note that x and w 1 are knowns, and t is unknown So if we nd t, then we are done This implies that y = v 2 x w 1t + w 2 t = v 2x + If w 2 w 1 v 2 0, then t = (y v 2 x/ ) / Suppose instead that w 2 w 1 v 2 = 0 Then w 2 = w 1 v 2 Since w 1 = w 1, therefore w 1 [, v 2 ] = [w 1, w 2 ] ( w 2 w 1 v 2 ) t ( w 2 w 1 v 2 ) and we are done This means that v w, a contradiction to our earlier assumption Hence it must be that w 2 w 1 v 2 0, in which case we know the claim holds QED 6