Mathematics > EXAM REVIEW > final exam review Ohio State University MATH 2568 (All)

final exam review Ohio State University MATH 2568

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LINEAR ALGEBRA REVIEW SOLUTION 7. Let ~v be the sum of all the basis vectors found in the previous problem. (a) Show that B = {~v, A~v, A2~v} is a basis for R 3 . Solution: We already know, at th... is point, that the eigenvalues of A are λ = 0, 2, 5. Moreover, we have that w~ λ is a basis for the Eλ-eigenspace where w~ 0 =   3 2 −3   , w~ 2 =   −1 0 1   , w~ 5 =   0 −1 1   . So ~v = w~ 0 + w~ 2 + w~ 5 =   2 1 −1   A~v = Aw~ 0 + Aw~ 2 + Aw~ 5 = 2w~ 2 + 5w~ 5 =   −2 −5 7   A 2 ~v = 2 2w~ 2 + 5 2w~ 5 =   −4 −25 29   . Since det   2 −2 −4 1 −5 −25 −1 7 29   = 2((29)(−5) + (25)(7)) + 2(29 − 25) − 4(7 − 5) 6= 0, it follows that B is a linearly independent set of 3 vectors in R 3 . So B is a basis for R 3 . Alternatively, we we could have done the following. Assume x1~v + x2A~v + A 2 ~v = ~0. Write f(t) = x1 + x2t + t 2 = (t − η1)(t − η2). Then show f(A)~v = x1 +~v + x2A~v + A 2 ~v = (−η1)(−η2)w~0 + (2 − η1)(2 − η2)w~ 2 + (5 − η1)(5 − η2)w~ 5 and use the fact that w~ 0, w~ 2 and w~ 5 are linearly independent to show that f(t) = 0, i.e., x1 = x2 = x3 = 0. (b) Suppose A = TB where B = {~e1,~e2,~e3} is the standard basis. Find TB, the matrix for T with respect to B. Solution: TB = A means T(~x) =  A   x1 x2 x3   B   B . We want to find the matrix TB where TB is the matrix such that T(~x) =  TB   x1 x2 x3   B   B . We know TB has columns ~ci where ~c1 =   ↑ T(~v) ↓   B , ~c [Show More]

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