HW8.pdf - HW 8 Solutions MATH 110 5.4#1(a False the zero...

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HW 8 Solutions, MATH 1105.4 #1 (a) False; the zero subspace and the entire vector space are always invariant.(b) True.Theorem 5.21. (c) False. Take the one-dimensional vector space andT=I. Then any twononzero vectors generate the entire space, but may be different. (d) False. TakeA=0100.The cyclic subspace generated bye2is all ofR2, but the cyclic subspace generated byT(e2) =e1is just the span ofe1. (e) True. Cayley-Hamilton theorem, takegto be the characteristicpolynomial. (f) True. See Exercise 19. (g) True. Take bases of each invariant subspace, andthen take the union.5.4 #6ab (a)z= (1,0,0,0),T(z) = (1,0,1,1), andT2(z) = (1,-1,2,2).These three vectors arelinearly independent.T3(z) = (0,-3,3,3) = 3T2(z)-3T(z). So{z, T(z), T2(z)}={(1,0,0,0),(1,0,1,1),(1,-1,2,2)}is a basis forT-cyclic subspace generated by the vectorz.(b)z=x3,T(z) = 6x, andT2(z) = 0.So, the cyclic subspace generated byzhas basis{x3,6x}.5.4 #13 Ifw=g(T)(v) for some polynomialg(t) =a0+a1t+...+antn, thenw=g(T)(v) =a0v+a1T(v) +...+anTn(v)span({v, T(v), ..., Tn(v)})W. AssumewW. SinceW=span({v, T(v), T2(v), ...}),wis a linear combination of a finite subset of{v, T(v), T2(v), ...}.

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