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Sol3 - SOLUTION TO ASSIGNMENT 3 MATH 251 WINTER 2007 1 Let...

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SOLUTION TO ASSIGNMENT 3 - MATH 251, WINTER 2007 1. Let T : V V be a nilpotent linear operator. Prove that if n = dim( V ) then T n 0. Show that for every n 2 there exists a vector space V of dimension n and a nilpotent linear operator T : V V such that T n - 1 6≡ 0. Proof. We prove the result by induction on n . Suppose that n = 1, we then have a linear map T : F F . T cannot be surjective (else it is an isomorphism), hence its image must be the zero subspace, which shows that T = T 1 = 0. Now for the induction step. Again, T : V V cannot be surjective, because it would then be an isomorphism and in particular T n is an isomorphism for every n . Thus, W = Im( T ) is of dimension at most n - 1. Note that T ( W ) T ( V ) = W so we may view T as a map from W to itself. Now T n - 1 ( W ) = { 0 } by induction and so T n ( V ) = T n - 1 ( T ( V )) = T n - 1 ( W ) = { 0 } . To show this result is optimal consider the linear transformation on F n given by T ( α 1 , . . . , α n ) = (0 , α 1 , . . . , α n - 1 ) . One finds that T n - 1 ( α 1 , . . . , α n ) = (0 , . . . , 0 , α 1 ) and so T n - 1 is not the zero transformation, though T is nilpotent. (Clearly T n = 0.) / 2. (a) Find a linear map T : R 3 R 3 whose image is generated by (1 , 2 , 3) and (3 , 2 , 1). Here ‘find’ means represent by a matrix with respect to the standard basis. (b) Find a linear map T : R 4 R 3 whose kernel is generated by (1 , 2 , 3 , 4) , (0 , 1 , 0 , 1). Proof. (a) Let us define T by T (1 , 0 , 0) = (1 , 2 , 3) , T (0 , 1 , 0) = (3 , 2 , 1) , T (0 , 0 , 1) = (0 , 0 , 0), or in terms of a matrix T ( x 1 , x 2 , x 3 ) = 1 3 0 2 2 0 3 1 0 . Clearly the image is spanned by (1 , 2 , 3) , (3 , 2 , 1). (b) We claim that B = { (1 , 2 , 3 , 4) , (0 , 1 , 0 , 1) , (0 , 1 , 0 , 0) , (0 , 0 , 1 , 0) } is a basis for R 4 . It is enough to show that S spans R 4 (then it must be a minimal spanning set). For that, it is enough to show that e 1 , . . . , e 4 are in Span( B ). For e 2 , e 3 we have it by construction. We also e 4 as (1 , 2 , 3 , 4) - 4(0 , 1 , 0 , 1) + 2(0 , 1 , 0 , 0) - 3(0 , 0 , 1 , 0). Once we know e 2 , e 3 , e 4 are in the span it also follows that (1 , 2 , 3 , 4) - 2(0 , 1 , 0 , 0) - 3(0 , 0 , 1 , 0) - 4(0 , 0 , 0 , 1) is in the span.

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