** Please use EQUATIONS to justify the two points. ** Please type the answer. Le
ID: 3111498 • Letter: #
Question
** Please use EQUATIONS to justify the two points.
** Please type the answer.
Let O: R^n rightarrow R^n be the zero map, and let T: R^n rightarrow R^n be a linear map. Show that it T^R = O for some k, then T^n = O. The solution is: we have given T^k = 0 then T must be singular map i.e. there exist a non zero element which map to zero. then nullity of T must. be greater than 1 i.e. dimension of image of T at most n-1. we have a descending chain img (T) Supersetequalto img (T^2) Supersetequalto img (T^3) Supersetequalto which indicates that if two consecutive subspace are equal then chain become constant at point on then T^n = 0 if T^n is non zero then chain img (T) Supersetequalto img (T^2) Supersetequalto img (T^3) Supersetequalto must be descending with all the dimension greater than zero which is impossible. hence there are n terms and dimension of img(T) is (n-1) which implies T^n = 0 **Could you please justify the following: In the list range(T) range(T^2), range(T^3) why is each space contained in the previous one? If the list fails to get smaller at some point, if range(T^K) = range(T^(k + 1), why must it be constant from this point forward?Explanation / Answer
range is set of images of elements in R^n. since T and O, are linear transforms from R^n -> R^n therefore [T] = [O] but as given T^n=O this shows that range will lies in O but as given in derivation img(T) img(T^2) img(T^3)....
and range is set of images, therefore, range will lie in O.
2) range (T^k)=range (T^(k+1)) (given) and range (T^k)=O (from (i)) therefore range (T^k+1) will also O hence this show contradiction therefore there exist some constant.
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