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 January 30th, 2010, 12:49 AM #1 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 linear transportation: one-to-one and onto T,U: V ? V by T(a1, a2,....) = (a2, a3,....) and U(a1,a2,...) =(0,a1,a2...) T and U are called left shift and right shift operators on V, respectively. a) Prove that T is onto, but not one-to-one b)Prove that U is one-to-one, but not onto. thanks
 January 30th, 2010, 02:56 AM #2 Newbie   Joined: Jan 2010 Posts: 4 Thanks: 0 Re: linear transportation: one-to-one and onto I assume V does not have a finite dimension. We have to show that every vector in V is in the range of T. Let Alpha = (a1, a2, a3, ...) be a vector in V. Let Beta = (a1, a1, a2, a3, ...) be a vector in V. T(Beta) = Alpha -> Alpha is in the range of T. In order to show that T is not one-to-one we can show that there is a non zero vector which is mapped by T to zero. Since T is a left shift we have T(x, 0, 0, 0, ...) = (0, 0, 0, ...) so T is not one-to-one. In order to show that U is one-to-one we can show that the only vector which is mapped by U to 0 is the zero vector. Equivalently we can show that if Alpha is not zero then U(Alpha) is not zero either. Let us assume there is a non zero vector (a1, a2, a3, ...) we'll refer to it as Alpha. if it is a non zero vector then there is an i such that ai is not zero. Since U is a right shift the (i + 1) position in U(Alpha) is not zero. Since U(a1, a2, a3, ...) = (0, a1, a2, a3, ...) all the vectors in the range of U have a zero in the first position. So there is no Alpha in V such the U(Alpha) = (1, 1, 1, 1, ...). Hope it helps...
 January 30th, 2010, 03:50 PM #3 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 Re: linear transportation: one-to-one and onto thanks

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# T(a1, a2) =(a1 a2, a1) one - one, onto

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