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January 29th, 2010, 11:49 PM   #1
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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.

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January 30th, 2010, 01:56 AM   #2
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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...
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January 30th, 2010, 02:50 PM   #3
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Re: linear transportation: one-to-one and onto

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