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Math Help - Linear Transformation problem

  1. #1
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    Linear Transformation problem

    Let T:R^2 \rightarrow R^3 be defined by T(x,y)=(x+y,0,2x-y).
    Find bases for N(T) and R(T), compute nullity and rank of T, verify the dimension theorem. Determine if T is one-to-one or onto.

    Proof.

    T is linear. By a theorem, R(T) = span{T(b)} for a basis b of R^2

    Let b = {(1,0),(0,1)}, then span{T(b)} = span {(1,0,2),(1,0,-1)} = R(T), so I know that rank of T is 2, then nullity of T must be 0.

    But now, I'm working this backward, the problem stated that I should find the bases for N(T) and R(T) first, well, I found the bases for R(T), and I use the theorem to find that N(T) = {0}. So do I need to use another way to prove that N(T) = 0? I meant, it is pretty straight forward that there is no point in R^2 that would make T zero.

    Now, N(T) = {0}, so T must be one-to-one. But now, how do I know if T is onto or not?

    Thank you very much!
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  2. #2
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    By definition, a function is onto if the range is the same as the co-domain. So in this case, can the function take every possible value in R^3?
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  3. #3
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    No, I don't believe so, since dim(R(T)) is 2 and the co-domain is 3.

    Now, if the rank (T) is equals to the dimension of the co-domain, can I say it is onto? (Probably not, right?)
    Last edited by tttcomrader; February 1st 2008 at 08:19 AM.
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  4. #4
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    No, I don't believe so, since dim(R(T)) is 2 and the co-domain is 3.
    Excellent. Exactly what I would have done.

    Now, if the rank (T) is equals to the dimension of the co-domain, can I say it is onto? (Probably not, right?)
    Yes, you can. The only subspace of a vector space V with the same dimension as the vector space is V.
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