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Onto linear transformation

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Say I have a matrix:

    [3 -2 1]
    [1 -4 1]
    [1 1 0]

    Is this matrix onto? One to one?

    2. Relevant equations


    3. The attempt at a solution
    I know it's not one to one. In ker(T) there are non trivial solutions to the system. But since I've confirmed there is something in the ker(T), does this indicate that it is also not onto as well? I know that being an onto transformation is the Im(T) where it represents all transformed vectors.

    The reduced matrix I got was this:
    [ 1 0 1/5 | c-b/5]
    [0 1 -1/5 | b/5]
    [0 0 0 | a - b ]

    Can 0 = a-b ever? If I put in random values for a,b,c the system will be inconsistent usually.
     
  2. jcsd
  3. Apr 18, 2017 #2
    You should be aware that a linear transformation ##T: V \rightarrow V## is one-to-one if and only if it is onto.
     
  4. Apr 18, 2017 #3
    Let's assume that it was from R4 -> R3. What now? It should never be one-to-one in that case, but can it still be onto?
     
  5. Apr 18, 2017 #4
    Yes it can. Take for example, ##T(x,y,z,w) = T(x,y,z,0)##.

    But in any case, your transformation is from ##\mathbb{R}^3## to ##\mathbb{R}^3##
     
  6. Apr 18, 2017 #5

    PeroK

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    Homework Helper
    Gold Member

    Every function is onto its image, as by definition the "image" is precisely the set mapped to by the function. You have to careful, therefore, when talking about whether a function is onto. Onto refers to whether the function maps to all the "range" of the function. So, you can change a function from onto to not onto or vice versa, simply by changing the nominated range.

    A linear transformation from a vector space into itself is onto iff it is one-to-one. But, if you change the nominated range, this is no longer true.

    You could, therefore, have phrased your question better. You could have asked:

    Say I have a matrix:

    [3 -2 1]
    [1 -4 1]
    [1 1 0]

    Does this matrix represent an onto mapping from ##\mathbb{R}^3## to iself?

    Note that the property of being "one to one" can also be gained or lost by changing the domain of the function. Most often this is used, for example with trig functions, by restricting a function to a smaller domain to make it one-to-one and hence to have an inverse on that domain.
     
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