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Linear Transformations

  1. Jul 13, 2012 #1
    write P for the vector space of all polynomials, a[itex]_{0}[/itex]+a[itex]_{1}[/itex]x+a[itex]_{2}[/itex]x[itex]^{2}[/itex]+...+a[itex]_{n}[/itex]x[itex]^{n}[/itex], , a[itex]_{0}[/itex], a[itex]_{1}[/itex],...,a[itex]_{n}[/itex][itex]\in[/itex]R, n=0,1,2...

    1. Find a linear transformation P->P that is onto but not one-to one
    2. Find such a linear transformation, that is one-to-one but not onto

    I have been thinking about this question for a long time, but still could not come up with some thoughts, can someone give me some hints?

    Thanks in advance!
     
  2. jcsd
  3. Jul 13, 2012 #2
    So if it is onto, the linear combination should have infinitely many solutions?
    If it is one-to-one, then the linear combination should have only one solution?
     
  4. Jul 13, 2012 #3

    vela

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    No. First, it doesn't make sense to say that a linear combination has solutions. What I think you are referring to are the solutions to the equation ##a_0+a_1 x+ a_2 x^2 + \cdots a_n x^n = 0##. But this has nothing to do with the problem at hand.

    What you have is P, the set of all polynomials, and you're trying to find functions that map a polynomial f(x), which is an element of P, to another polynomial g(x), which is also an element of P. For example, say function A: P→P maps a polynomial f(x) to the polynomial f(x)+1. Then
    \begin{align*}
    A(1) &= 2 \\
    A(10x-4) &= 10x-3 \\
    A(x^2) &= x^2+1
    \end{align*} for instance. You can show that A is one-to-one and onto, but it's not linear. You need to figure out some functions from P to P which will meet the specified criteria.
     
  5. Jul 13, 2012 #4
    Thanks very much for your input, vela.
    But how did you get A(1)=2?
     
  6. Jul 13, 2012 #5
    Say function T:A-->B, if onto but not one-to-one, then every element from B from get mapped to from A. If one-to-one but not onto, then some or every element from B get mapped to from A? But how can I come up with the function then?:confused:
     
  7. Jul 14, 2012 #6

    vela

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    1+1 = 2

    Here is a simple example. Let f:ℤ→ℤ be the mapping that takes n to 2n. This function is one-to-one; however, it does not map onto ℤ because its image, f(ℤ), doesn't contain the odd integers.
     
  8. Jul 16, 2012 #7
    Hmmm, let f: P-->P be the mapping that takes x to x[itex]^{2}[/itex], this function is onto but not one-to-one? (Is this correct for this question?)

    I still cannot think of a function that is one-to-one, but not onto:confused:
     
  9. Jul 16, 2012 #8

    vela

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    I don't think you understood the point I made in my first reply to you. You need to get that down before you can go onto the rest of the problem.
     
  10. Jul 16, 2012 #9
    I thought I understood what u meant:redface:
    For my question, say a function A: P→P maps a polynomial f(x) to the polynomial f(x). Then
    A(x)=x[itex]^{2}[/itex]
    A(x+2)=(x+2)[itex]^{2}[/itex]
    So this function is onto, but not one-to-one?:uhh:
     
  11. Jul 16, 2012 #10

    vela

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    Note that x and x2 are elements of P. So what you wrote above doesn't mean the same thing as what you wrote below.

    How do you get that A is onto? For example, the function f(x)=x is in P, right? What polynomial when squared is equal to x? There isn't one, so A can't be surjective.
     
  12. Jul 16, 2012 #11
    Hi vela,

    Can you give me hints on this question? I have been thinking about it over and over again, but still can not figure it out:frown:
     
  13. Jul 16, 2012 #12

    vela

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    Let's step back for a second. Consider functions from ℝ2 to ℝ2. Can you give an example of a function that is 1-1 but not onto and one that is onto but not 1-1?
     
  14. Jul 16, 2012 #13
    f(x)=2x+1 is 1-1 but not onto?
     
  15. Jul 16, 2012 #14
    no, I don't think I got the idea, so confused:cry:
     
  16. Jul 16, 2012 #15

    vela

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    What are the definitions of a one-to-one function and an onto function? It doesn't seem like you know those.
     
  17. Jul 17, 2012 #16
    Let T: V[itex]\rightarrow[/itex]W be a linear transformation.

    1. T is said to be onto if im T=W.

    2. T is said to be one-to-one if T (v)=T(v[itex]_{1}[/itex]) implies v=v[itex]_{1}[/itex].

    I think I understand the definitions, it just seems so hard to apply them into questions:frown:
     
  18. Jul 17, 2012 #17

    vela

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    OK, suppose f:ℝ→ℝ is given by f(x)=2x+1. This isn't a linear function, but that doesn't matter here. It turns out this function is both one-to-one and onto.

    To show it's one-to-one, suppose f(x)=f(y). That means 2x+1=2y+1, which implies x=y.

    To show it's onto, let ##y \in \mathbb{R}##. You want to show there is an ##x \in \mathbb{R}## such that f(x)=y. That's easy enough to do. Just solve 2x+1=y for x.

    Does this make sense?
     
  19. Jul 17, 2012 #18
    yes, but back to the original question, how do you approach that question? I mean I know the definitions, but how to apply them to that question?
     
  20. Jul 17, 2012 #19

    vela

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    Can you answer the question I asked in post 12?
     
  21. Jul 17, 2012 #20
    I tried to answer it in post 13, as I mentioned, I know the definitions, but have no clue how to use them.
     
    Last edited: Jul 17, 2012
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