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I Aren't all linear operators one-to-one and onto?

  1. Mar 12, 2016 #1
    Let W be a vector space and let A be a linear operator W --> W. Isn't it the case that for any such A, the kernel of A is the zero vector and the range is all of W? And that it is one-to-one from linearity? I ask because an author I am reading goes through a lot of steps to show that a certain operator is one-to-one and onto yet I thought it was a given for any linear operator.
     
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  3. Mar 12, 2016 #2

    BiGyElLoWhAt

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    What if ##A = \frac{d}{dx}##? One to One is bijective, right?
    I don't believe that that operator is one to one, if you consider it acting on ln(x) in W. If you differentiate, you get 1/x, and the domain of those two functions are different (assuming no absolute value sign on ln(x)). ln(x): (0, inf) and 1/x: (-inf,0) U (0,inf). This is an injective mapping, which is not 1 to 1.
     
  4. Mar 12, 2016 #3

    BiGyElLoWhAt

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    So I think it depends on W, is what I'm gathering. If W is a vector space that includes all real functions, then no, it's not 1 to 1. I'm not 100% sure, though. Someone will likely step in and correct me on this =/
     
  5. Mar 12, 2016 #4

    WWGD

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    T(a):=0 is linear: T(a+b)=0=T(a)+T(b); T(ca)=0=cT(a). Projection operators are not 1-1: P(a,b,c)=P(a,b,d) ; ## c \neq d ## For P(x,y,z):=P(x,y,0). Or, take a matrix with linearly-dependent rows/columns. EDITIt is not injective (independently of the choice of basis).
     
    Last edited: Mar 12, 2016
  6. Mar 12, 2016 #5
    Projection operators! Yes. thank you for that counter example.
     
  7. Mar 12, 2016 #6

    PeroK

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    A linear operator (from finite dimensional W to W) is 1-1 iff it is onto. That may be what you were thinking.
     
  8. Mar 12, 2016 #7
    Yes. I just didn't remember that not all linear operators are one-to-one.
     
  9. Mar 12, 2016 #8

    mfb

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    "finite dimensional W" is important here. It is not true if you drop that requirement.
     
  10. Mar 12, 2016 #9

    mathwonk

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    the most important linear operator is D = differentiation, on the (infinite dimensional) linear space W of smooth functions. Then notice that D of a constant function is zero, so it is not 1-1. (it is onto however.)
     
  11. Mar 12, 2016 #10

    mfb

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    Add integration with the requirement f(0)=0 and you have the example for 1-1 but not onto.
     
  12. Mar 12, 2016 #11

    mathwonk

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    nice remark.

    here we have the subspace Z of those smooth functions that equal zero at 0, and the complementary (real) line R of constant functions. then the space of all smooth functions is a direct product of Z and R, and D is 1-1 on Z and onto all smooth functions. the inverse of this restriction therefore, i.e. integration with f(0) = 0, is an injection from all smooth functions onto the subspace Z.

    So differentiation D is essentially projection of all smooth functions onto the subspace Z, followed by an isomorphism from Z to all smooth functions. And the special integration above is essentially the inverse isomorphism of all smooth functions onto Z, followed by inclusion of Z into all smooth functions.

    In general any linear surjection which has a kernel, restricts to an isomorphism on any subspace complementary to that kernel. Then the inmverse isomorphism is injective and not surjective.
     
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