Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving a Linear Transformation is Onto

  1. Mar 8, 2013 #1
    There's this theorem:

    A linear map T: V→W is one-to-one iff Ker(T) = 0

    I'm wondering if there's an analog for showing that T is onto? If so could you provide a proof?

    I'm thinking it has something to do with the rank(T)...
     
  2. jcsd
  3. Mar 8, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The definition of "onto" is b
    "f: U-> V is said to be "onto" V if and only if f(U)= V"

    Now, for linear transformations, f(U) is always a subspace of V so f is "onto" V if and only if range(f)= dim(V).

    Do you know that "range-nullity" theorem? For linear transformation f:U-> V nullity(f)+ range(f)= dim(V). In particular, if f is one-to-one, so that nullity(f)= 0, we must have range(f)= dim(V) so that f is also "onto" V. Conversely, if f is "onto" V, then it is also one-to-one. That is, a linear transformation is "one-to-one" if and only if it is "onto".
     
  4. Mar 8, 2013 #3

    WannabeNewton

    User Avatar
    Science Advisor

    Halls I believe you left something out. A linear map [itex]T:U\rightarrow V[/itex] is injective if and only if it is surjective provided that [itex]dimU = dimV[/itex]. In particular, as you said if [itex]T[/itex] is injective then [itex]nullity(T) = 0[/itex] thus [itex]rank(T) = dim(T(U)) = dimU[/itex] by rank nullity but we then need that [itex]dimU = dimV[/itex] because [itex]dim(T(U)) = dimV,T(U)\subseteq V\Rightarrow T(U) = V[/itex] and similarly for the converse. I believe you had a typo because it should be [itex]rank(T) + nullity(T) = dimU[/itex] as opposed to [itex]dimV[/itex].
     
  5. Mar 8, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, thank you for the correction.
     
  6. Mar 8, 2013 #5
    HallsofIvy, just to be sure, you mean that rank(f) because range(f) is a set, and if f is onto then range(f) = V


    I am familiar with the rank-nullity theorem. Let f:U-->V is a linear map, rank(f) + nullity(f) = dimU. I think that's what you meant because I am familiar with the statement that T (A linear map) is 1-1 iff nullity(T) = 0

    WannabeNewton: You guys both made a similar statement that a linear map T: U-->V is 1-1 iff range(T) = V and that dim(U)=dim(V)

    So T is an isomorphism if dim(U)=dim(V) and if T is either 1-1 or onto?
     
  7. Mar 9, 2013 #6

    WannabeNewton

    User Avatar
    Science Advisor

  8. Mar 9, 2013 #7

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Just a little LaTeX tip: Use something like \operatorname or \mathrm to get ##\operatorname{dim}V## instead of ##dim V##.
     
  9. Mar 9, 2013 #8

    WannabeNewton

    User Avatar
    Science Advisor

    Thanks broski =D
     
  10. Mar 9, 2013 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, not one of my better days is it!

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook