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Injective and Surjective linear transformations

  1. Dec 28, 2011 #1
    I was struck with the following question: Is there a linear map that's injective, but not surjective? I know full well the difference between the concepts, but I'll explain why I have this question.

    Given two finite spaces [itex]V[/itex] and [itex]W[/itex] and a transformation [itex]T: V→W[/itex] represented by a matrix [itex]\textbf{A}[/itex], we know that [itex]T[/itex] is invertible if and only if there is a one-to-one mapping between spaces, which means, [itex]\textbf{A}[/itex] is invertible.

    On the other hand, we know that [itex]T[/itex] is surjective if its image space can generate [itex]W[/itex]. The image space is related with the column space of [itex]\textbf{A}[/itex], so let's consider the system

    [itex]\textbf{A v} = \textbf{w}[/itex]

    where [itex]\textbf{v}[/itex] and [itex]\textbf{w}[/itex] are the coordinates of an element of [itex]V[/itex] and [itex]W[/itex], respectively. If [itex]\textbf{A}[/itex] is invertible than this system is always possible, meaning we can obtain any element of [itex]W[/itex] from an element of [itex]V[/itex], so [itex]T[/itex] must also be surjective.

    Perhaps I didn't formulate this right and I may be doing some mistake somewhere, but it's just a simple curiosity.
    Last edited: Dec 28, 2011
  2. jcsd
  3. Dec 28, 2011 #2
    I'm guessing you are using the following Alternative theorem:

    Let V and W be vector spaces of the same dimension and let [itex]T:V\rightarrow W[/itex] be linear, then the following are equivalent
    1) T is an isomorphism
    2) T is injective
    3) T is surjective

    This only holds for spaces of the same dimension!! For spaces not of the same dimension, this is clearly false! (indeed, because the matrix A in that case is not even rectangular)
  4. Dec 28, 2011 #3
    Yes I assumed they were of the same dimension, otherwise it wouldn't be invertible, but I have a problem with your statement. Consider for instance the projection

    [itex]T(x,y) = (x,0)[/itex]

    [itex]T[/itex] recieves a bidimensional vector and also returns one, so[itex]T:R^{2}→R^{2}[/itex]. The spaces are of the same dimension, but don't satisfy any of 1)-3).
    Last edited: Dec 28, 2011
  5. Dec 29, 2011 #4
    "T is invertible if and only if there is a one-to-one mapping between spaces, which means, A is invertible."

    Not so, actually. There exist one-to-one mappings from ℝ2→ℝ3, just not continuous ones. The point of interest isn't whether one exists, but whether T is that one-to-one mapping. You may have known that and just put the wrong word though.

    "The spaces are of the same dimension, but don't satisfy any of 1)-3)."

    He said the three are equivalent, not that they are true. What he means is that if any one of the three hold, then so do the other too. In your example, none of them hold, which is okay.

    No back to the original question. Try the map T:V→W, where V is dimension 2 and W is dimension 3, defined by

    T(x,y) = (x,y,0).

    It's essentially the identity map into the bigger space, so it's injective. However, since W is higher dimension, there's no way for it to be a surjection.
  6. Dec 29, 2011 #5


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    Right. But to say that the statements 1-3 are equivalent is to say that either they're all true or they're all false. So your example doesn't contradict what micromass said.
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