1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How to check if a transformation is surjective and injective

Tags:
  1. Dec 18, 2017 #1
    1. The problem statement, all variables and given/known data
    I have attached the question. Translated: Suppose T: R^4 -> R^4 is the image so that: ......

    2. Relevant equations
    So I did this question and my final answers were correct: 1. not surjective 2. not injective. My method of solving this question is completely different than the answerbook thoug. Is my method correct too?

    3. The attempt at a solution
    1. If i put the transformation inside a matrix the result would be:
    0 1 1 0--------------------------------------0 1 0 1
    0 0 3 1 which can be reduced t----0 0 1 -1
    0 -1 0 0-------------------------------------- 0 0 0 1
    0 -1 0 1--------------------------------------- 0 0 0 0

    1. learned that a condition for a surjective transformation is that there has to be a pivot position in each row, which is not true: pivot positions are in row 1, 2 and 3, but not 4. So not Surjective.

    2. Secondly I learned that a condition for a injective transformation is that there can be no free variables. In the matrix above, there is one column with only zeros. This means that X1 is a free variable, it doesn't matter what value you give to it, it will not affect the final outcome.

    Is what i did correct? Thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Dec 18, 2017 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    You are correct, but the methods you describe are more of cooking recipes than providing understanding for what is actually going on.

    In order for a transformation to be surjective, there needs to be at least one ##x## in the domain such that ##f(x) = y## for every ##y## in the codomain.

    In order for a transformation to be injective, each ##x## in the domain must be mapped to a unique element ##y## in the codomain.

    In the case of linear transformations, it is helpful to think about these concepts in terms of linear independence.
     
  4. Dec 18, 2017 #3

    FactChecker

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    You need to be careful. The concepts of surjective and injective are very basic and general. They are used in situations where pivot elements and matrices are not applicable. So if your methods are different, you may not be learning the basic definitions and methods that you should be learning.

    So the question is: How did the book do it and do you understand it? If their method looks more basic and general, you should be using their methods.
     
  5. Dec 19, 2017 #4
    My advice would be to look at surjective and injective from a Set Theory perspective. Then see how this definition fits into what you are doing.
     
  6. Dec 25, 2017 #5

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    These concepts can be tricky to describe. Technically I feel it is important to begin the statement with the element y in the target space, so that the quantifiers are in the right order. One way I like is to say that f is injective if for every y in the target space, there is at most one x in the domain, or source space, such that f(x) = y. Then one can say analogously that f is surjective if for every y in the target space, there is at least one x in the source space such that f(x) = y.

    then the OP might check that in case every row has a pivot, that no matter what column vector y he puts on the right, he will be able to find an x such that T(x) = y. Thus T is surjective.

    And if every column has a pivot, no matter what column y he puts on the right, he will not be able to find more than one x with T(x) = y. In this last case, injectivity, it will be sufficient to show that if he puts the zero column y on the right, then the only solution is the zero vector x.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted