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Linear Transformation 1-1/onto

  1. May 22, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]T: R^n \rightarrow R^m[/tex] is a linear transformation.

    a) Calculate Dim(ran(T)) if T is one to one.

    b) Calculate Dim(ker(T)) if T is onto.


    3. The attempt at a solution

    a) I need to calculate the dimension of range of T if it's 1-1.

    So there is a property that: ran(T) = col(A) (where A is the standard matrix). And hence Dim(ran(T)) = Dim(col(T)). Since T is 1-1 Ax=0 has the trivial solution.

    I'm not sure if I'm on the right track & I don't know where to go from here...

    b) I need to "calculate" the dimension of kernel of T if it's onto.

    I know that Dim(ker(T)) = Dim(null(A)) and of course Dim(null(A)) (that is the dimension of the null space of standard matrix A) is the nullity(A).
    If the linear transformation T is onto then the linear system Ax=b must be consistent for every b in Rn.

    Again I'm stuck here but this is my attempt so far...
     
  2. jcsd
  3. May 23, 2009 #2
    Recall the dimension theorem. That is, for T: V -> W,
    nullity(T) + rank(T) = dim(V)

    How can you tell if a linear transformation is 1-1 and if it is onto. It has something to do with the kernel and range, respectively.
     
  4. May 23, 2009 #3

    Yes I know that theorem but I don't know exactly what to do with it. Please correct me if I'm wrong:

    nullity(T) + rank(T) = n

    rank(T) = dim(row(A)) = dim(ker(T))

    Since ker(T) = null(A) and null(A) = row(A)

    So, dim(ker(T)) + dim(ker(T)) = n

    n is the dim(domain)

    Well, since the transformation T is 1-1 ker(T)={0} (I mean x=0 is the only vector for which T(x)=0).

    Btw, this is all for part a)
     
  5. May 23, 2009 #4
    You know that if T is 1-1, then ker(T) = {0}. So what is the dimension of ker(T)? You know the dimension of the domain, so plug them into the equation.

    If T is onto, then what is range(T)? Once you found that, you can calculate dim(range(T)) [also called rank] and plug them into the equation.
     
  6. May 24, 2009 #5
    The dimension of the domain is n, kernel of T is a subspace of the domain and it's 0 if T is 1-1. So, do I need to plug 0 into:

    nullity(T) + rank(T) = dim(domain) which becomes:
    => dim(ker(T)) + dim(ker(T)) = dim(domain)

    dim(0) + dim(0) = 0

    So, the Dim(ran(T)) if T is 1-1 is 0 ??


    b) Calculate Dim(ker(T)) if T is onto.

    I don't really know what you mean but if T is onto if the range is the entire codomain, Rm.
     
  7. May 24, 2009 #6

    matt grime

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    And what is the dimension of R^m? (In reply to your very last line.)
     
  8. May 24, 2009 #7
    The dimension of Rm is m.

    What about part a, did I calculate it correctly (in my last post)?
     
  9. May 24, 2009 #8

    matt grime

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    No, you didn't. You just said that an injective linear map always maps onto the zero vector, i.e. is the zero matrix.

    Try looking at the definitions and the rank nullity formula.
     
  10. May 24, 2009 #9

    HallsofIvy

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    How did you get that both nullity(T) and rank(T) are equal to "dim(ker(T))"?

     
    Last edited: May 25, 2009
  11. May 24, 2009 #10
    For rank(T):

    rank(T) = dim(row(A))

    Since row(A) = null(A) and null(A) = ker(T)

    dim(row(A)) = dim(ker(T))

    And for nullity(T):

    nullity(T) = dim(null(T))

    Since null(T) = ker(T)

    So, nullity(T) = dim(ker(T))

    Hence "nullity(T) + rank(T) = n" can be re-written as: "dim(ker(T)) + dim(ker(T)) = n". That's how far I got, I don't see how exactly to derive the "rank-nullity" theorem.

    Is "dim(Ran(T))+ dim(Kernel(T))= n" the rank-nullity theorem you mean? As you see above I tried it but I don't know how to get this result from "nullity(T) + rank(T) = n".

    If T is 1-1 then ker(T)={0}. I think this means that dim(Kernel(T))= 0 and dim(Ran(T))=m, so using the theorem:

    dim(Ran(T))+ dim(Kernel(T))= n
    m + 0 = n
    ?
     
  12. May 24, 2009 #11
    Forget about row/column stuff.
    The dimension theorem that is important here is that for a linear transformation T:V -> W, rank(T) + nullity(T) = dim V

    You found that ker(T)={0} when T is 1-1. Now you want to calculate nullity(T), which is the dimension of ker(T). Look back on the definition of dimension. How did you obtain that nullity(T)=0?

    You found that rank(T)=dim(W) if T is onto. In your case, dim(W)=dim(R^m)=m.
     
  13. May 25, 2009 #12

    matt grime

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    The row space of A is not its kernel. The kernel is the set of vectors mapped to zero. I hope you can see that that is not the row space. Your conclusion that 2*dim(ker(T))=n should have made you want to check what went wrong in your definition.
     
  14. May 25, 2009 #13
    dim(Ran(T))+ dim(Kernel(T))= n

    m + dim(Kernel(T)) = n

    dim(Kernel(T)) = n-m

    Does this answer part b?

    The definition is that the dimension of a subspace V is the number of vectors in a basis for V.
    ker(T)={0} when T is 1-1, so dim(ker(T)) = nullity(T) = 1

    1 + dim(ran(T)) = n

    dim(ran(T)) = n-1

    Hope this is right :smile:


    Yes, you're right but I would like to know how you can prove that the following two theorems are equivalent (or you can use one to prove the other):

    dim(Ran(T)) + dim(Kernel(T))= dim(domain)

    RankT + nullity T = dim(domain)
     
  15. May 25, 2009 #14

    matt grime

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    They are tautologically equivalent - the rank is the dimension of the space spanned by either the rows or columns of the matrix, and one of those is the range (depending on whether you use row vectors and right multiplication or column vectors and left multiplication).
     
  16. May 25, 2009 #15
    Ok, thanks. :smile:

    Anyway, did I finally calculate Dim(ran(T)) (T 1-1) and Dim(ker(T)) (T onto) correctly in my last post?
     
  17. May 26, 2009 #16

    matt grime

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    One of them yes. But you got the dimension of a zero dimensional vector space wrong.
     
  18. May 26, 2009 #17
    Isn't it 0? :biggrin:

    dim(Ran(T)) + 0 = n
    dim(Ran(T)) = n
     
  19. May 26, 2009 #18

    matt grime

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    You wrote dim({0})=1, didn't you?
     
  20. May 26, 2009 #19
    Ops!...lol... but I corrected it now. I know that {0} is (the only) vector space with dimension zero.
     
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