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Im(T) = ?

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data
    T: V --> W is a linear transformation where V and W are finite dimensional.

    If dim V is less than or equal to dim W, then T is one-to-one. True or false?


    2. Relevant equations



    3. The attempt at a solution
    First of all, I'm assuming that im(T) = W. Is that correct? If so, dim(im T) = dim(W).

    Dimension thm says dim(V) = dim(ker T) + dim(im T).

    So if dim(V) is less than or equal to dim(im T), then dim(ker T) = 0. Which means ker T is is as small as it can be, so T should be one-to-one.

    But this is wrong. The answer is supposed to be false. Can anyone help?
     
  2. jcsd
  3. Mar 4, 2010 #2
    W is the target space for the transformation, so the image of T is a subset of W. The image is equal to the target space if the function is onto.
     
  4. Mar 4, 2010 #3
    Hmm okay, and we don't know if this is onto. So my approach doesn't work =\

    Can anyone give me a hint of what I should instead?
     
  5. Mar 4, 2010 #4
    If you already know it's false then try to construct a counterexample.
     
  6. Mar 4, 2010 #5
    Well I don't know it's false, I looked at the answers :) Haha. And a counterexample is given there... but is there a way to show that it's false without finding a specific counterexample?

    (Just because on a test I wouldn't automatically know if it was true or false first, so I want to be able to understand why it's one or the other w/o a counterexample).
     
  7. Mar 4, 2010 #6
    A proof to show it's false might not work on all levels because there can be functions that satisfy the criteria. You could, as you pointed out, use the dimension theorem as well as that T is 1-1 implies that nullity T = 0 and try to reduce the problem into proving something simpler.

    A good way to prove something is false is to find a counterexample. Can you find a linear transformation where dim V is less than or equal to dim W but T is not one-to-one?

    For example, consider the mapping f(x,y) = x + y. Is f 1-1?
    And consider g(x,y)=(f(x,y), 0, 0)
     
    Last edited: Mar 4, 2010
  8. Mar 4, 2010 #7
    All we know about T is that it's a linear transformation.

    T = 0
    is a linear transformation, and it's about as far from being 1-1 as you can get...
     
  9. Mar 4, 2010 #8
    How are you getting T = 0? I understand that T = 0 is definitely not 1-1.

    But if T = 0, I don't get why dim(V) is less than or equal to dim(W)...

    I'm obviously missing something here =\
     
  10. Mar 4, 2010 #9

    Dick

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    Mathnerdmo isn't 'getting' T=0. It's proposing that T=0 is a counterexample for the proposition that T:V->W is necessarily 1-1 if dim(V)<dim(W).
     
  11. Mar 4, 2010 #10
    oh, okay. I still don't get what dim(V) and dim(W) actually are though.

    V could have many vectors in it, but if T=0 then W only has one vector, 0, doesn't it? So dim(W) = 0. So wouldn't that mean that dim(V) > dim(W)? Obviously not, but I don't understand why.
     
  12. Mar 4, 2010 #11

    Dick

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    Put V=R^2 and W=R^3 and T:V->W=0. dim(V)=2 and dim(W)=3, you can agree with that, right? What T is doesn't affect what dim(V) and dim(W) are. Does it? And T is not 1-1.
     
  13. Mar 4, 2010 #12
    Ah I think I get it. Let me rephrase to make sure.

    T: R^2 --> R^3. So T would like like T(x,y) = (0, 0, 0) if T = 0, right? (I'm a bit unsure about what T = 0 actually means, but it just means it takes any vector to zero, correct?)

    So then dim(R^2) = 2 and dim(R^3) = 3, but T is not 1-1 because every vector in R^2 that map to (0,0,0), not just one.
     
  14. Mar 5, 2010 #13

    Dick

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    Yes, that's it.
     
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