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Linear Algebra; Isomorphism

  1. May 19, 2015 #1
    Pre-knowledge

    If V and W are finite-dimensional vector spaces, and dim(V) does not equal dim(W) then there is no bijective linear transformation from V to W.

    An isomorphism between V and W is a bijective linear transformation from V to W. That is, it is both an onto transformation and a one to one.


    1- Question
    Let W be a proper subset of an vector space V, and let T be the projections onto W. Prove that T is not an isomorphism.

    2- Answer

    Since T is a projection onto W then, T(v)=w, therefore dim(V) > dim(W)

    However since W is a proper subset of the vector space V, W is missing an element of V and therefore dim (w) is smaller then dim (V) and is not equal. Therefore it is not a one to one, and so is not an isomorphism.

    This is my answer, however I am unsure if that is correct. It makes sense in my mind that a subset of a vector space is smaller then the vector space itself in dimension, however I am uncertain. Can you help me please?

    Thank you,
     
  2. jcsd
  3. May 20, 2015 #2

    Ssnow

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    Gold Member

    Hi,
    only a few suggestions, in general a projection is from a vector space ##V## in a subspace (subset that is a vector space) ##W##, so as you said we have that ## dim(W) \leq dim(V) ##, in the ##=## case your projection is the identity and is bijective but if ## dim(W) < dim(V) ## the two dimensions are different so is not a bijection...
    The second question is a conseguence of the first, in general projections are surjective and linerar operators so you can define ##T:V\rightarrow W## specifing where you send the basis of ##V## in ##W##, for example ##\{(x,y,0):x,y\in\mathbb{R}^{3}\}## is a subspace of dimension ##2## of ##\mathbb{R}^{3}## you can project from ##\mathbb{R}^{3}## sending ##e_{1}=(1,0,0)\mapsto (1,0,0),e_{2}=(0,1,0)\mapsto (0,1,0)\mapsto (0,1,0),e_{3}=(0,0,1)\mapsto (0,0,0)##.

    I hope to clarify something but in general your answer was correct,
    by
    Simone
     
  4. May 20, 2015 #3

    WWGD

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    Sure, for one, an n-dim space does not have non-trivial n-dimensional subspaces.
     
  5. May 20, 2015 #4

    WWGD

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    Maybe Rank-nullity will work: Given two bases with n vectors , the nullity will be 0 , so the dimension of the image is the entire space.
     
  6. May 22, 2015 #5

    HallsofIvy

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    Since you are given that W is a proper subset of V, it follows that there is some vector in V that is NOT in W and therefore T is not surjective.
     
  7. May 22, 2015 #6
    This is what I had assumed, I just waanted to verify that this is correct. Thanks all
     
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