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Checking for linearity

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-07-19at23145AM.png

    this is theorem 4.5
    Screenshot2012-07-19at23441AM.png


    3. The attempt at a solution

    The book says that condition A fails. How can you even know? Condition A has a w in it and there is no W in the above question.
     
  2. jcsd
  3. Jul 19, 2012 #2
    The lowercase [itex]u, v, w[/itex] are arbitrary vectors in some vector space. In the statement of the theorem, they are placeholders. In fact, so are the "vectors" you've been given. Indeed, you could've been given

    It would not change the meaning of the problem in the slightest because all these symbols are being used for is to denote what an element of this vector space is and how the inner product is defined between elements.

    You should interpret lowercase [itex]u[/itex], [itex]v[/itex], and [itex]w[/itex] as meaning arbitrary vectors in the space, and the theorem should hold for any such vectors in the space. All the problem statement has told you is that, well, vectors in this space are 2x2 matrices.
     
  4. Jul 19, 2012 #3
    But if there are only two vectors then how can it satisfy the condition for linearity

    (u+v)*w = u*w + v*w

    which is a condition that applies to 3 vectors
     
  5. Jul 19, 2012 #4
    They are two examples of vectors; they are not the only two vectors that exist in that space. The space in question is that of [itex]M_{22}[/itex], the space of all 2x2 matrices. Any 2x2 matrix is a valid vector here. Think of what they gave you as more of a statement about the form a vector in this space can take. It should be clear that you can come up with any third vector, any third 2x2 matrix, just with four arbitrarily-labeled components.

    In fact, once they told you that the space was that of 2x2 matrices, they didn't need to give you explicit forms for the vectors at all except to be able to easily write a formula for the inner product.
     
  6. Jul 19, 2012 #5
    I think I understand
     
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