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Linear operators & dimension

  1. Dec 14, 2009 #1
    Right so I've had an argument with a lecturer regarding the following:

    Suppose you consider [tex]P_4[/tex] (polynomials of degree at most 4): [tex]A(t)=a_0+a_1t+a_2t^2+a_3t^3+a_4t^4[/tex]

    Now if we consider the subspace of these polynomials such that [tex]a_0=0,\ a_1=0,\ a_2=0}[/tex], I propose that the dimension of of this subspace is 2 (versus the dimension of [tex]P_4[/tex] which is 5. Am I incorrect in saying this?

    Based on the answer to this I have a follow up question regarding a linear operator on [tex]P_n[/tex]
     
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  3. Dec 14, 2009 #2

    HallsofIvy

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    Yes, any such polynomial can be written as [itex]0(1)+ 0(x)+ 0(x^2)+ a_3 t^3+ a_4t^4[/itex] and so [itex]\{t^3, t^4\}[/itex] is a basis.

    Fire away!
     
    Last edited: Dec 15, 2009
  4. Dec 14, 2009 #3
    (Cheers)

    Suppose we consider an operator [tex]B: A(t) \rightarrow tA(t)[/tex] where [tex]A(t) \in P_4[/tex]

    Right, so my argument is that the result of this operation does not change the dimension, ie that [tex]dim(A(t))=dim(tA(t))[/tex] (basically I was trying to convince him that this was a linear operator). Am I missing something here?

    He seemed to argue that since [tex]tA(t)[/tex] now belongs to [tex]P_5[/tex] it now has an extra dimension, but I argue that since [tex]tA(t)[/tex] has the form [tex]a_0t+a_1t^2+a_2t^3+a_3t^4+a_4t^5[/tex] and since a polynomial in [tex]P_5[/tex] would be [tex]A(t)=a_0+a_1t^1+a_2t^2+a_3t^3+a_4t^4+a_5t^5[/tex] that it does indeed have the same dimension... :uhh:
     
    Last edited: Dec 14, 2009
  5. Dec 14, 2009 #4
    You are correct, the range of B has the same dimension, however, B is not a linear operator, because a linear operator is a linear transformation from a space to itself, i.e. the same space, it must take P_4 -> P_4, but tA(t) takes P_4->P_5, although to a 2 dimensional subspace in P_5 it is no longer the same space. (E.g. now we have t^5 in the range of B, but by definition that does not exist in P_4, so the spaces are different).
     
  6. Dec 14, 2009 #5
    Ahha, I had (oddly in retrospect) thought that a linear operator was one that mapped from a space to another such that both spaces had the same dimension.

    One final question, I also made the argument that a polynomial acted on by the following operator "loses" a dimension: [tex]\displaystyle P: A(t) \rightarrow \frac{A(t)-A(0)}{t}[/tex]
    Is that incorrect?
     
  7. Dec 14, 2009 #6
    if A(t)-A(0) is a constant then you are out of the space again...
     
  8. Dec 14, 2009 #7
    [tex]A(t)-A(0)[/tex] is never constant...

    [tex]\displaystyle \frac{A(t)-A(0)}{t}=\frac{a_0+a_1t+a_2t^2+a_3t^3+a_4t^4 - a_0}{t}=a_1+a_2t+a_3t^2+a_4t^3[/tex]
     
  9. Dec 14, 2009 #8
    Oops... I don't know what i was thinking. That is correct.
     
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