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Linear algebra - transformations questions.

  1. Dec 24, 2012 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I wish to pose a few questions I have concerning transformations:
    (1) I am trying to disprove the following statement:
    Let T: V->U be a linear transformation between vector spaces V and U, and let {v1,...,vn} be a set of vectors in V.
    If {Tv1,...,Tvn} spans U, then {v1,...,vn} spans V.
    (2) I would like to find the kernel and image of the following linear transformation:
    T:R2[x]->R2[x], T(p(x)) = xp(1) - xp'(1)


    2. Relevant equations



    3. The attempt at a solution
    (1) I came up with the following counter example, which I am not sure is correct:
    R2[x]->R2[x], T(p(x)) = int(p(x))dx
    I'd appreciate some feedback on this counter example.
    (2) I got that the ker(T) would have to be zero, and Im(T) = Sp{x,0,-x}. May someone kindly confirm please?
     
  2. jcsd
  3. Dec 24, 2012 #2
    Moreover, will Im(T) of the following transformation T: Rn->R, T(x1,x2,...,xn) = (x1+x2+...+xn) be Sp{(1,1,...,1)} where 1 is listed n times?
     
  4. Dec 24, 2012 #3

    pasmith

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    Are U and V of the same dimension? If not, and dim V > dim U = n, then clearly {v1,...vn} will not span V, even if {Tv1, ...,Tvn} spans U.

    There are a number of problems with that example.

    Firstly, you need to make that a definite integral, otherwise T(p) is not well-defined. For example,
    [tex]T(p)(x) = \int_a^x p(t)\,\mathrm{d}t[/tex]
    for [itex]a \in \mathbb{R}p[/itex]

    But then [itex]T(0) = 0[/itex] and [itex]T(x^n) = \frac{1}{n+1}(x^{n+1}- a^{n+1})[/itex] so there is no p for which T(p) = 1.

    Moreover, [itex]T(x^2) = \frac13(x^3- a^{3}) \notin \mathbb{R}_2[x][/itex] so T is not a map from [itex]\mathbb{R}_2[x][/itex] to itself.

    Looking at the definition, it should be obvious that if [itex]p(1) - p'(1) = 0[/itex] then [itex]p \in \ker(T)[/itex]. This is the case in particular if [itex]p(x) = x[/itex], so the kernel is not trivial.

    To work out the image, substitute a general [itex]p \in R_2[x][/itex] and collect up powers of [itex]x[/itex].
     
    Last edited: Dec 24, 2012
  5. Dec 24, 2012 #4

    pasmith

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    No; Im(T) cannot be Sp{(1, ..., 1)}, which is a subspace of [itex]\mathbb{R}^n[/itex]. Instead Im(T) = Sp{1}. (Being a 1-dimensional space, the real numbers are spanned by any non-zero real number.)
     
  6. Dec 24, 2012 #5
    (1) May I then use (0,1,2x) to corroborate my counter example above and thus disprove the claim.
    (2) By equating xp(x) - xp'(x) to zero, I got that either x would have to be zero, or a_0 would have to be equal to a_2. Does that mean that ker(T) = Sp{(1,0,1),(0,1,0)}.
    Is that correct?
     
  7. Dec 24, 2012 #6

    pasmith

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    Actually, the proposition you are asked to disprove is true if dim V = dim U is finite (and no finite set will span U unless dim U is finite). So there's no hope in looking for a counter-example where V = U.

    It is [itex]T(p)(x) = xp(1) - xp'(1)[/itex] not [itex]xp(x) - xp'(x)[/itex]. Setting [itex]p(x) = a_0 + a_1 x + a_2 x^2[/itex] gives [itex]p(1) = a_0 + a_1 + a_2[/itex] and [itex]p'(1) = a_1 + \frac12a_2[/itex], so [itex]T(p)(x) = (a_0 + a_1 + a_2)x - (a_1 + 2a_2)x = (a_0 - a_2)x[/itex]. This will be zero if [itex]a_2 = a_0[/itex] so the kernel is Sp(1 + x^2) and the image of T is Sp(x).
     
    Last edited: Dec 24, 2012
  8. Dec 24, 2012 #7
    I did mean to write xp(1) - xp'(1), and I believe you've made an airthmetic error there, namely p'(1) = a1 + 2a2, and not a1 + 1/2a2. Wouldn't you agree?
    Regarding the counter example and the statement which seemed (to me) to be false: I am not told that V and U have the same dimension. Actually, I am not given any information on U, other than that its dimension must be less than or equal to dim V. How did you deduce they had the same dimension?
     
  9. Dec 24, 2012 #8

    pasmith

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    You're right, I've now corrected it.

    Because your counterexample involved a map [itex]R_2[x] \to R_2[x][/itex], which means that [itex]V = U = R_2[x][/itex], so trivially dim V = dim U.

    As I originally pointed out, if V and U do not have the same dimension then the statement cannot be true: if dim V > dim U and {Tv_1, ... Tv_n} is a basis of U, then {v_1, ..., v_n} cannot possibly span V because it isn't large enough.
     
  10. Dec 24, 2012 #9
    (1) Okay, let us suppose, then, that dim V > dim U. Could you possibly help me come up with a counterexample to disprove that statement?
    (2) Won't ker(T), where T: Rn->R, T(x1,x2,...,xn) = (x1+x2+...+xn), be 0? But then how will that satisfy dim(ImT) + dim(KerT) = dim(V), unless dim(V) were 1, but isn't the dimension of Rn=n?
    (3) For finding Im(T), one normally chooses a base and then finds the transformation of each of its vectors. In your answer for T(p)(x)=xp(1)−xp′(1), you stated that Im(T)=Sp{(x)}. If we take 1+x+x2, however, we get 3x - 3x which is zero! How would you account for that? Alternatively, could you please explain why necessarily Im(T)=Sp{(x)}?
    (4) Let us please examine the following transformation: T(A) = 1/2(A-AT). I found kerT to be Sp{(0 1 1 0)}. Supposedly this is correct. Then I got ImT to be Sp{(0 -1/2 1/2 0),(0 1/2 -1/2 0)}. Which doesn't make sense as then dim(Im) + dim(ker) would not be equal to dim V. Would you please state the source of my error?
     
  11. Dec 24, 2012 #10
    PS my apologies for bombarding with questions. Hopefully you could continue assisting in this.
     
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