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Linear Algebra-Linear Transformations

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Define a function T: Psub3-->R3 by:

    T(p)=[p(3),p'(1), integral from 0 to 1 of p(x)dx]
    for p a polynomial in P sub3, the polynomials of degree less than or equal to 3.

    a. Show that T is a linear transformation

    b. Identify Psub3 with R4 in the usual way and write T in matrix form, i.e. T(x)=Ax for x<-R^4 and A is a 3 x 4 matrix.

    c. How do you know that ker(T), i.e.ker(A) is not just {0}?

    d. How do you know that there is a nontrivial polynomial, p, of degree less than or equal to 3 with p(3)=0, p'(1)=0 and integral from 0 to 1 of p(x)dx=0?


    Part 2:
    Expand the set {(1 -2) , (1 1)} to a basis for M2,2.
    {(3 -4) (0 1)}


    2. Relevant equations



    3. The attempt at a solution
    I've been working at this for awhile now and can't seem to make any headway. I think maybe its the notation throwing me off, but regardless any help is appreciated!
     
  2. jcsd
  3. Apr 23, 2012 #2

    HallsofIvy

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    Since you have "been working at this for awhile now", you surely must have some work to show. For (a), what are the propertied as linear transformation must have?
     
  4. Apr 23, 2012 #3
    I know in order to be a linear transformation, the following must be true:

    f(u + v)=F(u) + f(v) and f(cv)=cf(v) I believe.

    I guess I don't understand the notation and what I am supposed to be proving. Am i supposed to be proving how both the derivative and the integral are linear transformations in this case?
     
    Last edited: Apr 23, 2012
  5. Apr 23, 2012 #4

    micromass

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    OK, so you must show for arbitrary polynomials p and q in [itex]P_3[/itex] that

    [itex]T(p+q)=T(p)+T(q)[/itex]

    and

    [itex]T(cp)=cT(p)[/itex].

    Can you replace T(p+q), T(p) and T(q) by their definition??
     
  6. Apr 23, 2012 #5
    Well I guess I'm just missing something here or overanalyzing it perhaps. The way T(p) is written with the three terms: the p(3) and then its derivative and integral is throwing me off I think. I haven't seen it written like that before. It was assigned as a take home problem and the professor always make those more difficult or writes them in a different way than we are used to because we have the benefit of taking them home, using outside resources, etc. For some reason I'm just really having trouble working through this one.
     
  7. Apr 23, 2012 #6

    micromass

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    Just use the definitions given to you!!

    By definition

    [tex]T(p)=(p(3),p^\prime (1),\int_0^1 p(x)dx)[/tex]

    There is no trick here!!

    Now, apply the definitions to

    T(q) (just replace p above by q)

    and

    T(p+q) (just replace p above by p+q)
     
  8. Apr 23, 2012 #7

    HallsofIvy

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    What micromass said!

    Specifically, if [itex]p(x)= x^3- 2x^2+ x- 1[/itex], then [itex]p(3)= 3^3- 2(3^2)+ 3- 1= 27- 19+ 3- 1= 10[/itex], [itex]p'(x)= 3x^2- 4x+ 1[/itex] so [itex]p'(1)= 3(1^2)- 4(1)+ 1= 3- 4+ 1= 0[/itex], and [itex]\int_0^1 x^3- 2x^2+ x- 1 dx= \left[(1/4)x^4- (2/3)x^3+ (1/2)x^2- x\right]_0^1= -7/12[/itex] so T(p)= (10, 0, -7/12).
     
  9. Apr 24, 2012 #8
    Please forgive me if I'm asking a stupid question here, but where did you come up with that equation for p(x)? Or was that just an example?
     
  10. Apr 24, 2012 #9

    micromass

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    That was just an example.
     
  11. Apr 24, 2012 #10
    Okay, that's what i thought but just wanted to make sure. So for proving that it is a linear transformation would it be something like:

    [p(3),p'(1),integral of p(x)] + [q(3),q'(1),integral of q(x)] = [p+q(3),(p+q)'(1),integral of p+ q(x)]

    That just doesn't like right to me at all, so am I completely wrong here? And I apologize for not using the integral notation. I'm still trying to figure out how to use all the symbols and such.
     
  12. Apr 24, 2012 #11

    micromass

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    Yes, that would be correct. So, does that equality hold??
     
  13. Apr 24, 2012 #12

    Ray Vickson

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    Are you saying you don't believe that [itex] (p+q)(3) = p(3) + q(3), [/itex] or maybe you don't believe that [itex] (p+q)'(1) = p'(1) + q'(1), [/itex] or maybe you don't believe that [itex] \int_0^1 (p+q)(x) \, dx = \int_0^1 p(x)\, dx + \int_0^1 q(x) \, dx \, ?[/itex] Which of these do you not believe, and why do you not believe it?

    RGV
     
  14. Apr 24, 2012 #13
    I really am just having a hard time figuring out where to begin showing these proofs. I know everything I need is obviously given, but without an equation or something to plug numbers into, I can't figure out where to start. This whole concept is just hard for me to wrap my brain around I guess.
     
  15. Apr 24, 2012 #14
    As a proof, could I just use some arbitrary third degree polynomial like Hallsofivy did earlier, and plug the numbers into that? Or am I totally headed in the wrong direction with that idea?
     
  16. Apr 25, 2012 #15

    HallsofIvy

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    No, that was just an example because you indicated that you were not clear on exactly what the transformation was.

    To show that it is a linear transformation, you must show that the definition of "linear transformation" works for any polynomials.

    If p and q are such polynomials, T(p+ g)= ((p+g)(3),(p+ g)'(1), [itex]\int_0^1 (p+ g)dx[/itex]). Is that equal to T(p)+ T(g)= (p(3)+ g(3), p'(1)+ g'(1), [itex]\int_0^1 pdx+ \int_0^1 gdx[/itex])?

    If p is such a polynomial and a is a number, T(ap)= ((ap)(3), (ap)'(1), [itex]\int_0^1 (ap)dx[/itex]). is that equal to aT(p)= a(p(3), p'(1), [itex]\int_0^1 p dx[/itex])?
     
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