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Homework Help: Linear Transformation / Kernel Question

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data

    L(p(t)) = t*dp/dt + t^2*p(1)

    If p(t) = a*t^2 + b*t + c, find a basis for the kernel of L.

    2. Relevant equations

    None.

    3. The attempt at a solution

    I know that L(a*t^2 + b*t + c) = 0, so that would mean that the derivative needs to be zero and p(1) needs to be zero. This would lead me to believe a = b = c = 0, however {0,0,0} is not a basis. I am a bit confused here...
     
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  3. Apr 12, 2010 #2

    lanedance

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    substitute the given form for p(t) into L(p(t)). If L(p(t))=0 for all t, what can you say about the constants a,b,c. Try grouping powers of t...
     
    Last edited: Apr 12, 2010
  4. Apr 12, 2010 #3

    Dick

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    If you want to solve L(at^2+bt+c)=0, you might want to actually write out what L(at^2+bt+c) is, before you jump the conclusion that the only solution is a=b=c=0.
     
  5. Apr 12, 2010 #4
    Substituting p(t) in, I get:

    L(p(t)) = t*(2*a*t + b) + t^2*(a+b+c) = 3*a*t^2+b*t^2+c*t^2 + b*t = t^2*(3*a + b + c) + t*(b)

    so: t^2*(3*a + b + c) + t*(b) = 0.

    It appears as if b must be 0 at the least. If 3*a + b + c = 0 => a = -c/3 and c = -3*a will make this term zero, so would a basis be something like:

    {-3, 0, -1/3}?
     
  6. Apr 12, 2010 #5

    Dick

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    That's pretty sloppy. Shouldn't a and c at least be opposite signs? Can you try that basis vector again?
     
  7. Apr 12, 2010 #6
    After looking it over again (and writing my basis more carefully), I come up with a basis of:

    ker(L) = {(-c/3)*t^2, 0, -3*a}

    I tested a few values and this seems to check out fine...although knowing me I'm probably overlooking something again. In the same vein, is a basis for the range something like:

    range(L) = {t, t^2}?

    My book has both kernel and range defined, albeit in a very abstract way without examples, so needless to say I am quite confused on exact syntax for these two. Thanks for the continued help.
     
  8. Apr 12, 2010 #7

    Dick

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    A basis of the kernel is a set of polynomials {p1(t),p2(t),...,pn(t)} which spans the kernel where the p's are linearly independent. Let's start out easy. Can you give me one nonzero polynomial that's in the kernel?
     
  9. Apr 13, 2010 #8
    If a = 1, b = 0, and c = -3, then a*t^2 + b*t + c => t^2 -3 and:

    L(t^2-3) = t*(2*t) + t^2*(1-3) = 2*t^2 - 2*t^2 = 0.

    So t^2 - 3 is one non-zero polynomial in the kernel.
     
  10. Apr 13, 2010 #9

    lanedance

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    following on form Dick i just though i would add, you should probably define the meaning of your vector representatino of the polynomial
    for example i would write it as (c,b,a) repesents the polynomial a*t2 + b*t + c, so for example
    (1,0,0) = 1
    (0,1,0) = t
    (0,0,1) = t2

    and so
    (c,b,a) = a*t2 + b*t + c

    if you do it as above, i don't think there should be any t's in the vector expression
     
  11. Apr 13, 2010 #10

    Dick

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    That's good. Now are there any other polynomials in the kernel that aren't multiples of (t^2-3)? And lanedance is right, you should be more clear on what you mean if you want to write that is as a 'vector'. (t^2-3) is (-3,0,1), right?
     
  12. Apr 13, 2010 #11
    Using this, any vector of the form y * (-c/3, 0, -3a) (y is a scalar) will fall in the kernel of L. I cannot seem to find other polynomials that are not multiples of the vector above...so is that a basis for the kernel? I feel I may be getting close, however I am having a few syntax issues.
     
  13. Apr 13, 2010 #12

    Dick

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    (-c/3,0,-3a) isn't a vector. It's a lot of different vectors depending on the values of c and a. And it's not even in the kernel unless 3*(-c/3)+(-3a)=0 or c=(-3a) which is the condition you posted earlier. Can you rethink this a little?
     
  14. Apr 14, 2010 #13
    I'm not sure how to think of this then. I tried thinking of the kernel as a nullspace:

    t^2(3a,b,c) + t(0,b,0) = 0

    If I do this though, I obviously get that t^2 and t must both be zero. I thought it would suffice to say that any scalar multiplied by the vector (-1/3,0,-3) would yield a vector that would fall into the kernel if the vector representation (c,b,a) = a*t^2+b*t+c is used. Apparantley this is not the case, and I am more than a bit confused.
     
  15. Apr 14, 2010 #14

    lanedance

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    hmmm what happened, in post #8 you had t^2 - 3 in the kernel?

    so going back there... evaluating L(p(t)) = 0 for arbitrary an 2nd order polynomial p(t) = a*t^2 + b*t + gave you 2 constrainst based on teh coefficients of t & t^2 being zero. S

    So we have 2 equations to satisfy.
    b = 0
    c = -3a

    Putting this all together shows the kernel of L is polynomials of the form
    p(t) = at^2 -3a = a(t^2-3)
    for any scalar a

    Now in vector form, which i hope didn't confuse the matter but may have.

    If we start with the 3D vector space spanned with basis 1,t,t^2 as per post 9. The kernel is the subspace spanned by the vector (-3,0,1). NOtcie there is no t dependence in the vector format

    So the kernel is the subspace defined by the line through the origin in the direction (-3,0,1)

    In effect we started with a 3D space of polynomials, with 2 unique constraints to satisfy, namely the coefficeints of t & t^2 being zero in the above equation. In effect each constraint represents a plane through the origin in the solution space & the intersection of the planes gives the final 1D line for the solution of the kernel (nullspace)
     
    Last edited: Apr 14, 2010
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