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Homework Help: Linear Algebra 2 - Representing Matrix

  1. Mar 31, 2010 #1
    Sry, this will be the last question^^

    10muavl.jpg

    Its a similiar to the problem i ve postet before. Maybe if i can solve the first problem i can solve this to. But what i dont understand is that notation. I ve circled it with a red line.
    Does anyone know what this means?

    Thx
    Mumba
     
  2. jcsd
  3. Mar 31, 2010 #2

    dx

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    It means the representation of L in the basis B = {1, x, x²}. They're just calling that matrix AB
     
  4. Mar 31, 2010 #3
    So this A is the representing matrix?
    But what does then Ap=q mean? Or how can i calculate L for 1, x and x^2?
     
  5. Mar 31, 2010 #4

    dx

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    Ap = p(x+1) defines the action of L.

    So for p = 1, you get L(1) = x + 1 and so on.
     
  6. Mar 31, 2010 #5

    dx

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    Actually, it looks like q(x+1) is not intended as multiplication of q with (x+1). The action of L on the function p(x) is given by p(x+1).

    So, if p = 1 + x, then L(p) = 1 + (x + 1) = 2 + x.
     
  7. Mar 31, 2010 #6
    But so i get for
    L(x) = x^2 + x,
    L(x^2)=x^3 + x^2

    So should i choose a new basis, for example {1+x,x^2,x^3} to get the repr matrix?
     
  8. Mar 31, 2010 #7

    dx

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    See my post above.

    L(x) will be x + 1 instead of x² + x.
     
  9. Mar 31, 2010 #8
    L(p)=q(p)=p(x+1)?

    wie kommst du da auf L(x)=1+x?
    Was hast du denn dann für L(1)?
     
  10. Mar 31, 2010 #9

    dx

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    Treat a polynomial as the function p(x). The action of L is to take that function and return the function q(x) = p(x+1)

    So, if p(x) = 1 + x, then q(x) = p(x + 1) = 1 + x + 1 = x + 2 (i.e. L[1 + x] = 2 + x)

    If p(x) = 1, then q(x) = p(x + 1) = 1. (i.e L[1] = 1 )
     
  11. Mar 31, 2010 #10
    So i get then
    L(1) = 1
    L(x) = x+1
    L(x^2)=x^2+1

    ?
     
  12. Mar 31, 2010 #11

    dx

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    L[x²] = (x+1)² = x² + 2x + 1
     
  13. Mar 31, 2010 #12
    and then for
    L(1) --> 1 0 0
    L(x) --> 1 1 0
    L(x^2) --> 1 0 1

    So my matrix would be
    1 1 1
    0 1 0
    0 0 1

    ??
     
  14. Mar 31, 2010 #13
    ahh ok so if u change this fpr L(x^2) --> 1 2 1
    matrix:
    1 1 1
    0 1 2
    0 0 1

    correct? ^^
     
  15. Mar 31, 2010 #14

    dx

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    Yes, thats correct.
     
  16. Mar 31, 2010 #15
    :D:D
    cool thx alooooot
     
  17. Apr 6, 2010 #16
    Should it be like that:

    If p(x) =1 then, q(x) = p(x+1) = p(x)+p(1) = 1 + 1 = 2, so L[1]= 2!?
     
  18. Apr 6, 2010 #17

    dx

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    No, p(x) is a constant function, i.e. it has the same value for any x. So p(x + 1) would still be 1.
     
  19. Apr 6, 2010 #18
    hmm, ok
    but why ist L(x) = x+1?

    --> p(x) = x
    if its a constant function and p(x)=x, shouldnt be P(x+1)=x too???
     
  20. Apr 6, 2010 #19

    dx

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    p(x) = 1 is a constant function, p(x) = x is not!

    For a given polynomial p, we get the L[p] by replacing every x with x + 1.

    So if p = 1, we do nothing since there is no x, and L[1] = 1

    If p = x, we replace x by x + 1, and we get L[x] = x + 1

    If p = x2, we replace x by x + 1 to get (x + 1)2, i.e. L[x2] = (x + 1)2 = 1 + 2x + x2
     
  21. Apr 6, 2010 #20
    oh man
    yes sure thx again ^^
     
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