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Linear Algebra

  1. Apr 24, 2009 #1
    I am trying to solve this,,,need help....
     

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  2. jcsd
  3. Apr 25, 2009 #2
    Here's a http://en.wikipedia.org/wiki/Dirac_delta_function" [Broken].
     
    Last edited by a moderator: May 4, 2017
  4. Apr 26, 2009 #3
    But the S* is linearly Independent, How can we prove that?
     
  5. Apr 26, 2009 #4

    HallsofIvy

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    Use the definition of "linearly independent". If [itex]a_1\phi_1(f)+ a_2\phi_2(f)+ a_3\phi_3(f)= 0[/itex] for all f, must [itex]a_1, a_2, a_3[/itex] all be 0?
     
    Last edited: Apr 28, 2009
  6. Apr 27, 2009 #5
    what is the values of [tex]\phi0[/tex] [tex]\phi1[/tex] [tex]\phi-1[/tex] ???
     
  7. Apr 27, 2009 #6

    HallsofIvy

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    Your attachment says that phi-1(f)= f(-1), phi0(f)= f(0), and phi1(f)= f(1)!

    a phi-1+ b phi0+ c phi1= 0 means that, for any function f, [a phi-1(f)+ b phi0(f)+ c phi1= af(1)+ bf(0)+ cf(1)= 0.

    Since P3 is the space of polynomials of degree 2 or less, ax2+ bx+ c. Take f above to be x2, x, and 1. What do a, b, and c have to be?
     
    Last edited: Apr 28, 2009
  8. Apr 28, 2009 #7
    What was the values of a b c ?
     
  9. Apr 28, 2009 #8

    HallsofIvy

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    That is the question that pnaik2008 needs to answer! IF he/she can show that a= b= c= 0, then S* is linearly independent.
     
  10. Apr 28, 2009 #9
    How could we show that a=b=c=0?,and if we proof that S is linearly independent, what are the basis of the S??
     
  11. Apr 28, 2009 #10

    HallsofIvy

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    I've already explained that in my response #6. af(-1)+ bf(0)+ cf(1)= 0 for all polynomials of order 2 or less. Take f to be x2, x, and 1 and you get three equations to solve for a, b, and c.
     
    Last edited: Apr 28, 2009
  12. Apr 29, 2009 #11
    Since it has taken already too much time, I will give a shot.

    If f = x^2, then
    f(0) = 0
    f(-1) = 1
    f(1) = 1
    => a*0 + b*1 + c*1 = 0

    If f = x, then
    f(0) = 0
    f(-1) = -1
    f(1) = 1
    => a*0 + b*(-1) + c*1 = 0

    If f = 1, then
    f(0) = 1
    f(-1) = 1
    f(1) = 1
    => a*1 + b*1 + c*1 = 0

    So we have three equations
    b + c = 0 (1)
    -b + c = 0 (2)
    a + b + c = 0 (3)

    We get from (2)
    b = c

    Thus, we get from (1)
    2b = 0
    b = 0 (4)

    Putting that to (1)
    c = 0 (5)

    Putting (4) and (5) to (3), we get
    c = 0

    Thus,
    a + b + c = 0

    This means that S* is linearly independent.

    ---

    I leave the rest for the questioner.
     
    Last edited: Apr 29, 2009
  13. Apr 29, 2009 #12

    HallsofIvy

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    So you are taking af(0)+ bf(-1)+ cf(1)= 0? That's not what I had before but perfectly legal.

    You mean a*0+ b*(-1)+ c*1= -b+ c= 0, right?

    No, -b+ c= 0.
    The three equations you got, b+ c= 0, b+ c= 0, a+ b+ c= 0, the first two being the same, would give b= -c so that a+ b+ c= a- c+ c= a= 0 but does NOT mean b and c must be 0. for example b= 1, c= -1, a= 0 satisfy b+ c= 1- 1= 0, a+ b+ c= 0+ 1- 1= 0. That would prove S* is NOT linearly independent.

    However, the correct equations are -b+ c= 0, b+ c= 0, a+ b+ c= 0. Adding the first two equations, 2c= 0 so c= 0. Putting c= 0 into the first equation -b+ 0= 0 so b= 0. Putting b= c= 0 into the third equation, a+ b+ c= a+ 0+ 0= a= 0. Since a= b= c= 0, S* is linearly independent.

    But I would have liked to have seen pnaik2008 at least try that calculation.

     
  14. Apr 29, 2009 #13

    I fixed the mistake.

    I can't stand in waiting for b).

    My attempt is:

    The basis of S of (P_3, R) is (x^2, x, 1).

    According to the definition of dual basis, for example the dual basis of the vector (1 0) is (1 0)^T.

    Thus, the dual basis of (x^2, x, 1) is (x^2, x, 1).
     
  15. Apr 29, 2009 #14

    HallsofIvy

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    The three given functionals are [itex]\phi_{-1}(f)= f(-1)[/itex], [itex]\phi_0(f)= f(0)[/itex], [itex]\phi)_1(f)= f(1)[/itex]. A "dual basis" would be functions

    [math]f-1(x)[/math] such that
    [math]\phi_{-1}(f_{-1})= f_{-1}(-1)= 1,
    [math]\phi_{0}(f_{-1})= f_{-1}(0)= 0,
    [math]\phi_1(f_{-1})= f_{-1}(1)= 0

    [math]f0(x)[/math] such that
    [math]\phi_{0}(f_0)= f_{0}(-1)= 0,
    [math]\phi_{0}(f_0)= f_0(0)= 1,
    [math]\phi_1(f_0)= f_0(1)= 0

    [math]f1(x)[/math] such that
    [math]\phi_{0}(f_1)= f_{1}(-1)= 0,
    [math]\phi_{0}(f_1)= f_1(0)= 0,
    [math]\phi_1(f_1)= f_1(1)= 1

    Since we are talking about polynomials of degree 2 or less, this reduces to
    1) Find [itex]f_{-1}(x)= ax^2+ bx+ c[/itex] satisfying
    a) [itex]f_{-1}(-1)= a- b+ c= 1[/itex]
    b) [itex]f_{-1}(0)= c= 0[/itex]
    c) [itex]f_{-1}(1)= a+ b+ c= 0[/itex]

    2) Find [itex]f_{0}(x)= ax^2+ bx+ c[/itex] satisfying
    a) [itex]f_{0}(-1)= a- b+ c= 0[/itex]
    b) [itex]f_{0}(0)= c= 1[/itex]
    c) [itex]f_{0}(1)= a+ b+ c= 0[/itex]

    3) Find [itex]f_{1}(x)= ax^2+ bx+ c[/itex] satisfying
    a) [itex]f_{1}(-1)= a- b+ c= 0[/itex]
    b) [itex]f_{1}(0)= c= 0[/itex]
    c) [itex]f_{1}(1)= a+ b+ c= 1[/itex]
     
  16. Apr 30, 2009 #15
    Fixing the formatting of HallOfIvy:
    Reformatting what HallsOfIvy says


    [tex]\begin{tabular}{ | f_-1(x) | f_0(x) | f_1(x) | }
    & f_{-1}(x) & f_0(x) & f_1(x) \\
    \hline
    \phi_{-1} & 1 & 0 & 0 \\
    \phi_0 & 0 & 1 & 0 \\
    \phi_1 & 0 & 0 & 1 \\
    ax^2 + bx + c: x=-1 & a -b + c = 1 & c = 0 & a + b + c = 0 \\
    ax^2 + bx + c: x =0 & - - 0 & - - 1 & - - 0 \\
    ax^2 + bx + c: x=1 & - - 0 & - - 0 & - - 1 \\
    \end{tabular}[/tex]

    I did not write variables to the last two rows again.

    I am not sure about the #4 line.
    a -b + c =1? # should it be 0.

    Could you, please, hallOfIvy more explain your results?
     
    Last edited: Apr 30, 2009
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