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Linear Algebra - Representing Matrix

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    The Question:

    The map is given: [tex]L\rightarrow \Re_{2} \rightarrow \Re_{3}, p \rightarrow[/tex] p' + q*p , with q(x) = x.
    Now i should find the representing matrix for L with respect to the bases {1+x, x+x2, 1+x2} for [tex] \Re_{2}[/tex] and {1,x,x+x2,1+x3} for [tex] \Re_{3}[/tex].


    3. The attempt at a solution

    I dont know what i should do here ^^. But i tried to calculate it.
    p is the polynomial, so i thought its like this:
    p = a0(1+x) + a2(x+x^2) + a3(1+x^2)

    So i can calculate [tex]p\rightarrow p' +q*p:[/tex]
    [tex]p\rightarrow a_{0}+a_{1} + 3a_{1}x + a_{0}x + 2a_{2}x^2 + 2a_{1}x^2 + 2a_{2}x^3[/tex]

    Then i just counted the coressponding values together, means
    a0 ~ 1+x --> 1 + 1 + 0 + 0
    a1 ~ x+x2 --> 1 + 3 + 2 + 0
    a2 ~ 1+x2 --> 0 + 0 + 2 + 2

    Puting this all together gives the repr. Matrix
    | 1 1 0 |
    | 1 3 0 |
    | 0 2 2 |
    | 0 0 2 |

    Is this somehow correct or completely wrong? ^^
    I'm missing the second base here..
    Does anyone knows a good website (linear algebra 1) ? I have exam in little more than week and still a lot to learn.

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

    Mark44

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    Staff: Mentor

    Your transformation is not a map from R2 to R3 - it's a map from P2 to P3, where P2 is the function space of polynomials of degree <= 2, and P3 is the function space of polynomials of degree <= 3.
     
  4. Mar 31, 2010 #3
    yes, we called it R.
    sorry i meant the same. A transformations from <=2 to <=3....
     
  5. Mar 31, 2010 #4

    Mark44

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    Staff: Mentor

    But R2 is not the same as P2, nor is R3 the same as P3. The dimension of R2 is 2, while the dimension of P2 is 3.

    For this problem I would advise you to evaluate L at each of the three basis polynomials. Then write those three output polynomials in terms of the second basis. That should give you an idea of what the matrix for this transformation is.
     
  6. Mar 31, 2010 #5
    But this is what i wanted to do ^^.
    But i dont know how. I ve never seen this before...
    Thats why i asked is that Polynomial correct the way i have written it down?

    Maybe you can give me an example, lets say for 1+x...
    What should i do with this?
    Sorry but i really dont know...:(
     
  7. Mar 31, 2010 #6

    Mark44

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    Your formula is L(p) = p' + xp, so L(1 + x) = 1 + x(1 + x) = 1 + x + x^2. Do the same thing for the other two basis polynomials.
     
  8. Mar 31, 2010 #7
    Thx!!!!!!!!!!!!!!!!!!
    then i get for:
    L(x+x^2)=1+2x+x(x+^2)= 1 +2x+x^2+x^3
    L(1+x^2)=2x+x(1+x^2)=3x+x^3

    You said: write those three output polynomials in terms of the second basis.

    so if i did this correct i get for
    1+x --> 1 0 1 0
    x+x^2 --> 0 1 1 1
    1+x^2 --> -1 3 0 1

    so my matrix is
    1 0 -1
    0 1 3
    1 1 0
    0 1 1

    is this ok?
     
  9. Mar 31, 2010 #8
    yeah its good
    thx mate!!! ;)
     
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