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Homework Help: Basis of vector

  1. Apr 11, 2006 #1
    Let B be the basis of R^2 consisting of the vectors

    [tex]\left(\begin{array}{c}3 & 1 \end{array}\right)[/tex] and [tex]\left(\begin{array}{c}-1 & 3 \end{array}\right)[/tex]

    Let R be the basis consisting of

    [tex]\left(\begin{array}{c}2 & 3 \end{array}\right)[/tex] and [tex]\left(\begin{array}{c}1 & 2 \end{array}\right)[/tex]

    find a matrix P such that [tex] [x]_R= P [x]_B[/tex] for all x in R^2[/tex]

    the answer should be a 2x2 matrix but I dont see how that is possible since [x] is only a column vector. I'm not sure how to solve this problem. any ideas?
     
  2. jcsd
  3. Apr 12, 2006 #2
    A nx2 times a 2x1 gives a nx1, so really P has to be a 2x2 matrix.

    Start by finding the components of each basis in terms of the other. A motivating question may be something like what is [tex]P(1, 0)^T[/tex]? Fool around with this idea to find the components of P.
     
  4. Apr 12, 2006 #3
    I'm not sure what is meant by "Start by finding the components of each basis in terms of the other."

    I understand that "A nx2 times a 2x1 gives a nx1, so really P has to be a 2x2 matrix." but what about the fact that [x] can be ANY numbers. [x] is not even dependent of the basis so I dont know why they would even give the basis.

    So if [x] can be anything, then so can P, so how do I find a specific 2x2 matrix?
     
  5. Apr 12, 2006 #4

    MathematicalPhysicist

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    your matrix could be regarded as a 'changing basis matix'.
    in other words you need to let one of them to be a linear combination of the other basis:
    2=a1*3+a2*1
    3=b1*3+b2*1
    1=a1*(-1)+a2*3
    2=b1*(-1)+a2*3
    and your P would be the matrix:
    a1 a2
    b1 b2
     
  6. Apr 12, 2006 #5
    I tried that already, that does not give the correct answer. Also, I believe you mean:
    2=b1*(-1)+b2*3

    this gives you
    a1=.5
    a2=.5
    a3=.7
    a4=.9

    which is not the correct answer
     
  7. Apr 12, 2006 #6

    MathematicalPhysicist

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    your textbook has different answers than these?
     
  8. Apr 12, 2006 #7
    yes, i entered them for answers on the web and it tells me that they are wrong
     
  9. Apr 12, 2006 #8
    ok, i did the exact same thing for the other matrix and got the right answer, thanks for the help!
     
  10. Apr 12, 2006 #9

    MathematicalPhysicist

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    which other matrix are you reffering to?
     
  11. Apr 12, 2006 #10
    3=a1*2+a2*3
    -1=a1*(1)+a2*2
    1=b1*(2)+a2*3
    3=b1*1+b2*2
     
  12. Apr 13, 2006 #11

    MathematicalPhysicist

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    but this gives you the solution of this:
    [x]_B=P[x]_R
     
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