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Homework Help: Linear Algebra - Change of Basis

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Let B & C be the following subsets of R^2
    B= {[3 1] , [2 2]} (the vectors should be in columns instead of rows)
    C= {[1 0] , [5 4]}

    Let T: R^2 -> R^2 be the linear transformation whose matrix with respect to the basis B is
    [2 1]
    [1 5] (the brackets should be joint, it's a 2 x 2 matrix)

    Find the matrix T with respect to C. Check your answer by finding the determinant and the trace of each matrix.

    3. The attempt at a solution
    I found the change of basis matrix from B to C to be
    [(7/4) (-2/4)]
    [(1/4) ( 2/4) ] (again a 2x2 matrix)

    I did this by multiplying the change of basis matrix from the standard basis to C, with the change of basis matrix from the B basis to the standard matrix like this:
    [1 (-5/4)] [3 2]
    [0 (1/4)] [1 2]

    I then multiplied the given matrix by the change of basis matrix from B to C to get
    [3 (-3/4)]
    [1 (11/4)]

    I thought my answer was correct but the trace for the matrix I got is 8.25 whilst the trace of the original matrix is 7. They should be the same, right??
    Can anyone see where I went wrong? I'm not sure if I got the change of basis matrices right :S. The columns of the set B should form the change of basis matrix from B to the standard basis, shouldn't they?

    Any help would be appreciated :)
     
  2. jcsd
  3. Sep 13, 2008 #2

    gabbagabbahey

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    You've actually done the exact opposite here:

    The matrix for going from the standard basis (S) to C is actually:

    [tex]\overleftrightarrow{P}_{C \leftarrow S} \leftrightarrow \left( \begin{array}{cc}
    1 & 5 \\
    0 & 4 \end{array} \right)[/tex]

    While the matrix for going from B to S is actually:

    [tex]\overleftrightarrow{P}_{S \leftarrow B} \leftrightarrow \left( \begin{array}{cc}
    \frac{1}{2} & \frac{-1}{2} \\
    \frac{-1}{4} & \frac{3}{4} \end{array} \right)[/tex]

    You can check these by operating on each of the basis vectors in S (i.e. {[1,0],[0,1]}) with
    [tex]\overleftrightarrow{P}_{C \leftarrow S}[/tex] , to make sure you get the basis vectors of C; and by operating on each of the basis vectors in B with
    [tex]\overleftrightarrow{P}_{S \leftarrow B}[/tex] , to make sure you get the corresponding basis vectors in S.

    You can then easily obtain the change of basis matrix from B to C by using :

    [tex]\overleftrightarrow{P}_{C \leftarrow B} = {\overleftrightarrow{P}_{C \leftarrow S}}{\overleftrightarrow{P}_{S \leftarrow B}}[/tex]
     
    Last edited: Sep 13, 2008
  4. Sep 14, 2008 #3
    Are you in MATH1116 at ANU? If you're not then my lecturer's been doing some hard core plagiarism. :P

    Sorry I don't actually have an answer for you... I got the same as you but the trace that I calculated was 5.75...

    :S
     
  5. Sep 14, 2008 #4

    gabbagabbahey

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    Muso, the answer for [tex]\overleftrightarrow{T_c}[/tex] should be:


    [tex] \overleftrightarrow{T_c} {\leftrightarrow} \left( \begin{array}{cc} \frac{83}{4} & \frac{-277}{16} \\ 17 & \frac{-55}{4} \end{array} \right) [/tex]

    If you show me your work, I can tell you where you went wrong.
     
  6. Sep 14, 2008 #5
    Okay, I got P(C<--B)=
    [7/4 -1/2]
    [1/4 1/2] (2x2 matrix)

    so [T]C=P(C<--B)[T]B=
    [7/4 -1/2][2 1]=
    [1/4 1/2][1 5]

    [3 -3/4]
    [1 11/4]
    Much appreciated. :)
     
  7. Sep 14, 2008 #6

    gabbagabbahey

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    Okay, your first error is that your [tex]\overleftrightarrow{P}_{C \leftarrow B}[/tex] Matrix is incorrect.

    Your second error is that

    [tex] \overleftrightarrow{T_C} =\overleftrightarrow{P}_{C \leftarrow B} \cdot \overleftrightarrow{T_B} \cdot \overleftrightarrow{P}_{C \leftarrow B}^{-1}[/tex]

    NOT just

    [tex] \overleftrightarrow{T_C} =\overleftrightarrow{P}_{C \leftarrow B}} \cdot \overleftrightarrow{T_B}[/tex]

    Why don't you show me how you got your [tex]\overleftrightarrow{P}_{C \leftarrow B}[/tex]?
     
  8. Sep 14, 2008 #7
    I used the [c1 c2 | b1 b2] ~ [ I | P(C<--B)] formula

    so I got

    [1 5 | 3 2] ~
    [0 4 | 1 2]

    [1 0 | 7/4 -1/2]
    [0 1 | 1/4 1/2]

    So P(C<--B)=
    [7/4 -1/2]
    [1/4 1/2]

    Thanks
     
  9. Sep 14, 2008 #8

    gabbagabbahey

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    Hmmm... I've never seen that formula before, but I can derive a correct version of it for you:

    [tex] \left( \begin{array}{cc} \vec{c_1} ,& \vec{c_2} \end{array} \right) = \overleftrightarrow{P}_{C \leftarrow B} \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right) [/tex]
    since, by definition, the change of basis matrix will change the basis vectors of B into those of C.

    And so,

    [tex] \left( \begin{array}{cc} \vec{c_1} ,& \vec{c_2} \end{array} \right) \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)^{-1} = \overleftrightarrow{P}_{C \leftarrow B} \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right) \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)^{-1} =\overleftrightarrow{P}_{C \leftarrow B} [/tex]

    Try this formula, and don't forget to take the inverse.
     
  10. Sep 14, 2008 #9
    Thank you! I think my textbook is wrong... :S
     
  11. Sep 14, 2008 #10

    gabbagabbahey

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    Your welcome, wht do you get for [tex]\overleftrightarrow{P}_{C \leftarrow B}[/tex] now?
     
  12. Sep 14, 2008 #11
    I got
    [-3/4 13/4]
    [-1 3]

    Is that right?
     
  13. Sep 14, 2008 #12

    gabbagabbahey

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    Yup, did you get the right [tex]\overleftrightarrow{T_c}[/tex] now?
     
  14. Sep 14, 2008 #13
    Yep. Or at least the traces and determinants agree now. :)
     
  15. Sep 14, 2008 #14
    Yes I am. Lol. Are u done with the assignment? I'm still stuck on a couple of questions :(

    Thanks, gabbagabbahey :)
     
  16. Sep 14, 2008 #15

    gabbagabbahey

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    WOW! what are the odds of that?!:bugeye::rofl: And you're welcome:smile:
     
  17. Sep 14, 2008 #16
    I've done everything but the first and last questions. Need help with anything?
     
  18. Sep 14, 2008 #17
    I'm stuck on the last one too! I just posted it actually...
    And question 3 and 5. I've tried expanding q3 but I'm not getting anywhere :(. For q5, I'm having a bit of trouble trying to explain b and c.

    U need help with the first question?
    I actually found the geometric series by trial and error. Then I just showed that the sum of my series from n=10 to infinity is less than 10^-6.
     
    Last edited: Sep 14, 2008
  19. Sep 14, 2008 #18
    yeah my tutor did an example for the first one and basically he used 1/n! from 10 to infinity but that's not a geometric series, so... :S

    With 3 expand and factor out x^4 on the denominator and numerator. You should get a limit of -10/3.

    And with 5, solve for r^3-9r^2-12r+20=0, you should get r=-2, 10, 1. Then yk is the solution with initial values as given in the question. yk=a(-2)^k+b(10)^k+c(1)^k and solve for a, b, c
     
  20. Sep 14, 2008 #19
    Ahh thanks :).. I finally got ques 3. Expanding functions is such a pain!
    Try using fractions for both the 'a' and 'r' in your geometric series.
     
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