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Matrix composition question

  1. Aug 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that T : R2 → R2 is linear and has matrix

    -2 1
    5 2

    with respect to the standard basis S of R2.

    B = 1 1
    5 6

    (B is another poorly constructed matrix)

    What is the matrix of T with respect to B?

    2. Relevant equations

    T[itex]_{C,B}[/itex] = (T[itex]_{C,S}[/itex])[itex]^{-1}[/itex]I[itex]_{B,S}[/itex]

    3. The attempt at a solution

    Please see the pdf called q6b

    Also to see the question better presented look at problems1.pdf. Go to Problem sheet 3 question 6b
     

    Attached Files:

    Last edited: Aug 20, 2011
  2. jcsd
  3. Aug 20, 2011 #2
    sorry guys, I should point out that

    the answer to the question should be

    3 7
    0 -3

    which I don't seem to be getting
     
  4. Aug 20, 2011 #3

    ehild

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    That is wrong. A matrix in a new basis B is IBS-1T IBS

    ehild
     
  5. Aug 20, 2011 #4
    When you say T, do you mean T with respect to the standard basis ?
     
  6. Aug 20, 2011 #5

    In terms of basis reference, does your formula do the following (in terms of matrix composition)

    1. You create the identity map of matrix B with respect to the standard basis

    2. You multiply the B identity map with T to create a new matrix with respect
    to the standard basis ?

    3. You multiply the result of step 2 with the inverse of the identity map in step 1 to get the answer with respect to B
     
  7. Aug 20, 2011 #6

    ehild

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    With respect to the standard basis T=Ts, and Tb=IBS-1Ts IBS is the matrix of the linear transformation in the new basis.

    Think: The new basic vectors b1, b2 are B times the standard basic vectors. Edit: With "B"I denoted the matrix with columns equal to the basic vectors b1 and b2. And you get beck the standard basic vectors by multiplying b1, b2 by B-1.
    T transforms the basic vectors of B into vectors Tb1 and Tb2, defined in the standard basis. You need to express these vectors in terms of the new basis, so apply B-1 to them.
     
    Last edited: Aug 20, 2011
  8. Aug 20, 2011 #7

    ehild

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    You explained it very well.:smile: Yes. that is what I wanted to do, but I could not express myself so well.

    ehild
     
  9. Aug 20, 2011 #8

    HallsofIvy

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    Just a word about notation: It does not really make sense to ask about the matrix of a linear transformation "with respect to" another matrix. In your attached pdf files, B is NOT a matrix, it is two vectors, a new basis for [itex]R^2[/itex].

    Another way to find the matrix of a linear transformation with respect to a given (ordered) basis is: Apply the linear transformation to the each basis vector in turn. Write the result as a linear combination of the basis vectors. The coefficients give the columns of the matrix.
     
    Last edited: Aug 20, 2011
  10. Aug 20, 2011 #9

    ehild

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    Thanks, HallsofIvy. I know that I cannot mix vectors with transformations...I just used that the matrix that transforms the standard basis into a new basis has columns equal to the new basic vectors.
    I hope, I am right... Being a physicist, I use Maths a bit sloppy way.
     
  11. Aug 20, 2011 #10

    Ray Vickson

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    If the standard basis is e1=[1 0]^t and e2=[0 1]^t (^t = transpose) and b1=[1 5]^t, b2 = [1 6]^t, you can express e1 and e2 as linear combinations of b1 and b2, just by solving the equations b1=e1+5*e2 and e2=e1+6*e2 for e1 and e2; you can do this just as though the e's and b's were real variables instead of vectors---the algebra does not care what they are. A vector v = x1*b1 + x2*b2 can thus be written in terms of basis {e1,e2}, say as y1*e1+y2*e2, with y1,y2 = known linear combinations of x1,x2. Applying the matrix A = [[-2 1],[5 2]]to v = y1*e1+y2*e2 gives (-2y1+y2)*e1+(5y1+2y2)*e2. Now back-substitute for the yi in terms of the xi and for the ej in terms of the bj. You should get the result Tv = (3x1+7x2)*b1+(-3x2)*b2. Therefore, the matrix representation of T in the basis {b1,b2} is [[3 7],[0 -3]]. Once you have grasped these concepts through some simple examples, done step-by-laborious-step, then you can more successfully see where the matrix formulae come from and what they mean.

    RGV
     
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