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Transformation matrix with respect to two bases?

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##S = \{1, e^x, e^{-x}, e^{2x}, e^{-2x}\}## and ##B = \{1, sinh(x),cosh(x), sinh(2x), cosh(2x)\}##. S spans the vector space V, and a linear transformation T: V -> V is defined by T(y) = y'' - 3y' - 4y.

    (a) Find the representation matrix of T with respect to the bases S and B.
    (b) Use the change of basis matrix (transition matrix) P, from S to B, to find the representation matrix of T with respect to the bases B and B.

    3. The attempt at a solution
    (a) I don't understand what it means to find a single transformation matrix for two different bases.

    (b) I already found the change of basis matrix P from S to B by representing the elements in S in terms of the ones in B and set them as columns of the matrix P. But again, I don't understand what it means to use this change of basis to find a transformation matrix for two different bases, but this time the two bases are the same bases?
     
  2. jcsd
  3. Dec 11, 2013 #2

    vela

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    You have two vectors related by ##\vec{y} = T(\vec{x})##. You're used to working with one basis to represent both ##\vec{x}## and ##\vec{y}## and find the matrix of T that relates the coordinate vectors, but you don't have to use the same basis for both. One vector could be represented relative to B while the other is relative to S. Since you'll have a different coordinate vector for one of the vectors, the matrix for T will be different.
     
  4. Dec 11, 2013 #3
    Then I really don't understand the questions because it's asking for a transformation matrix with respect to both S and B, and then another transformation matrix with respect to B and also B again?
     
  5. Dec 11, 2013 #4

    vela

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    You want to find matrices TBB and TSB (or TBS — the question isn't clear). They're related by P.

    TSB, for example, takes the coordinates of ##\vec{x}## wrt to basis B and gives you the coordinates of ##\vec{y}## wrt basis S.

    TBB takes the coordinates of ##\vec{x}## wrt to basis B and gives you the coordinates of ##\vec{y}## wrt basis B.

    Sorry...I have to run...I'll post a bit later if someone else already hasn't.
     
  6. Dec 11, 2013 #5
    Okay, for T(y) = y'' - 3y' - 4y, I found the matrix that takes S and transforms it into coordinates wrt S. I just took the derivatives of each element of S and plugged them into T(y) and found ##\{-4, -6e^x, 0, -6e^{2x}, 6e^{-2x}\}## so I just put each element in its respective positive in each column.

    So for part (a), I want to take S, transform the basis into B by using P, and then somehow get a transformation for this new basis B, but I only have a transformation that takes me from S to S. What should I do here?
     
  7. Dec 11, 2013 #6

    vela

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    Let me use a simple example with two basis vectors. The two bases are S={sinh x, cosh x} and B={ex, e-x}. Suppose we have the linear transformation T(y) = y'.

    cosh x = T(sinh x) would correspond to
    \begin{align*}
    \begin{pmatrix} 0 \\ 1 \end{pmatrix}_S &= T_{SS}\begin{pmatrix} 1 \\ 0 \end{pmatrix}_S \\
    \begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}_B &= T_{BS}\begin{pmatrix} 1 \\ 0 \end{pmatrix}_S \\
    \begin{pmatrix} 0 \\ 1 \end{pmatrix}_S &= T_{SB}\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}_B \\
    \begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}_B &= T_{BB}\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}_B
    \end{align*} depending on which basis you use for the domain and which basis you use for the range.
     
  8. Dec 11, 2013 #7
    For the 2nd transformation you listed, how do I obtain that ##T_{BS}##? Is it the same thing as changing the basis S into basis B, then applying the transformation with respect to B to obtain the same answer?

    I think I got part (b) by doing this: ##T_{BB} = PT_{SS}P^{-1}## where ##T_{SS}## is the transformation I listed above
     
  9. Dec 11, 2013 #8

    vela

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    You have
    $$\sinh x = (1) \sinh x + (0) \cosh x = \left(\frac{1}{2}\right) e^x + \left(-\frac{1}{2}\right) e^{-x}$$ so with respect to the two bases, the representation for sinh x is ##\begin{pmatrix} 1 \\ 0 \end{pmatrix}## and ##\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}##. You can do the same thing for cosh x. I just plugged in the various representations for sinh x and cosh x to get the four equations.

    I'm not sure if this answers your question.
     
  10. Dec 11, 2013 #9

    vela

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    If you're asking how to get the matrix itself, the second equation tells you the first column of TBS is ##\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}##. If you don't see this, just multiply in the vector ##\begin{pmatrix} 1 \\ 0 \end{pmatrix}## into a 2x2 matrix, and you'll see you get the first column. You can get the second column similarly, by multiplying by ##\begin{pmatrix} 0 \\ 1 \end{pmatrix}##.
     
  11. Dec 11, 2013 #10
    For the second equation, you have ##(1,0)T_{BS} = (1/2, 1/2)##. I was asking how to obtain the general matrix ##T_{BS}## to directly transform something in S directly to some coordinates in B.
     
  12. Dec 11, 2013 #11
    When something says a matrix is respect to 'B' how can you tell if it means with respect to 'S to B' or 'B to B'? It says that when T is a transformation with respect to 'S', ##PTP^{-1}## is = to the transformation with respect to B.
     
  13. Dec 11, 2013 #12

    vela

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    If only one basis is specified, you can assume both the domain and range are expressed in this basis.

    In your example, matrix ##P## maps the representation of a vector wrt to S to its representation wrt B. ##P^{-1}## goes in the opposite direction. So you have
    \begin{align*}
    \vec{x}_S &= P^{-1}\vec{x}_B \\
    \vec{y}_S &= T\vec{x}_S = TP^{-1}\vec{x}_B \\
    \vec{y}_B &= P\vec{y}_S = PTP^{-1}\vec{x}_B
    \end{align*} You can therefore identify ##PTP^{-1}## as the matrix of T with respect to B.
     
  14. Dec 11, 2013 #13
    I am still having trouble understanding how to get the transformations. From your example here: https://www.physicsforums.com/showpost.php?p=4599971&postcount=6

    The 3rd equation from the top. It means that some 2x2 transformation matrix multiplied on the left side of some arbitrary vector in B will equal to some arbitrary vector in S, correct? So my goal here is to find that 2x2 transformation matrix. However, in order to solve for this 2x2 transformation matrix, I need the inverse of a 2x1 matrix which is impossible. So how can I solve for the 2x2 matrix ##T_{SB}## in your example?
     
  15. Dec 12, 2013 #14

    vela

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    Do you know how to find the matrix for T(y)=y' if you're just dealing with one basis, say S={sinh x, cosh x}?
     
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