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Homework Help: Transition Matrix

  1. Feb 16, 2010 #1
    I came across this problem in one of my linear algebra books.
    A linear transformation T:R^3 ->R^3 has matrix

    2 3 0
    -1 1 2
    2 0 1
    with respect to the standard basis for R^3. Find the matrix of T with respect to the basis
    B={(1,2,1),(0,1,-1),(2,3,2)}

    The answer given is
    -28 -19 -43
    5 4 7
    18 11 28
    but i have no idea how to get to that answer as the book does not provide workings/steps. Any help would be appreciated thanks.
     
  2. jcsd
  3. Feb 16, 2010 #2

    Fredrik

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    I recommend that you read this post (the part above the quote) to make sure that you understand the relationship between linear operators and matrices.
     
  4. Feb 16, 2010 #3
    I'll use subscript B to indicate a vector or linear transformation expressed as a matrix in the new basis, thus

    [tex]\left ( Tx \right )_B = T_B x_B[/tex]

    Let B be a matrix whose columns are the basis vectors of the new basis, expressed in the standard basis. Then the inverse of B will convert the components of a general vector from the standard basis to the new basis:

    [tex]B^{-1}Tx = T_B B^{-1}x.[/tex]

    So [itex]B^{-1}T[/itex] has the same effect on [itex]x[/itex] as [itex]T_B B^{-1}[/itex]. Now all we have to do is solve for [itex]T_B[/itex].

    [tex]B^{-1}T = T_B B^{-1}[/itex]

    [tex]B^{-1}TB = T_B.[/itex]
     
  5. Feb 16, 2010 #4

    Redbelly98

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    Moderator's note:
    Homework assignments or textbook style exercises for which you are seeking assistance are to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned coursework or just independent study.
     
    Last edited by a moderator: Apr 24, 2017
  6. Feb 16, 2010 #5

    vela

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    Say vectors [itex]\vec{x}[/itex] and [itex]\vec{y}[/itex] have the representations

    [tex]
    \vec{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2
    \hspace{0.5in}
    \vec{y}=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}_2[/tex]

    with respect to basis 1 and basis 2. You can construct a matrix P that will convert between the two representations:

    [tex]\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=P\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2[/tex]

    and its inverse P-1 will take you in the other direction:

    [tex]\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2=P^{-1}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1[/tex]

    If [itex]\vec{y}=T(\vec{x})[/itex], there are matrices A and B such that

    [tex]\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1
    \hspace{0.5in}
    \begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}_2=B\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2
    [/tex].

    It turns out that A and B are related by [itex]B=P^{-1}AP[/itex] because

    [tex]\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=P\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2[/tex]

    [tex]\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=AP\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2[/tex]

    [tex]\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}_2=P^{-1}\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=P^{-1}AP\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2[/tex]

    In your problem, you're given A, and you want to find B. So the problem boils down to finding P given the information you have about the two bases.
     
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